Physics Subject
Area Test
MECHANICS:
DYNAMICS
Motion in a circular path at
constant speed
Speed constant, velocity
changing continually
Velocity changing direction, so
there is acceleration
Called centripetal acceleration,
since it is toward the center of
the circle, along the radius
Value can be calculated by
many formulas, first is
ac = v2/r
*
A bicycle racer rides with constant speed around a circular track 25 m
in diameter. What is the acceleration of the bicycle toward the center
of the track if its speed is 6.0 m/s?
ac = v2 =__(6.0 m/s)2 = 36 (m/s)2 = 2.9 m/s2
r
12.5 m
12.5 m
Rotation and Revolution
Rotation-Around an Internal Axis-Earth
rotates 24 hours for a complete turn
Linear (tangential) versus rotational
speed
Linear is greater on outside of disk
or merry-go-round, more distance
per rotation
Linear is smaller in middle of disk,
less distance per rotation.
Rotational speed is equal for both
Rotations per minute (RPM)
Linear speed is proportional to both
rotational speed and distance from
the center
Rotation and Revolution
Revolution-Around an
External Axis-Earth revolves
365.25 days per trip around
sun
Same relationship
between linear and
revolutional speeds as
with rotational
Planets do not revolve
at the same
revolutional speeds
around the sun
Another important measure in UCM is period,
the time for 1 rotation or revolution
Since x=v0t , this implies that vT = 2r and thus
T= 2r/ v
Rearranging differently, v= 2r/ T and then
inserting it into the acceleration equation
ac = v2/r = 42r/T2
*
Determine the centripetal acceleration of the moon as it circles the earth,
and compare that acceleration with the acceleration of bodies falling on
the earth. The period of the moon's orbit is 27.3 days.
According to Newton's first law, the moon would move with constant
velocity in a straight line unless it were acted on by a force. We can
infer the presence of a force from the fact that the moon moves with
approximately uniform circular motion about the earth. The mean
center-to-center earth-moon distance is 3.84 x 108 m.
*
ac= 42r = 42(3.84 x 108)
T2
T= 27.3 da (24 hr/da)(3600 s/hr) = 2.36 x 106 s
(2.36 x 106)2
ac = 2.72 x 10-3 m/s2
The ratio of the moon's acceleration to that of an object falling near the
ac = 2.72 x l0-3 m/s2 =
earth is
g
9.8 m/s2
3600
1
*
The number of revolutions per time unit
Value is the inverse of the period, 1/T
Units are sec-1 or Hertz (Hz)
Inserting frequency into the ac equation
ac=42f 2r
*
An industrial grinding wheel with a 25.4-cm
diameter spins at a rate of 1910 rotations per
minute. What is the linear speed of a point on
the rim?
The speed of a point on the rim is the distance
traveled, 2r, divided by T, the time for one
revolution. However, the period is the
reciprocal of the frequency, so the speed of a
point on the rim, a distance r from the axis of
rotation, is
v = 2rf
v= (2)(25.4cm/2)(1910/1 min)(1min/60s)
v = 2540cm/s = 25.4 m/s.
*
Velocity can be defined in terms of
multiples of the radius, called radians
There are 2 radians in a circle, and so
the angular velocity w = v/r
In terms of period w= 2/T
In terms of frequency w=2f
*
At the Six Flags amusement park near Atlanta.
The Wheelie carries passengers in a circular path
with a radius of 7.7 m. The ride makes a
complete rotation every 4.0 s. (a) What is a
passenger's angular velocity due to the circular
motion? (b) What acceleration does a passenger
experience?
a) The ride has a period T = 4.0 s. We can use it to compute the
angular velocity as
2=
T
2 rad =  rad/s = 1.6 rad/s
4.0 s
2.0
(b) Because the riders travel in a circle, they undergo a centripetal
acceleration given by
ac= w2r = (/2 rad/s)2(7.7m) = 19m/s2.
Notice that this is almost twice the acceleration of a body in free fall.
Angular Velocity and Acceleration
Any real object that has a definite shape can be made to rotate – solid,
unchanging shape
Angular displacement -- q -- Radians around circular path
Angular velocity -- w --radians per second, angle between
fixed axis and point on wheel changes with time
Angular acceleration -- a -- increase of w , when angular
velocity of the rigid body changes, radians per seconds
squared
Rotational velocity, displacement, and acceleration all follow the
linear forms, just substituting the rotational values into the
equations:
q= wot + 1/2 at2
wf=w0 +at
q= x/r
a= a/r
wf2 = wo2 + 2aq
q=(w0 +wf) t/ 2
w= v/r
*
The wheel on a moving car slows uniformly from 70 rads/s to 42 rads/s in 4.2 s. If
its radius is 0.32 m:
a. a=
a. Find a
b. Find q
c. How far does the car go?
Dw = (42-70) rads/s = -6.7 rads/s2
Dt
4.2 s
b. q= wot + 1/2 at2 = (70)(4.2) + 1/2(-6.7)(4.2)2 = 235 rads
c. q = x / r in rads so x = q r = (0.32)(235) = 75 m
*
A bicycle wheel turning at 0.21 rads/s is brought to rest by the brakes in exactly 2 revolutions.
What is its angular acceleration?
q = 2 revs = 2(2) radians = 4 rads
wf=0 rads/s wo= 0.21 rads/s
Use angular equivalent of vf2 = vo2 + 2ax which is
wf2 = wo2 + 2aq
(0)2 = (0.21)2 + 2a(4)
a = -(0.21)2 = -1.8 x10-3 rad/s2
2(4)
Forces in Circular Motion
Centripetal Force
Force toward the center from an object, holding it in circular motion
At right angle to the path of motion, not along its distance, therefore
does NO work on object
Examples
Gravitation between earth and moon
Electromagnetic force between protons and electrons in an atom
Friction on the tires of a car rounding a curve
Equation is Fc=mac = mv2/r
*
Approximately how much force does the earth
exert on the moon? Moon’s period is 27.3 days
Assume the moon's orbit to be circular about a
stationary earth. The force can be found from
F = ma. The mass of the moon is 7.35 x 1022
kg.
Fc= mac = m 42r
T2
Fc = (7.35 x 1022 kg)42(3.84 x 108m)
((27.3 days)(24 hr/day)(3600 s/hr))2
Fc=2.005 x 1020 N.
*Forces in Circular
Motion
Centrifugal “Force”
Not a true force, but really the
result of inertia
“Centrifugal force effect” makes
a rotating object fly off in
straight line if centripetal force
fails
*
Imagine a giant donut-shaped space station located so far from all heavenly
bodies that the force of gravity may be neglected. To enable the occupants to
live a “normal” life, the donut rotates and the inhabitants live on the part of
the donut farthest from the center. If the outside diameter of the space station
is 1.5km, what must be its period of rotation so that the passengers at the
periphery will perceive an artificial gravity equal to the normal gravity at the
earth's surface?
The weight of a person of mass m on the earth is a force F = mg.
The centripetal force required to carry the person around a circle of radius r is
F =mac = m 42r
T2
We may equate these two force expressions and solve for the period T:
750m
r
2
mg =m4 r
g
9.81m / s2
T2 T=2
=2
*
“Banking” road curves makes turns without skidding possible
For angle q, there is a component of the normal force toward the center
of the curve, thus supplying the centripetal force. The other
component balances the weight force.
FN sin q = mv2/r
FN cos q = mg
tan q= v2/gr
thusly q = tan-1 (v2/gr)
This equation can give the proper angle for banking a curve of any
radius at any linear speed
*Banked Curve Example
A race track designed for average speeds of 240 km/h (66.7
m/s) is to have a turn with a radius of 975 m. To what angle
must the track be banked so that cars traveling 240km/h have
no tendency to slip sideways?
Determine q from
q = tan-1 (v2/ g r)
= tan-1 (66.72/9.81(975))= 24.9o
*
Newton’s first initiative for the Principia was
investigating gravity
From his 3rd law, he proposed that each object
would pull on any other object
He likewise noted differences due to distance
His final relationship was that Force was
proportional to masses and inversely proportional
to distance squared
Using a constant Fg = Gm1m2
r2
*
Newton found that his law would only work
when measuring from the center of both
objects
This idea is called the center of gravity
Sometimes it is at the exact center of the object
Sometimes it may not be in the object at all
All forces must be from the CG of one object
to the CG of the other object
*
G was elusive to find since gravity is a weak
force if masses are small
Cavendish developed a device which made
measurement of G possible
The value of G is 6.67 x 10-11 N m2
This puts Fg in Newtons
kg2
G can be used then to find values of many
astronomical properties
*
Consider a mass m falling near the earth's surface. Find its acceleration in terms
of the universal gravitational constant G. The gravitational force on the body
is
F = GmME
r2
ME = mass of the earth
r = the distance of the mass from the center of the
earth, essentially the earth's radius.
The gravitational force on a body at the earth's surface is F = mg.
mg= GmME
or g =GME
r2
r2
Both G and ME are constant, and r does not change significantly for small
variations in height near the surface of the earth. The right-hand side of this
equation does not change appreciably with position on the earth’s surface, so
replace r with the average radius of the earth RE
g = GME
RE2
*
Show that Kepler’s third law follows from the law of universal gravitation. Kepler’s third
law states that for all planets the ratio (period)2/ (distance from sun)3 is the same.
Make the approximation that the orbits of the planets are circles and that the orbital speed
is constant.
The sun's gravitational force on any planet of mass m is
F= GmM
r2
M =the mass of the sun. Because the mass of the sun is so much larger than the mass of
the planet, we can assume, as Kepler did, that the sun lies at the center of the planetary
orbit. The circular orbit implies a centripetal force. This net force for circular motion is
provided by the gravitational force. Equating these two forces, we get
Fc=GmM = 42mr
Rearranging gives T2 =
42
r2
T2
r3
GM
*Example
Use the law of universal gravitation and the measured value of the acceleration of gravity g to
determine the average density of the earth. The density, r of an object is defined as its mass
per unit volume: r = m/V where m is the mass of the object whose volume is V
From a previous example
g = GME
RE2
Substitute for M an expression involving r, r = ME/V.
If we take the earth to be a sphere of radius RE. Then
r = ME
and ME= 4/3RE3 r
4/3RE3
The equation for g can then be rewritten in terms of the density as
g = G(4/3RE3 r) = 4/3GRE r
RE2
*
Upon rearranging, we find the density to be
r=
3g
4REG
Inserting the numerical values, we get
r =
3(9.81 m/s2)
4(6.38 x 106 m)(6.67 x 10-11 N m2/ kg2
r = 5.50 x 103 kg/m3
*Escape Velocity
Earth spacecraft must get entirely away from the earth to go on to other planets
This requires giving a spacecraft enough energy to overcome the gravitational
potential energy of earth
This gives an equation such that
Where M and R vary
according to the
vesc
2GM
=
R
celestial object involved
*
If the escape velocity is equal to the speed of light, gravity will
keep even light from escaping--the idea behind the black hole
Conjecture due to observations from space
Theory is a supergiant star collapses in on itself creating super
strong gravity at a small point
*Black Holes
Gravity is great due to small
distance with huge mass
Gravity only great near the
object, at distance gravity is
no different
*Keplers Laws
First Law: Each planet travels ina an elliptical path around the
sun, and the sun is at one of the focal points
*Keplers Laws
Second Law: An imaginary
line is drawn from the sun to
any planet sweeps out equal
areas in equal time intervals.
*Keplers Laws
Third Law: The square of a
planet’s orbital period (T2) is
proportional to the cube of the
average distance (r3) between
the planet and the sun or
T 2 ∝ r3
*
ac=v2/r
f=1/T
ac=42r/T2
Fc=mac
ac=42rf2
Fc=mv2/r
T2 = 42
r3
GM
ac= w2r
w=v/r
Fg=GMm/r2
w=2/T
w=2f
v=2r/T = 2rf
T=2
r
g
*
*Let F be a force acting on an object, and let r be
a position vector from a rotational center to the
point of application of the force. The magnitude
of the torque is given by
 = rF sin q
*q = 0° or q = 180 °:
torque are equal to zero
*q = 90° or q = 270 °: magnitude of torque attain
to the maximum
*
*The SI units of torque are N.m
*Torque is a vector quantity
*Torque magnitude is given by
 = rF sin q
*Torque will have direction
* If the turning tendency of the force is counterclockwise, the
torque will be positive
* If the turning tendency is clockwise, the torque will be
negative

The work done by the torque is given by
W = q
*
* In rotation problems, its not
only the mass of an object
that is important but also its
location
* The spacial distribution of
the mass of an object is
called the
* Moment of inertia
(I)
I =1/2 MR 2
*
*When a rigid object is subject to a net torque (≠0),
it undergoes an angular acceleration
 = Ia
*The angular acceleration is directly proportional to
the net torque
*The angular acceleration is inversely proportional
to the moment of inertia of the object

*The relationship is analogous to
F = ma
*
*The work-energy theorem tells us
*When Wnc = 0,
Wnet = DKE + DPE
KE i + PE i = KE f + PE f
*The total 
mechanical energy is conserved and remains
the same at all times

1 2
1 2
mvi + mgyi = mv f + mgy f
2
2
*Remember, this is for conservative forces, no
dissipative forces such as friction can be present
*
*A ball is rolling down a ramp
*Described by three types of energy
*Gravitational potential energy
*Translational kinetic energy
*Rotational kinetic energy
PE = Mgh
KE t =
1
Mv 2
2
1 2
KEr = Iw
2

*Total energy of a system
E=
1
1
Mv 2 + Mgh + Iw 2
2
2
*
* Conservation of Mechanical Energy
(KEt + KEr + PEg )i = (KEt + KEr + PEg )f
* Remember, this is for conservative forces, no dissipative
forces such as friction can be present
1
1 2 1
1
2
2
2
mvi + mgyi + Iw i = mv f + mgy f + Iw f
2
2
2
2

*
*The work done on the body by the external
torque equals the change in the rotational
kinetic energy W = DK = D 1 Iw 2
1 22 1 2
W = K 2 - K1 = Iw 2 - Iw1
2
2
*The work equals the negative of the change in
potential energy

1
1 2
2
-U2 + U1 = Iw 2 - Iw1
2
2
*Conservation of Energy in Rotational Motion
1 2
E = U + Iw = constant
2

*
*Choose two points of interest
*One where all the necessary information is given
*The other where information is desired
*Identify the conservative and non-conservative
forces
*Write the general equation for the Work-Energy
theorem if there are non-conservative forces
*Use Conservation of Energy if there are no nonconservative forces
*Use v = w to combine terms
*Solve for the unknown
*
* The “upthrust” or buoyancy is equal to the weight of the
displaced fluid
* V = volume of liquid displaced
* G = the acceleration of gravity
Fbouyant = rVg
Objects that are less dense than water will float
To find where an object sits in the water, find the ratio of its density to the
density of water
roak
rwater

= percent
submerged
*
Bernoulli's Principle: slower moving air
below the wing creates greater pressure
and pushes up.

*
Q = Av cosq
* Q = volumetric flow rate
* A = cross sectional area
* v = fluid velocity
* q = the angle between the direction of the fluid flow and a vector normal to A
m = rvA = rV
*
Mass flow rate (m)
*
When a pipe is constricted, the mass flow rate is conserved


v1A1 = v 2 A2
*
1 2
rv + rgh + p =*constant
2
Where
v = fluid velocity
g = acceleration due to gravity
h = height
p = pressure
r = fluid density