St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1 (Mechanics), Note 1: Speed and Velocity
Mechanics: study of motion. Either kinematics (how things move) or dynamics (why things move). Motion
rotational (circular) or translational (linear). Motion in 1-direction rectilinear. Definitions: (1) Distance: how far
something travels or moves (scalar). (2) Displacement: total change in position (vector). If car goes 100 km west then
returns 100 km east, distance travelled is 200 km but total displacement is zero. (3) Speed: ratio of distance per unit
time. Distance d in meters, time t in seconds, speed s in m/sec. Distances, speeds always presume some frame of
reference (where you’re measuring from). If you’re in a train going at 50 km/hr, and you walk to the front of the train
at 5 km/hr, your speed with respect to the train is 5 km/hr, but with respect to the ground it’s 50 + 5 = 55 km/hr.
speed s = dx/dt, or s = d/t
Example 1: If you’re travelling at 25 km/h, how far can you go in 45 min?
Solution 1: 45 min/(60 min/hr)=0.75 hr. s=d /t so d = st. d = (25 km/h) * (0.75 h) = 18.75 km.
(4) Velocity: not the same as speed. Velocity is speed with a direction. Same formula, but a vector. Math note:
quantities can be scalars (just magnitude) or vectors (magnitude plus direction, usually indicated by arrows/hats over
the quantities). Few difficulties in one-dimension problems; but major importance in 2, 3-D work. In train example,
imagine you walk from the port side (left) to the starboard; now you need Pythagorean Theorem to solve from
velocity with respect to ground: (50)2 + (5)2 = c2; c2 = 2500 + 25 = 2525; now speed c = 50.24 km/hr!
velocity v = dx/dt, or v = d/t
This time, d = displacement (distance with direction, a vector with a line, ie 35 km East).
Example 2: A person jogs a circular ½ mile track in 6 minutes. Find average (a) speed and (b) velocity.
Solution 2: Begin by conversions: 6 min = 6*60 = 360 seconds. ½ mile = 800 m (approx). Now (a) speed s = d/t =
800 m/360. Speed s = 2.2 m/sec. (b) velocity v = displacement/time. Now if you go around a circle, your
displacement N is counteracted by your displacement S; displacement E counteracted by displacement W; total
displacement is zero! v = 0 m/360. Velocity v = 0 m/sec!
Example 3: A car goes east from Toronto to Kingston (200 km) in 2.2 hours. Find average (a) speed, and (b) velocity.
Solution 3: Use formulas speed vav = d/t and vav = dr/t.
Speed is distance over time, so speed vav = 200 km/2.2 hr = 90 km/hr.
Velocity is displacement over time, so velocity vav = 200 km [East]/2.2 hr; velocity vav = 90 km/hr East.
Example 4: The car turns returns to Toronto in 2.1 hours. Find the average (a)speed and (b)velocity for the whole trip.
Solution 4: Again, speed = distance/time, or vav=d/t. Now d=200*2=400 km. t = 2.2 + 2.1 hr = 4.3 hr. So speed
vav = 400 km/4.3 hr = 93 km/hr Velocity vav = dr/t. t still 4.3 hr, but what about d? d = 200 km E + 200 km W
= zero! So vav = dr/t = 0 km/4.3 hr = 0 km/hr Average Velocity = zero!
Example 5: A plane flies N at 300 km/hr; wind blows (a)W and (b)NW 45 o at 400 km/hr. Find the plane’s velocities.
Solution 5: (a) Draw a vector diagram to scale (1 cm=100 km/hr)
and measure the resultant vr = 5 cm. So vr=5*100 = 500 km/hr NW.
Or use Pythagoras: (vr)2 = (300)2 + (400)2 = 90000 + 160000 =250000
(vr) = 250000 = 500 km/hr NW (53o West of North by tangent rule).
(b) c2 = a2 + b2 – 2abcos(<c), so v2 = (300)2 + (400)2 – 2(300)(400)cos(135);
v2 = 90,000 + 160,000 + 169705; v2 = 419,705; v = 647 km/hr
sinb/b = sin c/c, so sinb/400 = sin(135)/612; sin b = (0.7071)(400)/647;
sin b = 0.4371; <b = 26 degrees West of North
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1 (Mechanics), Note 2: Acceleration
Recall velocity definition: change in displacement (distance) divided by time. If this ratio constant, you have constant
velocity:
v = dx/dt or
v = d/t
Acceleration: suppose velocity changing? You get acceleration, which is change in velocity divided by time.
Acceleration: rate at which velocity changes. Since velocity is the derivative of distance x with respect to time t,
acceleration would be the derivative of the derivative, or second-order derivative. Measured in (m/sec)/sec = m/sec 2.
a = (d/dt)(dx/dt), or a = d2x/dt2 = dv/dt
or a = v/t
Instantaneous acceleration: acceleration at one instant of time: given by limit a = lim v/t
t->0
Acceleration Equations: Acceleration can be uniform (constant speeding up or slowing down) or nonuniform (rate of
speeding up irregular, changing). Most problems (and their equations) assume uniform acceleration. 5 important
constant acceleration formulas, used everywhere:
Derivation 1: Now
But
So final velocity:
Acceleration
Multiply by t:
Add starting velocity vo both sides:
v = x/t.
x = (xf-xo), xo=starting, xf =finish.
v f = (xf-xo)/t.
a = v/t = (vf-vo)/t.
at = (v f-vo).
vf = vo + at (Equation 1)
Derivation 2: Now average two numbers:
Average velocity v:
nav = (n1 + n2)/2
vav = (vo + vf)/2 (Equation 2)
Derivation 3: If v changes put vav in v= d/t
Cleaning up
d = vt; d = ((vo + vf)/2)t
d = (1/2)(vo + vf)t (Equation 3)
Derivation 4: Combine Equations 1 and 3:
But vf = vo + at, so:
Clean it up:
d = (1/2)(vo + vf)t
d = (1/2)(vo + (vo + at))t.
d = vot + (1/2)at2 (Equation 4).
Derivation 5: Begin with Equation 3:
Now recall Equation 1, solve for t:
Put t in Equation 3
Recall your algebra:
So tidy up:
Multiply by 2a:
d = (1/2)(vo + vf)t
v f = vo + at
at = vf - vo
t = (vf - vo)/a
d = [(v f + vo)/2](vf - vo)/a
(v f + vo)(vf - vo) = vf2 - vo2
d = (v f2 - vo2)/2a
2ad = (v f2 - vo2) 2ad + vo2 = vf2 (Equation 5).
Example 1: A driver sees a stop sign and brakes; his distance x or d x from the sign after time t is d x = 432–72t+3t2.
Find (a) distance from the sign; (b) velocity at time t; (c) time to stop; (d) his decelleration; (e) his original speed.
Solution 1: (a) at beginning, t=0, so x = 432 – 72(0) + 3(0) 2; d = 432 m. (b) v = dx/dt; recall derivative of f(x) = x n is
(x) = nxn-1; so dx/dt = -72 + (2)3t; v = 6t – 72 (c) if he stops, v=0, so 0 = 6t – 72; 6t = 72; t = 12 sec; (d) a = d2x/dt2, or
a = dv/dt. dv/dt = 6; a = 6 m/sec2. (e) Run problem backwards, vf = vi + at, so vf = 0 + (6)(12); v = 72 m/sec.
Without Calculus: (a) same d = 432 m. (b) 2a(xf - xo) + vo2 = vf2 so v2= 2a(x – 432) + vo2 (so we still need a, v, or x
to go further); (c) set x=0, solve quadratic 3t 2 – 72t + 432 = 0; t = 12 sec; (d) Run problem backwards, v i = 0, so d = vit
+ (1/2)at 2; 432 = (0)t + (1/2)a(12)2; so a = 6 m/sec2 (e) Going backwards: vf2 = vi2 + 2a(d); vf2 = 02 + 2(6)(432); v2 =
5184; v = 72 m/sec.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1 (Mechanics), Note 3: Graphing and Gravity
Recall Acceleration Equations: v = d/t and a = v/t (Definitions)
vf = vo + at (Equation 1)
vav = (vo + vf)/2
(Equation 2) d = (1/2)(vo + vf)t (Equation 3)
d= vot + (1/2)at2 (Equation 4)
2ad + vo2 = vf2
(Equation 5)
Let's Use these to solve some real problems:
Example 1: How long does it take a car to travel 30 m if it accelerates at 4 m/sec 2? How fast will it be moving?
Solution 1: d = vot+(1/2)at2 vo zero, so d = (1/2)at 2. t2 = 2d/a = 2*30/4 = 15; t = 3.9 sec. For speed, v f= vo + at, so vf
= 0 + (4)(3.9); vf = 15.6 m/sec. To get km/hr, multiply by 3.6; v = 56.2 km/hr.
Example 2: Find a baseball's acceleration if the pitcher's wind up is 3 m and the speed is 50 m/sec. How far will the
ball go in 2.5 seconds?
Solution 2: 2ad+vo2=vf2 vo zero, so 2ad = vf2 Rearrange: a = vf2/2d
a = (50)2/2(3-0)=2500/(2*3)=2500/6 = 416 m/sec2! Assuming no resistance, v = d/t, so d=vt; d=(50)(2.5)=125 meters.
Graphing: If you can't solve physics problem by math, use graphs. If you draw position-vs-time graph, slope of line
x/t = v, velocity. If you draw velocity-vs-time graph, slope v/t = a, acceleration. And area under a v-t curve is
basically v*t = (x/t)*(t) = x = distance!
Example 1: A car accelerates at t=0 from 50 m/sec to 150 m/sec (t=10 sec).
V
How far did it go between t=3 sec and t=7 sec?
150Solution 1: Draw v-t graph at right. Area under curve
100basically square + triangle. Square = Length * Height = (7-3)(80) = 320 m 50Triangle = (1/2)Base*Height = (1/2)(7-3)(120-80)= 80 m. 320+80 = 400 m 00 2 4 6 8 10 t
Gravity: force of attraction between masses. Measured by Galileo (1564-1642) using inclined plane; he found the
distance ball falls along plane is proportional to square of time:
d = (1/2)at2
d=distance (m),
a=acceleration (m/sec 2),

t=time (sec)
Acceleration due to gravity on Earth constant g = 9.80 m/sec 2, downward. Now if d=(1/2)at2, a little math shows: a
= (2d/t2). The only acceleration down ramp is gravity g, but at angle ; this is less than gravity by following amount:
a = g*sin( So if we measure d,t,  on inclined plane, we can determine gravitation constant:
g = a/sin() = 2d/t2sin()
Galileo's Free fall hypothesis: all objects on Earth fall with the same uniform acceleration (9.8 m/sec 2), in absence of
air resistance (feather vs. cannonball!).
Example 2: A ball is thrown off a 70 meter tower at a downward speed of 3 m/sec. How far will it have fallen after
(a) t = 1 sec and (b) t = 2 sec? (c) How long will it take to hit the ground? (d) How fast will it be moving?
Solution 2: (a) dy = vot + (1/2)at2, so y = 3(1) + (1/2)(9.8)(1)2; y(1 sec) = 7.90 meters.
(b) For t=2sec, y = 3(2)+(1/2)(9.8)(2)2; y(2 sec) = 25.6 meters
(c) 70 = 3t + (1/2)(9.8)t2, solve for t. t = 3.48 sec, -4.10 sec (impossible); so t = 3.48 sec.
(d) vf = vo + at; so v = 3 + (9.8)(3.48); v = 37.1 m/sec. By calculus, v=dy/dt; now dy = 3t + (1/2)(9.8)t2, so derivative
dy/dt would be: v = 3 + 9.8t; v = 3+9.8(3.48); so v = 37.1 m/sec
Example 3: How far would a ball be travel down a 10o ramp after 2 seconds of time?
Solution 3: g = 2d/t2sin(), so 9.8 = 2d/(2)2sin(10); 9.8 = 2d/0.6257; 2d = 6.13; d = 3.066 meters.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1 (Mechanics), Note 4: Vectors
Vector vs. Scalar: scalar has magnitude (size); vector has size + direction. To add scalars, use arithmetic (8 km + 5
km = 13 km). To add vectors, take direction into account: add if same direction, subtract if opposite: 8 km E + 5 km E
= 13 km E, but 8 km E + 5 km W = 3 km E. Of course, can also be done by scale drawing/graphing.
Total 13 km E
Total 3 km E
8 km E
8 km E
5 km E
8 km E
5 km W
What if vectors at right angles? Use Pythagorean Theorem: c2 = a2 + b2. Put tail of one to tip of the other ("tip to
tail method") to get magnitude. Use tangent rule (tan = opposite/adjacent) to get direction.
80 km/hr N


450 km/hr E
Example 1: A plane flies E at 450 km/hr and a wind blows N at 80 km/hr. Find the resultant.
Solution: c2 = a2 + b2. c2 = (450)2 + (80)2 = 202,500 + 6400 = 208900. c= 457 km/hr. Direction: tan  = 80/450 =
0.1778.  = 10 degrees N of E. Velocity 457 km/hr, 10o N of E.
What if they're not at right angles? Again, scale drawing/graphing may help. Or break up into right angle triangles
(component method) and add in order.
B
A
Example 2: A plane flies 75oE of N at 450 km/hr; an 80 km/hr wind blows 45oE of N. Find the total speed (resultant).
Solution: Solve triangle A first. 450 km/hr 75 oE of N hypotenuse. Now tan =opp/adj, sin =opp/hyp, cos =adj/hyp
x-component VAx, y-component VAy. sin(15)=VAy/450, cos(15)=VAx/450
VAy=116 km/hr
VAx=434 km/hr.
similarly for B: sin (45) = VBy / 80, cos(45) = VBx /80.
+ VBy = 56 km/hr
VBx= 56 km/hr.
Add both sets of components to get your totals:
V y =172 km/hr
Vx =490 km/hr
Now Add Vx and Vy (Pythagoras): (Vy)2 + (Vx)2 =(VT)2; (VT)2 = (172)2 + (490)2
(VT)2 = 29,584 + 240,100 = 269684. VT = 520 km/hr.
Direction: tan  = opp/adj = 172/490 = 0.351  = 19 degrees. Resultant 520 km/hr, 19o N of E.
Alternate method by cosine rule: c2 = a2 + b2 - 2abcos(<c).
c2 = (80)2 + (450)2 - 2(80)(450)cos(150)
c
2
c = 6400 + 202500 + 62354 = 271254; c = 520 km/hr
Direction by sine rules: sin(<a)/a = sin(<b)/b = sin(<c)/c
b
sin(<c)/c = sin(<a)/a; sin(150)/520 = sin(<a)/80; 0.5/520 = sin(<a)/80; sin(<a) = 0.07692
<a = 4 degrees. Already component 15 N of E, so total 15+4 = 19o N of E.
a
Subtracting Vectors: to subtract two vectors, just add negative of the one being subtracted; A – B = A + (-B).
Multiplying/Dividing Vectors by Scalars: product of vector A and scalar c is just cA; a vector in the same direction
as A but c times bigger/smaller; if scalar is negative –c, product –cA is vector in opposite direction to A.
Example 3: Subtract the vector B = 3 m/sec @ 45o W of N from A = 5 m/sec @ 75o E of N.
Solution 3: A = B = A + (-B), so resultant C = 5 m/sec @ 75o E of N + 3 m/sec @ 45o E of S
Cosine rule c2 = a2 + b2 - 2abcos(<c); c2 = (5)2 + (3)3 – 2(3)(5)cos(120); c2 = 25 + 9 +15;
c2
= 49; c = 7 m/sec. Sin b/b = sin c/c; sin b/3 = sin(120)/7; sin b = 0.371; b = 21 o
75 E of N + 21 degrees = 96 E of N, or 6 S of E; So resultant C = 7 m/sec @ 6o S of E.
a=5
b=3
c
St. Joseph’s Secondary School
Grade 12 Physics University (SPH4U1)
Unit 1 (Mechanics), Note 5: Vectors and Problem Solving
Relative Velocity: Useful application of vectors is finding relative velocity (velocity measured in a particular
reference frame). Usual convention: two subscripts, first is for object in motion, second for reference frame used.
So if a Boat is travelling upstream, say, in a river of Water carrying it downstream, its true velocity of the boat as seen
from the Shore would be:
VBS =
boat velocity
seen from shore
VBW
+
boat velocity
in water
VWS
water velocity
with respect to shore
Example 1: A boat can travel at 25 km/hr. If it wants to travel directly across a river (speed downstream = 5 m/sec),
at what angle upstream must it head?
V WS
Solution: Convert 25 km/hr = 25,000 m/hr = 6.94 m/sec.
Draw vector diagram, solve for angle between V BS and VBW.
VBS
River
o
sin() = opp/hyp = 5/6.94 = 0.72.  = 46 .
 VBW
v=5 m/sec
Example 2: A plane (speed 200 km/hr) heads N, but meets
a 100 km/hr SW wind (wind blowing 45o W of S). Find the
plane's speed w.r.t. (with respect to) the ground.
Solution: Make a vector diagram, use (a) components or (b) sine rule.
VPG = VPA + VAG.
VAGy= -100(sin45)= -70 km/hr
VAG (air w.r.t. ground): VAGx = -100(cos45)= -70 km/hr
VPA (plane in air):
+ VPAx =
0 km/hr
V PAy = 200 km/hr.
Total
VPGx =
-70 km/hr
V PGy = 129 km/hr
Sum by Pythagoras: (VPG)2 = (VPGx)2 + (VPGy)2 = (-70)2 + (129)2 = 4900 + 16641 = 21541
VPG = 147km/hr. Direction: tan  = opp/adj = -70/129 = 0.54;  = 28o
100
45 o
200

Example 3:Two cars approach a corner, the first at 12 m/sec, the second at 24 m/sec. Find the velocity of car 1 as
seen by car 2.
(a)
(b)
CAR 2
Solution: In case (a) earth is
24 m/sec
VE2
fixed; in (b), car 2 fixed; so
Earth moves to car 2 at 24 m/sec (VE2)
CAR 1
V1E
Car 1 moves w.r.t. Earth 12 m/sec (V1E)
12 m/sec

Total V12 = V1E + VE2 (V12)2 = (12)2 + (24)2 =144+576 = 720 V12 = 26.8 m/sec. tan  = opp/adj = 24/12 = 2; = 63o.
Example 4: A boat moves at 1.5 m/sec; a passenger inside walks up a flight of stairs (angle 45 degrees) at 0.5 m/sec.
Find the velocity of the passenger w.r.t. the water.
Solution: Draw diagram and solve. Use cosine rule (easier).
c2 = a2 + b2 - 2abcos(c). Angle c = 180 - 45 = 135 cos135 = -0.7071
c
b
c2
= (1.5)2 + (0.5)2 - 2(1.5)(0.5)(-0.7071) = 2.25 + 0.25 + 1.06 = 3.56
c = 1.88 m/sec. For angle, sine rule or complete triangle, tan = opp/adj.
a = 1.5
Sx=0.35
Sx = (0.5)cos45 = (0.5)(0.7071) = 0.35. Sy = 0.35 (symmetry). tan = 0.35/1.85;  = 11o.
Multiplying Vectors: Two kinds of multiplication; dot product (scalar) and cross
product (vector). Dot product of vectors A * B defined to be Abcos, =angle
between the two vectors.
Example: work W is dot product of Force F and
displacement d, so W = Fdcos. Cross product of vectors A X B is third vector at right
angles to both A and B; so A X B = Csin, angle between A and B, C is in direction
perpendicular to both and given by the right hand rule (more on this in later notes).
Example: Torque  is vector product of Force F and radius r, so  = Frsin.
F
d

St. Joseph’s Secondary School
Grade 12 Physics University (SPH4U1)
Unit 1 (Mechanics), Note 6: Vectors and Projectile Motion
Projectile Motion: motion of object that is projected into air at some angle (cannonball, football, etc.) Air resistance
usually ignored; only acceleration after object leaves thrower's hand is gravity g=9.81 m/sec2, downwards. Done by
components: x-motion separate from y-motion, and both can be calculated separately.
Two Dimensional General Equations:
Projectile Motion (y+up; a x=0; ay=9.8 )
x-component (Horizontal) y-component (Vertical)
x-component
y-component
vyf = vyo + ayt
vx = vxo
vy = vyo - gt
vxf = vxo + axt
dx = vxot + (1/2)axt2
dy = vyot + (1/2)ayt2
dx = vxot
dy=vyot –(1/2)gt2
2
2
2
2
2
2
vx = vxo + 2axdx
vy = vyo + 2aydy
vx = vxo
vy2 = vyo2 – 2gy
Note: If you take y as +ve downwards, then – becomes + for Vertical Projectile Equations.
Also: If you choose as initial kickoff angle, vxo = vocos, vyo = vosin.
Example 1: A rock is thrown horizontally from a 200 m cliff. It strikes the ground 90 m from the cliff base. Find the speed
of the throw, and the time it takes to fall.
v xo
2
Solution 1: First, fnd the time to fall: y=(1/2)gt .
t2 = (2y/g) = (2*200)/9.8 = 400/9.8 = 40.8; t = 6.3 sec
Use the time to find the speed: x=vxot; vxo = x/t = 90/6.3 = 14.28 m/sec
Example 2: A softball is thrown at angle t = 30o with a 20 m/sec velocity.
Find (a) the maximum height, (b) the time in the air, (c) the range, or how
far it goes before hitting the ground, (d) the velocity vector at maximum
height, and (e) the acceleration vector at maximum height.
Solution 2: Find vxo and vyo; vxo = vocos30 = (20)(0.866) = 17.3 m/sec; v yo = vosin30 = (20)(0.5) = 10 m/sec. (a) Maximum
height dy = vyot – (1/2)gt2. But vy = 0 at maximum height; yo = 0. Use: vy2 = vyo2 + 2ady; (0)2 = (10)2 + 2(-9.8)dy; 0 = 100 + 19.6(dy); dy = 5.1 m
(b) Time: dy = vyot – (1/2)gt2. Set dy=0 and yo = 0. 0 = (10)t – (1/2)(9.8)t2; -10t = -4.9t2; -10 = 4.9t; t = 2.04 sec.
(c) Total distance: dx = vxot = (17.3)(2.04) = 35.3. dx = 35.3 m.
(d) At highest point: no y-component to velocity; only constant horizontal, so vxo = 17.3 m/sec
(e) Acceleration vector: Same at top as everywhere, a = 9.81 m/sec, downwards.
Example 3: A ball is thrown sideways from a 50-m roof; it lands 45 m from the base. What was the ball's initial speed?
Solution 3: Find the time to fall: dy = vyot – (1/2)gt2 Set dy=50 and vyo = 0. So we can say 50 = 0 – (1/2)(-9.8)t 2; so 50 =
4.9t2; t2 = 10.2; t = 3.19 sec. Now vx independent of fall, and constant. So vx = dx/t; vx = 45/3.19; vx = 14.1 m/sec.
Example 4: A baseball is thrown upwards with an initial speed v=15 m/sec; it rises to its maximum height, then falls back
into the pitcher's hand. Find the (a) time it takes for the ball to reach its maximum height; (b) maximum height; (c) speed of
the ball when it returns to the pitcher; (d) time when the ball passes 8 meters above the ground.
Solution 4: (a) vyf = vyo + ayt, so at the top, vy = 0; 0 = 15 + (-9.8)t; t = -15/-9.8; t = 1.53 seconds. (b) dy=vyot –(1/2)gt2 so
dy=(15)(1.53)–(1/2)(9.8)(1.53)2; dy = 22.95 – 11.47; dy = 11.48 meters. (c) vy = vyo + ayt, so to get to the top and back again,
t = (1.53)*2 = 3.06 seconds.; so vy = 15 + (-9.8)(3.06); vy = 15 – 29.98; vy = -15 m/sec; or argue by symmetry, v= 15 m/sec
up, therefore v=15 m/sec down. (d) dy=vyot –(1/2)gt2, so 8 = (15)t – (1/2)(9.8)t2; or 4.9t2 – 15t + 8 = 0 (quadratic); so t=(b+(b2 –4ac))/2a; or t=(-b-(b2–4ac))/2a; t=0.69 sec or t=2.37 sec (2 solutions because ball passes 8 m twice; once going
up, then again going down).
Example 5: A ball is thrown upwards so that its height after time t is h = 20t – 5t 2. Find the (a) initial speed of the ball; (b)
the velocity after 1 sec; (c) the time of the maximum height of the ball; (d) the maximum height.
Solution 5: (a) Easiest using calculus. Velocity v = dy/dt, so d y = 20t – 5t2 becomes v = 20 - (2)5t(2-1); v = 20 – 10t; at t = 0,
v = 20 m/sec. (b) after t=1 sec, v = 20 – 10(1); v = 10 m/sec. (c) at maximum height, v = 0, so 0 = 20 – 10t; 10t = 20; so
then t = 2 seconds. (d) h = 20t – 5t2, so h = 20(2) – 5(2)2; h = 40 – 20; h = 20 meters.
Without Calculus must rewrite equation for h = 0; 0 = 20t – 5t 2; 5t2 = 20t; 5t = 20; t = 4 sec up and down, so 2 seconds to
reach maximum height; then use vyf = vyo + at to find initial speed, v after 1 second; then vy2 = vyo2 – 2gh for h.
St. Joseph’s Secondary School
Grade 12 Physics University (SPH4U1)
Unit 1 (Mechanics), Note 7: Vectors and Parabolic Motion
Review: Once again, let's review the general equations of motion (we'll need 'em!):
Two Dimensional General Equations:
Projectile Motion (y+up; a x=0; ay=9.8 )
x-component (Horizontal) y-component (Vertical)
x-component
y-component
vyf = vyo + ayt
vx = vxo
vy = vyo - gt
vxf = vxo + axt
dx = vxot + (1/2)axt2
dy = vyot + (1/2)ayt2
dx = vxot
dy=vyot –(1/2)gt2
2
2
2
2
2
2
vx = vxo + 2axdx
vy = vyo + 2aydy
vx = vxo
vy2 = vyo2 – 2gy
Note: If you take y as +ve downwards, then – becomes + for Vertical Projectile Equations.
Also: If you choose as initial kickoff angle, vxo = vocos, vyo = vosin.
Proof that Projectile Motion is Parabolic: Equation of a parabola is y = ax – bx 2. Can we derive parabola equation from
the general equations of motion? Seems reasonable, seeing as we already know that d x = vxot + (1/2)axt2 and dy = vyot +
(1/2)ayt2. But let's prove it:
Step 1: Begin with projectile motion's x and y equations:
x = xo + vxot
y = yo + vyot – (1/2)gt2
Step 2: Set xo = 0, yo = 0:
x = vxot
y = vyot – (1/2)gt2
Step 3: Resolve for t:
x = vxot; so therefore t = (x/vxo)
Step 4: Put t into y-equation:
y = vyo(x/vxo) – (1/2)g(x/vxo)2
Step 5: Use conventional forms: vxo = vocos, vyo = vosin.
Step 6: Put forms into y equation: y = (vosin)(x/vocos) – (1/2)g(x/vocos)2
Step 7: Clean up the algebra:
tan= sin/cos
y = (tan)x –(1/2)(g/vo2cos2)x2
Step 8: In any particular problem: angle , gravity g, and vo are fixed so you can make them constants.
Step 9:
So set: a = tan; b = (1/2)(g/vo2cos2)
Step 10: Therefore:
y = ax – bx2 Q.E.D.
Example 1: Tricky Parabola Problem: A rescue plane drops supplies to mountain climbers 200 m below. (a) If the
plane travels at 69 m/sec, how far before the climbers will it have to drop the goods? (b) If the plane drops the supplies
400 m before the ridge, what velocity should they be given to get there? (c) In this case, with what speed do the supplies
arrive?
Solution 1: (a) Basic parabola problem, so dy=vyot –(1/2)gt2 Set vyo = 0; so dy = –(1/2)gt2
(200) = -(1/2)(-9.8)t2; t2 = 40.8; t = 6.38 sec.
Now: dx = vxot + (1/2)axt2, vxo = 69, ax = 0. So dx = (69)(6.38) + 0; x = 440 m.
(b) Drop 400 m, so they'll overshoot climbers, drop with v downwards.
x = vxot, so (400) = (69)t; t = 5.79 sec, not 6.38 sec! So y = vyot – (1/2)gt2;
(200) = vyo(5.79) – (1/2)(-9.8)(5.79)2; 200 = vyo(5.79) + 164.67; 35.33 = 5.79 vyo;
vyo = 6.1 m/sec, downwards.
(c) What speed do the supplies land with?
vyf2 = vyo2 – 2gy; vyf2 = (6.1)2 – 2(-9.8)(200); vyf2 = 37.21 + 3920 = 3957;
vyf = 63 m/sec; vxf = 69 m/sec; v2 = (vxf)2 + (vyf)2; v2 = 4761 + 3957; v = 93 m/sec.
Example 2: A ball is tossed upwards from the ground at a speed of v yo m/sec. Find (a) the height h of the ball after time t;
(b) the greatest height reached by the ball; (c) the time of its return to ground level.
Solution 2: (a) Use dy=vyot –(1/2)gt2; h = dy so h = vyot –(1/2)gt2; h = vyot –(1/2)(9.8)t2; h = vyot–4.9t2.
(b) Now vy2 = vyo2 – 2gy; and at the top, vy = 0, so 0 = vyo2 – 2gy; vyo2 = 2gy; y = vyo2/2(9.8); so h = vyo2/19.6
(c) We know in general v y = vyo – gt, and again at top v y = 0; so 0 = vyo – gt; vyo = 9.8t; so t = vyo/9.8 (time up). Time up and
down twice as much, or t = 2vyo/9.8.
St. Joseph’s Secondary School
Grade 12 Physics (SPH4U1)
Unit 1 (Mechanics), Note 8: Newtonian Dynamics
Force: A Vector; magnitude plus direction; defined roughly as any kind of push or pull; stretches or compresses item. Can
be measured by stretch/compression in spring scale. Aristotle (384-322 BC) erroneously argued force needed to keep object
in motion; if you remove force, motion stops. The greater the force, greater the speed (didn't realize effects of friction).
Italian physicist/astronomer Galileo (1564-1642) argued body would stay at rest or constant speed, and a force would
change it. So objects in constant motion slow down because of another force (friction).Newtonian Mechanics: English
physicist/mathematician Isaac Newton (1642-1727) formulated laws of motion and universal gravitation, 1666, during black
plague (stays at Cambridge). Hides work because of fear; publishes after comet discoverer Edmund Halley (1656-1742)
visits, pays for 1687 Principia Mathematica. 3 famous laws of motion:
Law One: (Galilean Law of Inertia): Every body continues in a state of rest or constant velocity unless acted upon by
some force. In real world, friction provides force that makes bodies slow down/stop.
Law Two: Acceleration is proportional to force, and inversely proportional to mass. Force (Newtons N or dynes) = mass
(kg or g) * acceleration (m/sec 2 or cm/sec2). 1 kg defined by platinum-iridium cylinder kept at International Bureau of
Weights and Measures (Paris); or can be defined atomically: Atomic Mass Unit u defined as mass of 1 Carbon atom = 12
atomic mass units, or 12 amu; this way, 1 u = 1.6605*10-27 kg.
F = ma
Law Three (Law of Reaction): Every action has equal/opposite reaction; whenever one object exerts a force F on a second
object, the second exerts an equal and opposite force –F on the first.
Notes: mass defined as measure of body's inertia (not same as weight, which depends on gravity). Objects have same mass
in space, weightless there; mass same on moon, weight (1/6). Weight really force due to gravity, so: Fg = mg. (g=9.81
m/sec2). Forces when two objects in contact (rock on table) called contact forces; if contact forces act perpendicular to
surface (ie downwards on horizontal table), they're normal forces.
Problem 1: Calculate the force needed to accelerate a car of mass m = 2000 kg at an acceleration of 4g.
Solution 1: 4g = 4(9.8) = 39.2; F = ma; F = (2000)(4*9.8); F = 78,400 N.
Problem 2: What (a) decelleration and (b) force do you need to stop the same car going at 100 km/hr within 50 m?
Solution 2: (a) 100 km/hr = 28 m/sec; vxf2 = vxo2 + 2ax; 02 = 282 + 2(a)(50); a = -7.84 m/sec2; (b) Now since F = ma;
F=(2000)(-7.84); F = -15,680 N.
Problem 3: A person jumps from a house roof 4 m high. When he hits the ground, he bends his legs so his body
deccelerates 0.7 m. If his upper mass is 50 kg, calculate (a) the final velocity before he hits the ground, (b) the decceleration
on his legs; (c) the average force on his legs.
Solution 3: (a) vf2 = vo2 + 2ax; vf2 = 02 + 2(9.8)(4); vf2 = 78.4; vf = 8.9 m/sec. (b) vf2 = vo2 + 2ax; 02= 8.92 + 2(a)(0.7); 1.4a =
-78.4; so a = - 56 m/sec2. (c) F =ma= (50)(-56+ -9.8); F=(50)(65.8); F = -3290 N.
Problem 4: A student has been asked to move a dolly full of
textbooks from one side of the room to another. He says, “by
Newton’s Third Law, no matter how much force I put on the
cart, it will pull back with the same amount of force; so it will
never move.” Explain the fallacy in the student’s reasoning.
Solution 4: The student is right that action/reaction forces are
equal and opposite; but he’s forgotten they act on different
objects. There are six forces altogether (see diagram). The
student moves forward when the ground pushes forward on the
student (F1) is greater than the cart pulls backwards (F 2); and the
cart moves when the students pulling force on it (-F 2) is bigger
than the frictional force of the cart on the ground (F3).
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1 (Mechanics), Note 9: Frictional Forces
Friction: resistive force caused by microscopic irregularities in object surfaces. Rolling friction: friction when a body rolls
across a surface (quite small). Kinetic friction: friction when a body slides (large). Frictional force proportional to the
normal force Fn of an object, which is equal to weight mg. So if you double weight, you double frictional force F f.
Proportionality between normal force and frictional force given by coefficient of friction k. Formula is:
Ffr = kFn (kinetic friction only).
Static friction: Frictional force of object that isn’t
moving (ie, a heavy filing cabinet; push it, it doesn’t
move. But if you push hard enough, you can overcome
this static friction, body moves, kinetic friction takes
over). Coefficient as above, s, and formula similar:
Ffr = sFn (static friction only)
Rolling Friction: Frictional force of object like ball
bearing, ball; usually so small it is neglected:
Ffr = rFn (rolling friction only)
Inclined Planes: common application of frictional forces. Choose x to be along plane, y perpendicular to plane; that way,
Ff has x-component only, Fn y-component only. W=mg still down, but find its projection along y-axis to get value of F n.
Then problem easier. Remember: acceleration of body down incline doesn’t depend on mass (as Galileo showed).
Ffr
Fn
motion
Fgrav(y)
Fgrav=mg
Example 1: (a) If the frictional coefficient between a floor and a 40 kg crate is 0.40, what force is needed to push it at a
constant speed across the floor? (b) What force is needed if the frictional coefficient is zero?
Solution 1: (a) Fn=W=mg=(40)(9.8)=392 N. Ffr = kFn Ffr = (0.4)(392) = 156.8 N. (b) If k=0, Ffr=0.
Example 2: A coffee cup slides off a dashboard if the car decelerates from 40 km/hr to 0 in 3.5 sec or less, but not if the
time is more. What is the static friction coefficient?
Solution 2: 40 km/hr = 11.1 m/sec. vf = vo+at; 0 = 11.1 + (a)(3.5); 11.1 = -3.5a; a = 3.2 m/sec2.
Ffr = sFn; Ffr = ma; Fn=W=mg; so ma = smg; m cancels; a = sg; 3.2 = s(9.8); s=0.32.
Example 3: A 7 kg block is placed on a 9 meter, 20 degree incline. Calculate its acceleration and velocity at the bottom if
(a) there’s no friction, and (b) if the frictional coefficient is 0.2.
Solution 3: (a) if there’s no friction, “rotate the plane” until it is level; F = ma x ; max = mgsin20 so you can cancel m, and a x
= gsin20 (recall relation a = gsin discovered by Galileo in Inclined Plane lab for ball, where F fr is essentially zero); a x =
(9.8)(0.34); so ax = 3.3 m/sec2. Now vf2 =vo2+2ax; vf2=(0)+2(3.3)(9); vf2 = 59.76; vf = 7.7 m/sec.
(b) If there’s friction, the normal sideways acceleration a = gsin will be made less by friction. But friction F fr = FN, so we
need the normal force to find the effect. Since normal force F N perpendicular to sideways forward force F applied = mgsin,
normal force must be FN = mgcos. Frictional force must then be F fr= mgcos. Take that away from sideways
acceleration ax. So ax = gsin20 – kgcos20 = (sin20 –kcos20)g. ax = (0.34 – (0.2)(0.93969))g = 0.16g = (0.16)(9.8) = 1.5
m/sec2. ax = 1.5 m/sec2. Now vf2 = vo2 + 2ax; vf2 = (0) + 2(1.5)( 9); vf2 = 27; vf = 5.2 m/sec.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1 (Dynamics), Note 10: Gravitational Forces
Laws vs. Definitions: Scientific laws general descriptions about the world; grounded in physical reality (ie, Newton’s
Second Law, F=ma). Definitions arbitrary, not necessarily based on nature; sometimes derived mathematically (ie,
kinematic equations, Newton’s first law, a definition of inertia). Some positivist philosophers (ie Comte, Popper; all
science measurable, falsifiable) claim even Newton’s 2nd “law” is just definition (of force). But almost all agree Newton’s
1st “law”is really definitional: only valid in Inertial Reference Frame; ie, frame at constant or zero velocity. Rotating
frames (like merry-go-round) and Accelerating frames (falling elevator) do not follow Newton’s 1st law; these are Noninertial Reference Frames.
Rotational Dynamics: Circular motion governs solar system (planetary motion) and galactic motion. If an object orbits a
central point at a constant speed, it is in uniform circular motion (speed stays the same, but direction, and hence velocity,
change). Consider planet orbiting sun. It has center-seeking, radial or centripetal acceleration ac (due to gravity) and
tendency to fly away from center (centrifugal motion).
Centripetal acceleration ac given by:
ac = v2/r, v =speed (m/sec)m, r=radius/distance (m)
Period of revolution: obvious; v = d/T, so T = d/v. Distance around circle is circumference 2r; so period T = 2r/v.
Orbital Dynamics: If ac = v2/r, you can put that in F=ma to get force laws for circular motion:
F=ma=m(v2/r)
F=force (N), m=mass (kg), v=speed (m/sec), r=radius or distance (m)
Example 1: Calculate the ac of the Earth around the Sun and the force the Sun exerts on it.
Solution 1: ac = v2/r; r=149.6*106 km = 149.6*109 m. T = 365 * 24 * 60 *60 = 3.153*10 7 sec.
v = d/T; v =
2*(3.14)*(149.6*109)/(3.153*107); v = 2.97*104 m/sec. ac = v2/r; So therefore ac = 8.878*108/149.6*109; ac = 5.9*10-3 m/
sec2 F = ma = (5.98*1024)(5.97*10-3) = 3.52*1022 N.
250g
1.6m
Example 2: A 250 g ball revolves in a 1.6 m radius circle twice a second. Calculate the v, a c.
Solution 2: v = 2r/T = (2)(3.14)(1.6)/(0.5) = 20.1 m/sec. ac = v2/r = 403/1.6 = 252 m/sec2.
Example 3: How large must the coefficient of friction be between the tires and the road if a
car is to safely round a level curve of radius 95 m with a speed of 90 km/hr?
Solution 3: 90 km/hr = 25 m/sec. a c = v2/r = (25)2/95 = 6.57 m/sec2. Now Ffr = sFn; ma =
s(mg); a = sg; (6.57) = s(9.8); s = 6.57/9.8; s = 0.67.
Example 4: A car travels at speed v around a freeway off-ramp of radius r. (a) Derive a formula for the angle at which the
off-ramp should be banked so that no friction is needed. (b) How big would the angle be for an off-ramp of radius 50 m and
an entry speed of 50 km/hr?
Solution 4: Since the car can’t move vertically, the car’s weight balances the normal force, or mg = F Ncos; resolve for FN
= mg/cos. Now the circular force going around the ramp is F = ma = m(v 2/r); since the ramp is banked, that inward force
is F = FNsin. So FNsin = m(v2/r). But FN = mg/cos; [mg/cossin= m(v/r); so mgtan = m(v2/r). cancel out m (note
this means it doesn’t matter how heavy the car is); gtan = (v2/r); or tan= (v2/rg). (b) If r = 50 m and v = 50 km/hr = 14
m/sec, tan= (14)2/(50)(9.8); tan = 196/490; tan = 0.4; = 22
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 11 (Mechanics): Gravity and Circular Motion
Nonuniform Circular Motion: Force is not directed towards center of circle (uniform), but at some angle, so you
have sum of two forces: FC, force towards center, and FT, the force tangent to the circle (which increases/decreases
speed, makes orbit smaller/bigger). By Pythagoras, the Total Force FTOT2=FC2+FT2. Since m=constant, F=ma, then
Total Acceleration aTOT2=aC2+aT2.
FTOT
Fc
FT
aTOT
aC
aT
Centrifugation: way of separating mixtures; if you spin test tubes at high speeds, heavier materials sink to the bottom
of the tube, light objects rise to the top. Often used in medicine to separate blood cells/plasma/etc.
Universal Gravity: According to legend (Stukely), Isaac Newton (1642-1727) got idea for gravity from apple falling
from tree (g=9.81 m/sec2). He calculated Moon’s acceleration around Earth (by a C=v2/r, aC=0.002 m/sec2), found that
aC=(1/3600)g. But Earth-Moon distance (384,000 km) 60 times Earth-apple distance (earth radius 3600 km). But if
you square it: 60*60 = 3600! So the acceleration of gravity (and hence, the gravitational force F G) decreases as the
square of the distance or radius r. Also, the more massive the objects m 1 and m2, the stronger the gravity. This leads
to 1687 Law of Universal Gravitation:
Every particle in the universe attracts every other particle with a force that is proportional to
the product of their masses and inversely proportional to the square of the distance between
them. This force acts along the line joining the two particles.
In formula form, Gravitational Force FG is:
FG = G m1m2/r2
G=Gravity constant 6.67*10-11 Nm2/kg2.
m1, m2=masses (kg) r=distance (m)
Gravity Near the Earth: If Universal gravity applied on earth, weight of object m1g = G m1m2/r2. Cancel mass of
object m1, you get g = G m2/r2. Rearrange for mass of the earth: m2 = gr2/G. So me= (9.8)(6.38*106 m)2/(6.67*10-11);
me = 5.98* 1024 kg. Now earth not a perfect sphere (mountains, oceans), so radius varies, but correct to 1 part in a
million. Geophysical satellites now measure “gravity anomalies” by orbital deviation when they pass overhead.
Example 1: A centrifuge rotates at 50,000 rpm. The top of a 4-cm long test tube is 6 cm from the center (axis of
rotation); the bottom is 10 cm. Find (a) centripetal acceleration in “g”s; (b) the force on the tube, if the mass is 12 g.
Solution 1: (a) Centripetal acceleration ac = v2/r, so at tube’s top d = 2r = (2)(3.14)(0.06) = 0.377 m per rev; t = 60
sec/50,000 rpm, so t = 0.0012 sec/rev. Speed v = d/t, v = 0.377 m/0.0012 sec; v = 314 m/sec. Acceleration a c = v2/r;
v = (314)2/0.06; Acceleration ac = 1.6*106 m/sec2; or divide by 9.8 to get 1.67*105 g’s. At tube bottom, v = (2r)/t =
(2*3.14*0.1)/(0.0012) = 523 m/sec, so ac = v2/r = (523)2/(0.1); ac = 2.74*106 m/sec = 2.79*105 g’s.
(b) Average a = 1.6*106 + 2.74*1206)/2 = 2.19*106; F= ma = (0.012)(2.19*106); F = 26,280 N (3 tons!)
Example 2: A racing car starts from rest in the pit, and accelerates uniformly to 55 m/sec in 11 sec, moving on a
circular track of radius 500 m. Calculate the a T and the aC, then calculate the total acceleration. If the mass of the car is
2500 kg, find the total force.
Solution 2: aT = v/t = 55/11 = 5 m/sec2. aC = v2/r = (55)2/500 = 3025/500 = 6.05 m/sec2.
Now, aTOT2 = aC2 + aT2 aTOT2= (5)2 + (6.05)2; aTOT2 = 25 + 36.6 = 61.6; aTOT = 7.84 m/sec2.
Total force FTOT = ma = (2500)(7.84) = 19,600 N.
Example 3: Find the force between the Earth and a 2000 kg spaceship orbiting 12,760 km above the earth.
Solution 3: F = Gm1m2/r2. F = (6.67*10-11)(2000)(5.98*1024)/(12.76*106)2 = 4900 N.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 12 (Mechanics): Gravity Applications
Recall: Formula for Gravitational Force FG is: FG = G m1m2/r2 (G=Gravity constant 6.67*10 -11 Nm2/kg2; m1,
m2=masses (kg) r=distance (m) Calculating “g” for Earth: g = Gme/r2 so g = (6.67*10-11)(5.97*1024)/(6.38*106)2; so
g = 3.98*1014/4.07*1013; g = 9.8 m/sec2 Note that g depends on mass of planet (obviously), and inverse-square on
radius; so as you climb a high mountain, gravitational force decreases.
Gravity Applications: Satellites orbit the earth, transmit TV, phones, Internet. But satellite must stay over same place
on Earth so dish antennas can be aimed. Satellite placed into “geosynchronous orbit”; its speed matches speed of
Earth’s rotation, so satellite appears fixed over same point. Idea first suggested by English science fiction author
Arthur C. Clarke; never patented! To find orbit, set (as above) F=ma equal to F=Gm 1m2/r2:
Now msa = msv2/r:
Cancel ms:
Velocity v = d/t, and d=2r, so
Put in line 3:
Cancel r on left side:
Tidy up:
Solve for r, orbital height:
msa = Gmems/r2
msv2/r = Gmems/r2
v2/r = Gme/r2
v = 2r/T v2 = 4r2/T2
(4r2/T2)/r = Gme/r2
4r/T2 = Gme/r2
4r3/T2 = Gme
r3 = GmeT2/4
Weightlessness: Satellites experience “weightlessness” because high speed of satellite (and centripetal acceleration
outwards) counteracts inward gravitational force. Weightlessness also possible on earth; quickly falling objects (like
planes, elevators). In falling elevator, F = ma on people inside, but acceleration a is g minus the elevator’s
acceleration. If elevator accelerates at g downwards, force on people inside will be zero, they will float. They are not
really weightless, as gravity still acts; they are accelerating as fast as the elevator is, so can float within it.
Example 1: Find the gravitational constant g on top of Mount Everest (8848 m altitude).
Solution 1: Earth radius 6380 km, so total r = 6380 km + 8.9 km = 6.389*10 3 km = 6.389*106 m g = G m2/r2 so g =
(6.67*10-11)(5.98*1024)/(6.389*106)2; g = 9.77 m/sec2. Higher the altitude, lower the constant g.
Example 2: Calculate the value of “g” on a mountain 6.5 km high.
Solution 2: As before, g = Gme/r2; g =(6.67*10-11)(5.97*1024)/(6.38*106 + 6500)2; g= 3.98*1014/4.078*1013. So g=9.76
m/sec2 (a bit less than 9.8, as expected).
Example 3: Calculate the speed and altitude of a geosynchronous satellite.
Solution 3: r3 = GmeT2/4 So r3 = (6.67*10-11)(5.97*1024)(86,400 sec)2/42; Now r3 = 2.97*1024/39.4384; r3 =
7.54*1022; r = 4.2229 * 107 meters. Next: velocity v = d/t, and d=2r, so v = 2r/T = 2(3.14)(4.2229*107)/(24
hrs*3600sec/hr). v = 3069 m/sec.
Example 4: Calculate the weight of a 75 kg man accelerating (a) downward and (b) upward in a a = 4.9 m/sec 2
elevator Acceleration a = (1/2)g.
Solution 4: (a) F = ma, so a = 9.8 – 4.9; a = 4.9 m/sec 2; F = (75)(4.9); F = 367.5 N. For (b) , a = 9.8 + 4.9 = 14.7; so
Force F = ma = 75)(14.7); F = 1102.5 N.
Example 5: Find the altitude of a satellite with a period of 90 minutes (US spy satellites).
Solution 5: r3 = GmeT2/4 so r3 = (6.67*10-11)(5.97*1024)(90 * 60 sec)2/42; r = 1.16*1022 /39.4; r3 = 2.94*1020
m; so r = 6654730 m; r = 6,654 km. Now earth radius is 6,380 km, so satellite just 274 km above the Earth (great
visibility for closeups, very high speed to elude detection).
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 13 (Mechanics): Kepler’s Laws
Johannes Kepler (1571-1630): German astronomer/physicist; inherited most of the astronomical observations of
Danish astronomer Tycho Brahe (1546-1601), who invented the Tychonic System (Earth at the center; the Sun
moves around the Earth, but all planets move around the Sun). Kepler used Brahe’s data to prove that the planets
followed elliptical, not circular paths. Kepler “sleepwalked” to his answers; books 1596 Mysterium Cosmographicum,
1607 Nova Astronomia full of mysticism, theories of music, Regular Solids, etc. Now Remembered for Kepler’s
Three Laws of Planetary Motion:
Law I: The path of each planet around the Sun is an ellipse, with the Sun at one focus. (Planet follows the path x 2/a2 +
y2/b2 = r2; a,b = semimajor/semiminor axes).
Law II: The planet moves so that a line connecting it with the Sun sweeps out equal areas in equal times (in other
words, the closer the planet gets to the Sun, the faster it moves).
Law III: The square of a planet’s period is proportional to the cube of its distance from the Sun. Mathematically we
say: r3/T2 = K, or (r13/T12) = (r23/T22).
y
x
*
*
AI
AII
AI = AII if
TI=TII
Proof of Kepler’s Laws: Comes from Newton. Just set F=ma equal to F=Gm 1m2/r2:
Therefore, for Sun ms and Planet mp:
mpa = Gmpms/r2
2
Now a = v /r:
mpv2/r = Gmpms/r2
Cancel mp:
v2/r = Gms/r2
Velocity v = d/t, and d=2r, so v = 2r/T v2 = 4r2/T2
Put in line 3:
(4r2/T2)/r = Gms/r2
Cancel r on left side:
4r/T2 = Gms/r2
Move r2 over from right to left:
4r3/T2 = Gms
Pass 4to right side:
r3/T2 = Gms/4
Seeing as G, ms, 4,  are constants:
r3/T2 = K (Kepler’s Third Law)
So Kepler’s Third Law really a consequence of Universal Gravitation and F = ma.
Example 1: If the year for Mars is 684 Earth days, find its distance from the Sun.
Solution 1: 684/365.25 = 1.87 years.
r13/T12 = r23/T22 So ((1.496*1011)3/12) = (r23/1.872).
3.348*1033 = r23/3.49; 1.168*1034 = r23; r2 = 2.27*1011 meters.
Example 2: Find the Sun’s Mass from the Earth-Sun distance, r=1.496*1011 meters.
Solution 2: 4r3/T2=Gms;4(3.14)2(1.496*1011)3/(3.1557*107)2=(6.67*10-11)ms; ms=1.98*1030 kg
Example 3: It takes Pluto 248 Earth years to orbit the Sun. Find its average distance from the Sun (Note: Pluto’s orbit
is rather eccentric; for portions of it fall inside the orbit of Neptune). How long would it take to send a radio signal
from a spaceproble (like Voyager) visiting the planet?
Solution 3: r13/T12 = r23/T22, so r13/(248)2 = (1.5*1011)3/(1)2; r13 = (3.375*1033)(248)2 = 2.075*1038; r = 5.9*1012 m
Now v = d/t, so t = d/v. Speed of light is same as speed of TV, radio; c = 3*10 8 m/sec; t = 5.9*1012/3*108; t = 19,666
sec = 5.46 hours.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 14 (Mechanics): Work and Energy
Work and Energy: Work is product of force and displacement. If displacement not parallel to force, must include
cos as correction factor, thus:
Work W = Fd
or
W = Fdcos
F = force (Newtons/dynes), d=displacement (m/cm), W=work (Joules in mks/ergs in cgs). Important: if you exert
force perpendicular to displacement (ie, holding bag of groceries aloft and carrying it across floor), you do NO
WORK ,as horizontal displacement at 90 degrees to vertical force.
Varying Force: Equation W=Fd assumes F=constant.
What if force varies? Simply plot changing force F (or
Fcos) vs. displacement d. Seeing as W=Fd, just find area
under curve; total area is F*d, so that’s your work.
F
distance d
Kinetic Energy: Energy is the ability to do work; same units as work. Energy can be from motion; if two cars have
identical speed, but one car is twice as heavy, it will have twice the kinetic energy. Energy also from gravitational
position; if you lift an object to the top of a tall building, it will have more gravitational potential energy, depending on
the mass and the height (if you doubt this, ask yourself: “Which would you rather be hit by...a penny from the 2nd
floor, or a piano from the 10th floor?”) Kinetic energy KE derived by:
Begin with work:
Substitute a:
“d” cancels:
Energy now:
W = Fd. But F=ma, so W=mad. Now vf2 = vo2 + 2ad, so a = (vf2-vo2)/2d
W = m[(vf2-vo2)/2d]d
W = m [(vf2-vo2)/2]
KE = (1/2)m(vf2 – vo2) or (if vo=0) Ek=(1/2)mv2
Gravitational Potential Energy: PE easy to derive:
Begin with Work W = Fd But F=ma, so
W=mad.
Now in gravitational field: a = g
W=mgd
Distance object falls in gravitational field is heighth: Ep=mgh
Other Potential Energies: chemical (stored in a battery), and elastic (stored in a compressed spring). English
physicist Robert Hooke (1635-1703) found force law, known as Hooke’s Law for Springs: F = -kx, F=force (N),
k=spring constant (n/m), x=compression distance of spring. Easy to derive Potential energy from this:
F = -kx when compressed, F=0 when left alone (uncompressed).
So average Force is:
Fav = (0x + kx)/2 = (1/2)kx
Work is Force times distance:
W = Fd = (1/2)kx*x
So Elastic Potential Energy:
Ep = (1/2)kx2
For Chemical Energy in a battery, you need to know Power in watts (Power P = VI, V=volts, I = current in amperes),
and time t to drain battery. Electrical Energy Ee = Pt
Example 1: How much work is needed to accelerate a 1000 kg car from30-40 m/sec?
Solution 1: KE = (1/2)m(vf2 – vo2); KE = (1/2)(1000)(402 – 302); KE = 500(1600-900); KE=350,000 Joules.
d = 20 m

h = dsin
Example 2: How much work is needed to carry a 10 kg backpack up a 20 m, 40 o inclined plane?
Solution 2: PE = mgh; But h= dsin; so PE = mgdsin; PE=(10)(9.8)(20)(0.643) = 1259 Joules.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 15 (Mechanics): Power and Energy Conservation
Work-Energy Principle: For body in motion, net work done on an object is equal to its change in kinetic energy. For
gravitation, net work done on an object is equal to its change in potential energy. For springs, net work done on an
object is equal to its change in elastic potential energy. So summing it up, we get General Work-Energy Principle:
Wtotal = Ek + Ep
Note: You could include other forms of energy in this principle; say a batttery (Ee=Pt) powers a spring (Ep=(1/2)kx2)
which fires a ball (Ek = (1/2)mv2) causing it to rise to a given height by its gravitational potential energy (Ep=mgh).
Then your work-energy principle has 4 changes in energy, not two. Normally we just use kinetic for motion and
potential for everything else.
Conservative vs. Nonconservative Forces: For conservative force, work depends only on the initial and final
positions of the object; how you get there (the detours you take) doesn’t matter (gravity, elastic). If you lift a 5 kg box
10 m, drop it 5, lift it 2, drop it 3, your total is 10-5+2-3 = 4, so work mgh = (5)(9.8)(4) = 196 Joules. Same result for
direct path: mgh = (5)(9.8)(4) = 196. For nonconservative force, the path you take does matter (friction); ie if you
push a box straight across a floor it takes less work than pushing it zig-zag across, because the zig-zag path is longer.
Conservation of Energy: Only works for conservative forces (ie, not friction). Basically, energy conservation says
energy can neither be created nor destroyed; it is just converted into a new form. Formal definition: “If only
conservative forces are acting, the total mechanical energy of a system neither increases nor decreases in any process.
It stays constant---it is conserved.”
Power: Defined as rate at which work is done/energy is transformed from one form into another. Normally defined as
work or energy over time. Units: [Joules/sec=Watts]. Old unit is horsepower, or hp; still often used in cars.
Conversion Unit: 1 horsepower = 746 watts.
Power = Work/Time
P = W/T
[Watts] = [Joules]/[sec]
Power versus Energy: not the same thing! Energy is force over a given distance (W=Fd); but power is rate of energy
transfer. A 100 kg person may be able to easily walk ten kilometers, if he takes enough time about it; but he may not
be able to cover 10 km in one dash, even though his muscles have the stored energy; they just can’t convert it to
motion fast enough (the power is insufficient).
Alternate Formulae: If P=W/T, some easy manipulations give us useful alternates, for power in terms of force,
distance, time, acceleration and velocity:
(1)
P = W/T
(2)
But W=Fd and F=ma, so:
(3)
P = Fd/T and P=mad/T
(4)
Velocity v=d/t, so:
(5)
P = Fv and P=mav
Example 1: A 1 kg cannonball is pressed against a spring of constant k = 250 N/m, compressing it 5 cm. When the
spring is released, calculate the speed of the shot cannonball.
Solution 1: EP=(1/2)kx2; EP=(1/2)(250)(0.05)2; EP=0.3125 J. Since by energy conservation, Ek=EP, 0.3125=(1/2)mv2;
0.3125=(1/2)(1)v2; v2 = 0.625; v=0.8 m/sec.
Example 2: Calculate the velocity needed for a 100 kg man to jump over a 5 meter fence.
Solution 2: Find the work to jump the fence: E P=mgh; EP=(100)(9.8)(5) = 4900 J. Now by energy conservation, E k=Ep
so 4900=(1/2)mv2; 4900=(1/2)(100)v2; v2=98; v=9.9 m/sec.
Example 3: A 1000 kg car has a maximum power output of 240 hp (179,040 Watts). How steep a hill can it climb at
70 km/hr (19.4 m/sec) if theres (a) no friction, and (b) the frictional forces add up to 6000 N?
Solution 3: P=Fv=mgsinv. (a=gsin on hill) 179040=(1000)(9.8)sin(19.4); sin = (179040)/(1000)(9.8)(19.4) =
0.94;  = 70 degrees if no friction. But there is friction, so  must be less. How much less? P=Fv; P=F fricv; P=(6000)
(19.4); Pfric=116,400 Watts to overcome before car goes forward. So total Power is now 179,040-116,40 = 62,640
Watts. P=Fv=mgsinv; 62,640=(1000)(9.8)sin(19.4); sin = 0.329;  = 19.2 degrees.
St. Joseph's Secondary School
Grade 12 Physics University (SPH4U1)
Unit 1, Note 16 (Mechanics): Momentum
Linear Momentum: Basically, product of object’s mass m and velocity v. Symbol is “p”; Measured in kg-m/sec:
p = mv
Force on object needed to change its momentum: in fact, Newton first defines 2 nd force law this way: “the rate
of change of the momentum of an object is proportional to the net force applied on it.” In other words, if you
have 2 identical cars, 1 has twice the velocity, it will also have twice the momentum. If you have motorbike and
truck going same speed, truck will have more momentum (more mass). In addition, if you have car going at 50
km/hr, and you add a force F to it (step on gas) and it increases to 60 km/hr, if you add 2F, it will go to 70 km/hr.
Rewriting Newton’s version of momentum law:
F = p/t, or, using calculus, F = dp/dt
Calculus Shortcuts: Begin with momentum formula to write equations of physics: p = mv; p = m(dx/dt); so
(dp/dt) = m(d2x/dt2). But (d2x/dt2) is acceleration a, so (dp/dt) = ma = Force F. So: derivative of momentum is
force.
Next, try antiderivative (or integral) of momentum: antiderivate of any function f(x) = ax n is
[(1)/(n+1)]ax(n+1) so antiderivative of p = mv is [(1)/(n+1)]mv (1+1), or mv)dt = mvdt = m[(1)/(1+1)v(1+1) =
(1/2)mv2 = Ek. So: Integral or antiderivative of Momentum is kinetic energy.
Conservation of Momentum: Under certain conditions (an elastic collision), momentum is conserved (it stays
the same) before and after. In other words, “the total momentum of an isolated system of bodies remains
constant.”. This allows us to write useful law:
momentum before = momentum after
m1v1 + m2v2 = m’1v’1 + m’2v’2
here, the primes m’, v’ mean mass, velocity after the collision; m,v on their own (no prime) are before the
collision.
Impulse: Seeing as F=p/t, then:
p = Ft.
Quantity p is “impulse”; it is the “change in momentum” of an object (same units, kg-m/sec).
Example 1: What is the momentum of a 22-g sparrow flying at 11 m/sec?
Solution 1: Easy p=mv; p=(0.022)(11); p = 0.242 kg-m/sec
Example 2: A resting atomic nucleus decays into an alpha particle (speed 3.8*10 5 m/sec) and a nucleus (mass 57
times larger). What is the speed of the nucleus?
Solution 2: m1v1+m2v2=m’1v’1+m’2v’2; 0+0=(3.8*105)(1)+(57)v;-57v=3.8*105; v = -6666 m/sec.
Example 3: Calculate the bat-ball force when a 0.145 kg ball at 39 m/sec is hit back at 52 m/sec (contact time
0.001 sec).
Solution 3: F = p/t; p=(mv); p=(0.145)(39--52); p=13.195; F=13.195/0.001; F = 13195 N.
Example 4: A 10,000 kg railroad car travelling at 24 m/sec collides with an identical car at rest. If the cars lock
together as a result of the crash, what is their common speed afterwards?
Solution 4: p = m1v1 + m2v2 = (10,000)(24) + (10,000)(0) = 240,000 kgm/sec. After collision, p = m’v’, so by
conservation of momentum 240,000 = (10,000 + 10,000)v’; v’ = 12 m/sec.
St. Joseph's Secondary School
Grade 12 Physics University (SPH4U1)
Unit 1, Note 17 (Mechanics): Elastic Collisions
Recall from last note: Momentum of an object P = mv; Conservation of Momentum says momentum before =
momentum after an elastic collision. Also Impulse p = Ft.
Energy also Conserved in Collision: If a collision is elastic (no energy lost as heat), then both momentum and kinetic
energy are conserved. This allows us to write a double equation that’s very useful for problem-solving, since it gives
us 2 equations in 2 unknowns:
m1v1+m2v2 = m’1v’1+m’2v’2
(1/2)m1v1 +(1/2)m2v22 = (1/2)m’1v’12+(1/2)m’2v’22
2
(Elastic)
(Elastic)
Elastic Collision in One Dimension: Easy; just apply conservation of momentum, energy:
Begin with momentum:
m1v1+m2v2 = m’1v’1+m’2v’2
Then energy:
(1/2)m1v12+(1/2)m2v22 = (1/2)m’1v’12+(1/2)m’2v’22
Rewrite momentum:
m 1(v1-v1’) = m2(v2’ – v2)
Now rewrite energy:
m 1(v12-v’12) = m2(v’22-v22)
Now algebraically (a2–b2)=(a+b)(a-b) so:
m1(v1-v1’)(v1+v1’) = m2(v’2-v2)(v’2+v2)
Divide Line 5 by line 3:
(v 1+v1’) = (v2+v2’)
Or:
v1-v2 = v’2-v’1
What it means: For any elastic head-on collision, the relative speeds of the two particles after the collision is the same
as it was before, no matter what the masses are.
before
collision
after
Elastic Collisions in Two/Three Dimensions: Much more challenging; billiard balls, atomic physics (particle
accelerators), car crashes, etc. Suppose an object m 1 heads towards another object m2, which is at rest. If they deflect
in 2 dimensions, we get these equations:
m1
m1
m2
’1
’2
(I) (1/2)m1v12 = (1/2)m’1v’12+(1/2)m’2v’22
(II) m1v1=m1v’1cos1’+m2v’2cos’2 (x-momentum cons.)
m2
(III) 0 = m1v’1sin’1 + m2v’2sin’2 (y-momentum conserved, and equal to zero)
Since we have 3 equations, if we have the masses and starting v 1 (v2=0), all we need is 1 of the other 4 variables (v’ 1,
v’2, ’1, ’2) to solve the equations; lots of math, true, but doable!
Example 1: A billiard ball of .25 kg moving with speed 4 m/sec collides with a second billiard ball of identical mass at
rest. What are the speeds of the two balls after the collision (assuming it is elastic)?
Solution 1: m1v1+m2v2 = m’1v’1+m’2v’2, but v2 = 0. Second ball at rest, so m1v1 = m’1v’1+m’2v’2. Now masses are
identical, so they cancel out: v 1 = v1’ + v2’. But v1 = 4, so v1’ + v2’ = 4. Now: v1-v2 = v’2-v’1, but v2 = 0 and v1 = 4; so
4 = v2’ – v1’. Add two equations together; 8 = 2v 2’; v2’ = 4, so v1’ = 0. The conclusion: the moving billiard ball is
stopped by the collision, and the one at rest moves off speed 4.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 18 (Mechanics): Inelastic Collisions
and Center of Mass
Elastic vs. Inelastic Collisions: Recall in elastic collisions, momentum p=mv and kinetic energy KE=(1/2)mv 2 both
conserved; ideal situation, no friction. But in real world, most collisions inelastic, some kinetic energy KE converted to
heat Q. Even so, total energy KE+Q always conserved, as is total vector momentum. Best example of inelastic
collision: blob of putty that stops and deforms itself (heat Q) when it collides with another; final KE zero.
Ballistic Pendulum: A device used to measure projectile/bullet speed. Basically, it is large block of mass M attached
to string, suspended from ceiling. When bullet of mass m, velocity v hits block, the two (m+M) rise together with
velocity v' to height h. As total energy constant, possible to calculate v of bullet:
Conservation of momentum:
mv = (m+M)v'
So:
v = [(m+M)/m]v'
Conservation of energy:
KE = PE
So:
(1/2)(m+M)v'2 = (m+M)gh
Cancel (m+M):
(1/2)v' 2 = gh
Multiply by 2, take root:
Now substitute line 2 for v':
v'2 = 2gh
so: v' = √ 2gh
v = [(m+M)/m]√2gh
(formula for Ballistic Pendulum)
Example 1: A 20 gram bullet strikes a 1 kg ballistic pendulum, making it rise a vertical distance of 0.5 m. Calculate
the speed of the bullet.
Solution 1: v2=[(m+M)/m]2(2gh); v2=(1.02/0.02)2(2*9.8*0.5); v2=(51)2(9.8)=25489; v= 159 m/sec.
Center of Mass: Critical idea in physics. How do you calculate motion of complex object (like an 8-ball that moves
forward and spins at same time, or diver that jumps up, goes down, and twists/turns as she dives )? You assume that
the object moves as though all its mass concentrated at one poinr, called center of mass (XCM). This can be proven, but
you need University Calculus to do it, so we'll just give working equations for now. Assume two masses, m 1 and m2,
along x-axis, at points x1, x2. Sum of two masses M=m1+m2. So Center of mass XCM defined to be:
Center of mass XCM = (mlxl + m2x2)/(ml+m2) = (mlxl + m2x2)/M
Example 2: Five particles of equal mass 1 kg lie along the following positions on the x-axis: x=l m, x=2 m, x=3 m,
x=4m, x=5 m. Find the center of mass.
Solution 2: Chosen because answer obvious. By symmetry, x=l and x=5 centered on x=3; also x=2, x=4 also centered
on 3. x=3 obviously centered on 3. So center of mass x=3.
Okay, Now To Prove it: XCM = [m1x1+...m5x5]/(5m); XCM = (1)(x1+...x5)/5(1); XCM = (1)(1+2+3+4+5)/5; XCM = 15/5;
So XCM = 3.
Example 3: A rocket is fired into the air; at its highest
point (half the range R, or D), it separates into two
equal mass stages, I and II. Stage I falls vertically
back to Earth. Where does stage II land?
Solution 3: The CM of the rocket follows a parabolic
path back to its landing position at Range R (R=2D).
Since stage I lands at distance D from the CM, stage II
must also land at a distance D from the CM, for a total
distance of 3D. Distance D = 3D.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 19 (Mechanics): Angular Momentum
Brief Note on Terms: A Rigid body is body with shape that doesn't change; all particles in same positions relative to
each other (as opposed to fluids). Rotational Motion: all points in body move in concentric circles with same center,
called Axis of Rotation. Because rotational motion circular, 1 revolution = 360 degrees. Mathematicians use radians,
defined 1 rad = 57.3 degrees (so 2 rad = 2*3.14*57.3 = 360 degrees).
Angle (in radians) = length around circle l/ radius r
Angular velocity () = Angle /time t
Angular Acceleration = /t
l
Now v =d/t = l/t = r/t; so v= r
Tangential acceleration aT=v/t = r/t; aT = r.
x
 2
2
2
Centripetal Acceleration aC = v /r =  r /r; aC=  r.
Frequency f = /2 Time Period T = l/f
Summary of Angular Momentum Definitions/Terms:
= l/r
aT = r.
(2)  = /t
(6) ac = 2r
(3)  = /t
(7) f
(4) v = r
(8) T = 1/f
Kinematic Equations for Uniformly Accelerated Motion: Just translate old equations into new ones, replacing (1)
linear displacement x by angle , (2) velocity v by angular velocity , and (3) acceleration a by angular acceleration .
Old Linear Equations
v = vo + at
x = vot + (1/2)at2
vf2 = vo2 + 2ax
vav = (vo + vf)/2
New Angular Equations
 = o + t
 = ot + (1/2)t2
f2 = o2 + 2
av = (o + f)/2
Example 1: A centrifuge is accelerated from 0-20,000 rpm in 4 min. Find angular acceleration.
Solution 1: =(20,000 rpm)(2radians/60 sec)=2100 rad/sec. = t = 2100/240; =8.75
Example 2: Through how many turns has the rotor turned?
Solution 2:  = 0 + (1/2)t2;  = 0 + (0.5)(8.75)(240)2;  = 252,000; N = 252,000/2 = 40107.
Example 3: What is the linear speed 1.5 m from the center of a circle rotating once every 4 sec?
Solution 3: f= 1/T = 1/4 = 0.25.  = 2f (Eq. 7) so = 2(3.14)(0.25) = 1.57 rad/sec. v = r so v = (1.5)(1.57);
therefore v= 2.355 m/sec.
Example 4: A bicycle wheel slows down from a speed of 8.4 m/sec to rest, over a distance of 115 meters. Each wheel
has a diameter of 68 cm. Find the (a) angular velocity of the wheel when it first starts decellerating: (b) the total
number of revolutions before the wheel stops; (c) the angular acceleration of the wheel, and (d) the time it takes to
come to a stop.
Solution 4: (a) If wheel diameter 68 cm, radius r = 0.34 m. v = r so v/r;  = 8.4/0.34 = 24.7 rad/sec (b) 2r
meters/rev, so number of revs = d/2r; N = 115/2(3.14)(0.34); N = 53.8 revs. (c) = l/r = 115/0.34 = 338 rad. Now
f2 = o2 + 2so = (f2 - o2)/= (0 – 24.72)/(2)338;  = -0.902 rad/sec2. (d)  = o + t, so t = ( - o )/; t =
(0 – 24.7)/-0.902; t = 27.4 seconds.
St. Joseph’s Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 20 (Mechanics): Torque
Recall: Summary of angular momentum terms:
= l/r
(2)  = /t
(3)  = /t
aT = r.
(6) ac = 2r
(7) f
Also: Old Linear Equations
v = vo + at
x = vot + (1/2)at2
vf2 = vo2 + 2ax
vav = (vo + vf)/2
(4) v = r
(8) T = 1/f
r
l
x
New Angular Equations
 = o + t
 = ot + (1/2)t2
f2 = o2 + 2
av = (o + f)/2
Torque: Try to open a door, apply same force to handle or close to hinge; handle force works better. Why? Effect of
force proportional to (a) its size and (b) perpendicular distance from axis of rotation (door hinge) to place where
force acts. This distance is called the lever arm (or moment arm). Product of force and lever arm is called the
torque  (unit N-m):
 = Fr,
= Frsin
F=force (N), r = lever arm (m) (if F at 90o to r)
(if F at any other angle)
Derivation of Force Law: Angular Acceleration proportional to F (because F=m). As tangential acceleration T=
r then the Force F and Torque  is given by:
F = mr
and
 = mr2
Moment of Inertia: Most rotating rigid bodies made of many particles, all revolving around one center. Apply above
torque equation to each particle, you get:
 = mr2mr2)
 = sum of all moments
Now mr2) is given a new name: it’s the Moment of Inertia of all the particles in the body put together. Moment of
Inertia mr2)= I, and is the sum of all moments, for each particle, together:
mr2)= I = m1r12 + m2r22 + m3r32 + ….mnrn2
So Sum of all Torques 
is:  = mr2) so 
This is rotational/angular equivalent of Newton’s Second Law, F=ma, and is valid for rigid body rotating around
fixed axis. Also works if body is moving forward with constant v, or even accelerating with constant a (so long as the
rotational axis doesn’t change direction).
Example 1: (a) Calculate the torque on the hinge of a door or width 1.3 meters if you exert a 85 Newton force on it.
(b) What happens if you exert the force at an angle of 45 degrees?
Solution 1: Easy,  = Fr, so  = (85)(1.3) = 110.5 Nm. If at a 45 degrees angle, = Frsin, so (85)(1.3)(0.7071)
= 78.13 Newtons
St. Joseph’s Secondary School
Grade 12 University Physics
Unit 1, Note 21 (Mechanics): Gravitational Potential Energy
Achieving Circular Orbit: With all our circular formulas, we can now determine how much energy it takes
for a spaceship to achieve orbit. The full derivation uses calculus, but we can use the estimation of the area
under a Force-displacement curve instead (recall that Energy is force times displacement, or, the area under
the Force-displacement curve).
Satellites: To lift satellite mass m to height h, potential energy formula E = mgh.
Now you’re lifting satellite away from the earth’s gravitational field.
Force between earth (Mass M) and satellite (mass m) given by F = GMm/r2.
Ordinarily: you’d get the work/energy from the Force by multiplying by
displacement d; W = Fdcos. But that supposes Force F doesn’t change! As you
get further from Earth (bigger d), Force decreases (inversely: bigger radius means
lower force). Three ways to do it: (1) use calculus (an integral or antiderivative, of
Fdr), (2) use the area under the Force-Distance graph, or (3) find the difference in
energies between two points, one at ground level r1, the other in outer space r2:
Energy E = Fd, so E = (GMm/r2)r; E = (GMm/r)
So the change in Energy from r2 to r1 is:
E = (GMm/r2) – (GMm/r1)
Now imagine the satellite is launched to a point far away from the earth’s gravitational field (say, a space probe). If r 1
= earth radius (where probe is launched), r 2 = infinitely far away, so (GMm/r2) = 0. Thus the energy needed to escape
Earth’s gravitation field (or gravity well) becomes:
E = -(GMm/r1)
Now suppose rocket wants to escape Earth’s gravitation. How fast does it need to go? Set the rocket’s kinetic energy,
Ek equal to the rocket’s gravitational energy.
Kinetic Ek=(1/2)mv2; Gravitation Ep=(GMem/re) (1/2)mv2 = (GMem/re)
Cancel out the rocket’s mass m:
(1/2)v 2 = (GMe/re)
Double both sides, then take square root:
v = (2GMe/re)
Me = 5.98*1024 kg, re = 6.38*106 m:
v = (2*6.67*10-11 * 5.98*1024/6.38*106)
Resolve for v:
v = 11,200 m/sec, or 11.2 km/sec (Escape velocity).
Example 1: Find the escape velocity of an astronaut returning from the Moon.
Solution 1: v = (2GMm/rm); Mass of moon Mm is 7.35*1022 kg, radius of moon rm is 1.74*106 m. So v =
(2*6.67*10-11 * 7.35*!022/1.74*106); v = 5635000; v = 2,378 m/sec, or 2.3 km/sec. So launching a rocket from the
Moon requires only 20% of the speed of Earth.
St. Joseph’s Secondary School
Grade 12 University Physics
Unit 1, Note 22 (Mechanics): Binding Energy
Recall: The Gravitational Energy needed to lift a satellite of Mass m from the ground (earth radius) to some orbital
height r is given by E = (GMm/r) – (GMm/re). The Gravitational Energy needed to escape Earth’s gravity field
(the gravity well) is E = -(GMm/r1). This implies an escape velocity for the Earth of v = (2GMe/re), or v = 11.2 km/
sec.
Orbiting Satellite: Now suppose a satellite of mass m is orbiting the Earth with orbital radius r. If the orbit is circular,
the force of gravity must balance the centripetal force (if gravity is larger, the satellite spirals down to the ground and
burns up. If the centripetal force is larger, the satellite flies away from the earth). Centripetal force F c is easy. F=ma,
but acceleration is circular, ac = v2/r, so the centripetal force is Fc = mv2/r. The gravitational force is Fg = GMm/r2.
Equate the two: mv2/r = GMm/r2. Now cancel radius r: mv2 = GMm/r (Equation A).
Energy of a satellite in orbit: If orbit is constant, there is kinetic energy Ek and potential energy Ep.
Kinetic energy is Ek = (1/2)mv2, and Gravitational Potential Energy Ep is -GMm/r.
So Total energy ET = Ek + Ep;
ET = (1/2)mv2 - GMm/r.
Equation A says mv2 = GMm/r, so:
Tidying up:
ET = (1/2)GMm/r – GMm/r
ET = -(1/2)GMm/r, or ET = -(1/2)Ep
In short: total energy of a satellite in orbit of radius r is negative and equal to half the gravitational potenrial
energy at that orbital radius. There are three possible energy states: E below zero (negative), E equal to zero, or E
above zero (positive). If you can increase the energy to zero (binding energy), the object will just escape the Earth’s
gravity field into space. If you can get it above zero, the extra energy E will give the satellite kinetic motion by Ek =
(1/2)mv2; but if the energy is less than zero, the satellite is trapped in earth’s gravity field forever.
Escape Velocity: For satellite to escape Earth’s gravity, speed must be at least escape velocity (for earth, v=11.2
km/sec). Fastest speed possible is speed of light (c=3.00*10 8 m/sec); if object is massive enough/small enough,
light/radiation can get “trapped” inside, resulting in a black hole from which nothing can escape. The radius of the
hole (a superdense collapsed star) is called the Schwatzchild radius (named after astronomer Karl Schwartzchild
(1873-1916)) or “Event Horizon”; if a spaceship approaches closer than this, it will be trapped forever.
Example 1: Calculate the radius of a black hole
made by a giant star of 30 solar masses.
Solution 1: Msun = 1.99*1030 kg, so 30M =
5.97*1031 kg. Now v = (2GM/r), so v2 = 2GM/r.
We want r, so r = 2GM/v2. r = 2*6.67*1011
*5.97*1031/(3*108)2; r = 7.96*1021/9*1016; r =
88,488 meters; r = 88.4 km. So: a star 30 times
bigger than the sun is packed into a space smaller
than radius 100 km, and if you get to within 88 km
of it, you are doomed!
St. Joseph' Secondary School
Grade 12 University Physics (SPH4U1)
Unit 1, Note 23 (Mechanics): Rotational Energy
Moments of Inertia: Symbol I = mr2; unit kg-m2. Recall Rotational analogue of Newton’s Second Law is  =
mr2)so If we know angular acceleration  and moment of inertia I, we can get torque  (in N-m). But
it’s a pain to calculate mr2) every time, so we use a table for the Moments of Inertia for common objects:
Object
Location of Axis
(a) Thin hoop of
Radius R
(b) Cylinder of
Radius R
Moment of Inertia
Through center
Through Center
(c) Sphere of
Through Center
Radius R
(d) Rod Length L Through Center
(e) Rod Length L Through End
MR
(1/2)MR 2
Radius of Gyration k
2
R
1.414R or (root 2)R
(2/5)MR2
0.632R or (root 2/5)R
(1/12)ML2
(1/3)ML2
0.288L or (root 1/12)L
0.577L or (root 1/3)L
Radius of Gyration: A sort of “average radius”; symbol is k. It is defined to be the point at which, if all the mass of
the object was concentrated there, it would have the same moment of inertia as the original object. So you can
redefine moment of inerta I of any object can be rewritten this way:
I = Mk2
I=moment of inertia (kg-m2), M=mass (kg), k=radius of gyration (meters m)
Kinetic Energy: Normal formula KE=(1/2)mv2. But we’ve made an analogy between linear velocity v and angular
velocity . We’ve also made analogy between mass m and moment of Inertia I. So rewrite the Kinetic Energy
Formula for Rotational Kinetic Energy:
Rotational KE = KER = (1/2)mv2
KER = (1/2)(mr22)
KER = (1/2)(mr2)2
2
Since mr = I = moment of inertia:
KER = (1/2)I2. (Body Rotating).
KE = (1/2)mv2
Now suppose body is rolling downhill (ie, going forward while it is rotating; such as a bowling ball going down a hill).
Then we must add the linear Kinetic Energy KE = (1/2)mv2 to get the total Kinetic Energy, like so:
KEtotal = (Rotational KE) + (Linear KE)
KEtotal = (1/2)I2 + (1/2)mv2 or KEtotal = (1/2)(I2 + mv2)
Example 1: What’s the formula for the speed of a solid marble of Mass m and Radius r if it rolls down an incline
plane of vertical height h?
Solution 1: Use conservaton of energy, but include rotational kinetic energy this time. E k = Ep; Ep = mgh; Ek =
(1/2)mv2 + (1/2)I2; so mgh = (1/2)mv2 + (1/2)I2; Now I for marble (sphere) is I = (2/5)mr 2; therefore mgh =
(1/2)mv2 + (1/2)(2/5)(mr2)2; now  = v/r, so 2 = v2/r2. Then mgh = (1/2)mv2 + (1/2)(2/5)(mr2)v2/r2). Cancel m and
v, you get gh = (1/2)v2 + (1/5)v2; gh = (7/10)v2; v2 = 10gh/7; therefore v = (10gh/7)