Circular Motion Kinematics
Centripetal Acceleration
v1
l
v
r

v = v2 + (-v1)
Triangles in the same
proportion
v2
v2
v
r

-v1
An object moves around a circle at
constant speed. By geometry the triangle
formed by the radii and l is in the same
proportion as the triangle formed by the
velocity vectors and v
v = l
v
r
Now… ac = v = v l = v2
t
r t
r
So… v = l v
r
Note that the centripetal acceleration
(v2/r) is directed towards the center of
the circle
Circular Motion Dynamics
Centripetal Force
According to Newton’s second law (Fnet = ma) an
object moving in a circle at constant speed must
have a net force in the same direction as the
acceleration….
This must be towards the center of the circle
Fc = m v2
r
The force that keeps a car moving around a
bend in the road at constant speed is friction
directed towards the center of the turn
The force that keeps the moon moving around
the earth at a constant rate is the force of
gravity directed towards the center of the
Earth.
Circular Motion Dynamics
Centripetal Force
A 1000 kg car rounds a curve on a flat road of radius 50m at a speed of 14m/s.
Will the car make the turn of will it skid if a) the pavement is dry (Ffr = 5880 N)
and b) the pavement is icy (Ffr = 2450 N )? c) Does the ability of the car to make
the turn depend on the mass of the car?
m = 1000 kg r = 50m v = 14 m/s
a) FFr = 5880 N
Force needed is… FC = m v 2
r
= (1000kg)(14m/s) 2
= 3920 N
(50m)
The maximum friction force that can be provided by the car’s tires is bigger than
the force needed to keep the car turning so it will be able to make the turn
b) FFr = 2450 N
The maximum friction force that can be provided by the car’s tires is now smaller
than the force needed to keep the car turning so it won’t be able to make the
turn
c) The more mass the car has, the more friction between the tires and the road
but the car has proportionally more resistance to a change in motion. This will
offset the increased friction, so bigger cars can’t turn quicker. (plus the higher
center of gravity makes them more likely to role!)
Circular Motion Dynamics
Centripetal Force
The Moon (m= 7.35 x 1022 kg) orbits the Earth at a distance of 384 x 106 m. a)
What is the moon’s acceleration towards the Earth? b) How much does the Earth
pull on the moon? c) Does the moon pull on the Earth more or less or the same
as the Earth pulls on it? (The moon moves around the Earth one orbit every 27.4
days)
mm = 7.35 x 1022 kg
r = 384 x 106 m
T = 27.4 days v = ? Fg = ?
v = d / t = 2r / T
Fc
v
v = 2 (384 x 106m) / (27.4D x 24 H/D x 3600 s/H) = 1019 m/s
ag = v 2
= (1019 m/s)2
= 0.0027 m/s2
(384 x 106 m)
r
Fg = mm v 2
r
= (7.35 x 1022kg) (1019 m/s)2
= 1.99 x 1020 N
(384 x 106 m)
The moon pulls equally back on the Earth according to Newton’s third law of motion. The
moon moves around the Earth because of its much smaller mass (inertia).
Circular Motion Dynamics
Universal Gravitation
We can define a field as an area of force that surrounds an object. Gravitational
fields exist around all masses.
The amount of gravitational force between two objects is due to the amount of
mass each object has.
It decreases in strength with distance in proportion to the inverse of the
distance squared.
M1
Fg
Fg
r
M2
Fg = G M1 M2
r122
G = Universal Gravitational Constant = 6.67 x 10-11 Nm2/kg2
M = Mass of objects (kg)
r = distance separating centers of masses (m)
Circular Motion Dynamics
Universal Gravitation
What is the force of gravity acting on a 65kg man standing on the surface of planet
Earth? (rearth = 6.38 x 106 m, MEarth = 5.97 x 1024 kg)
Fg = G ME MM
MM
rE2
Fg
rE
Fg =
6.67 x 10-11Nm2/kg2 (5.97 x 1024kg)(65kg)
(6.38 x 106m)2
ME
Fg = 635.9 N = 636 N
Check:
Fg = mM g = (65kg) (9.81N/kg) = 637.6 N = 638 N
Circular Motion Dynamics
Universal Gravitation
Tides are created by the gravitational attraction of the sun and moon on Earth.
Calculate the net force pulling on Earth during a) a new moon and b) a full moon
and c) a first quarter moon. (Mmoon = 7.35 x 1022 kg, MEarth = 5.98 x 1024 kg,
MSun = 1.99 x 1030 kg, rME = 3.84 x 108 m, rSE = 1.50 x 1011 m)
Earth
New Moon
Full Moon
Sun
1st Quarter
FSE = G MS ME = 6.67 x 10-11Nm2/kg2 (1.99 x 1030kg)(5.98 x 1024 kg)
(1.50 x 1011m)2
rSE2
FME = G MM ME
rME2
= 6.67 x 10-11Nm2/kg2 (7.35 x 1022kg)(5.98 x 1024 kg)
(3.84 x 108m)2
= 3.53 x 1022 N
= 1.99 x 1020 N
Circular Motion Dynamics
Universal Gravitation
Earth
New Moon
Full Moon
Sun
1st Quarter
FSE = 3.53 x 1022 N
FME = 1.99 x 1020 N
a) During a new moon
FE = FSE + FME = 3.53 x 1022 N + 1.99 x 1020 N
= 3.55 x 1022 N
b) During a full moon
FE = FSE + FME = 3.53 x 1022 N - 1.99 x 1020 N
= 3.51 x 1022 N
c) During 1st Quarter
FE = FSE + FME
FSE

FE
FE = (FSE2 + FME2) = ((3.53 x 1022 N)2 + (1.99 x 1020 N)2)
FME
FE = 3.53 x 1022 N
 = tan-1 (FME / FSE) = tan-1 (1.99 x 1022 / 3.53 x 1022 ) = 0.320
Circular Motion Dynamics
Gravitational Field Strength
Field strength is defined as the amount of force per quantity.
For a gravitational field around an object it is the number of Newton’s of force
acting on every kg of a second object placed in this field.
g = Fg
g
r122
M2
r12
M1
= G M1
What is the gravitational field strength on the surface of the
moon? (Mm = 7.35 x 1022 kg, rm = 1.74 x 106 m)
gM = G MM
r M2
gM =
kg
6.67 x 10-11Nm2/kg2 7.35 x 1022
= 1.62 N/kg
(1.74
106surface
m)2
(about 1/6 that
onxthe
of Earth)
Circular Motion Dynamics
Gravitational Field Strength
A typical white dwarf star, which once was an average star like our sun but is now in
the last stages of its evolution, is the size of our moon but has the mass of our sun.
What is the surface gravity (g) of this star? (Ms = 1.99 x 1030 kg)
g
gWD = G MS
r M2
rM
MS
gWD =
6.67 x 10-11Nm2/kg2 1.99 x 1030 kg
= 4.38 x 107 N/kg
(1.74 x 106m)2
(This is about 4.5 million times our surface gravity on Earth)
Circular Motion Dynamics
Gravitational Potential
Gravitational Potential Energy is the energy an object has due to its position in a
gravitational field.
The change in GPE (GPE) between two points at different distances from an
object we have already expressed as - M2gdv according to the work-energy
theorem
now….g = G M1
GP
r122
The gravitational potential energy of mass M2 at distance r12
can be defined as….
GPE = - G M1 M2
r12
M1
r12
Like field strength, the amount of potential energy per kg of
a second object placed at this position in the gravitational field
is called gravitational potential
GP = - G M1
r12
The GPE and GP get smaller the further you go from the mass M1. They both are reduced to
zero at infinity. It should also be noted that the work done to move a mass between two
points in a gravitational field is independent of the path taken. (proved by calculus)
Circular Motion Dynamics
Escape Speed
How much speed is needed to send a rocket (M2) soaring into space so that it
escapes the pull of a planet (M1)?
ve
If the rocket just makes it to infinity and slows down all the
way its final velocity (vf) will be zero and its final GPE will
be zero so the total energy will be zero
KEi + PEi = KEf + PEf = 0
1/2 M2 ve2 - G M1 M2 = 0
rp
r
M1
ve =  2G M1
rp
Determine the escape speed from the surface of earth
rearth = 6.38 x 106 m
MEarth = 5.97 x 1024 kg
G = 6.67 x 10-11 Nm2/kg2
ve =  2G ME =  (2(6.67 x 10-11 Nm2/kg2) (5.97 x 1024 kg ))
rearth
6.38 x 106 m
= 11.2 x 103 m/s
(25000 mi/h)
Circular Motion Dynamics
Escape Speed - Black Hole
When a huge star many times our own sun’s mass runs out of nuclear fuel gravity
causes it to collapse in on itself eventually resulting in an incredibly dense piece of
matter.
Light has a speed (c) of 3 x 108 m/s. If the escape speed is
greater than this not even light will escape. Karl Schwarzchild
calculated the radius around a black hole at which light
wouldn’t be able to escape. This boundary is called the event
horizon (R)
c =  2G MBH
R
R
MBH
So ….R = 2G MBH
c2
If a black hole has a mass of 3 times our own sun’s mass and
a radius of 2km determine its event horizon (Ms = 1.99 x 1030
kg)
c = 3 x 108 m/s
R =
MBH = 5.97 x 1030 kg
G = 6.67 x 10-11 Nm2/kg2
2G MBH = 2(6.67 x 10-11 Nm2/kg2) (5.97 x 1030 kg )
c2
(3 x 108)2 m
= 8850m
(8.9 km)
Circular Motion Dynamics
Kepler’s Third Law
Kepler’s third law relates the force of gravity using Newton’s Universal Law of Gravitation with centripetal
force.
This force of gravity that the earth exerts on the moon is the centripetal force that the moon requires to
orbit the earth. That means that…
Fg = Fc
G ME m / r2 = m vt2 / r
or…..vt2 = GME / r
With this equation, we can find the velocity of the moon around the earth given the mass of the earth and
the radius of the moon’s revolution. This relationship is the same for any planet revolving around the sun.
Since the period of a planet, T, can be related to velocity by v = 2r / T, we’re all set. Now we just solve for
T.
(2r / T)2 = GM / r
So…..T2 = (42 r3) / (GM).
or…..T2  r3
This is Kepler’s third law. Using this equation, we can find the period of a planet if we know the radius
of its revolution and the mass of the planet it revolves around.
Find the time for the earth to revolve around the sun, given that the sun’s mass is 1.991 • 10 30 kg and
that the earth is 1.496 • 1011 m from the sun.
T2 = [4 • 2 • (1.496x1011 m )3] / [(6.673x10-11 N•m2/kg2) • (1.991x1030 kg)]
T = 3.155 x 107 s = 365.14 days.
(as expected!!)
= 9.953 x 1014 s2.
Circular Motion Dynamics
Satellites and Weightlessness
Satellites are freefalling towards the earth just as a dropped ball falls towards earth but
they have such high tangential velocities that as they fall they follow the curvature of
the planet.
The earth’s surface drops about 5m for every 8km horizontally. How fast should a
projectile be fired to go into a low orbit around the earth?
A projectile falls about 5m in 1s so it should be fired at 8km/s (18000 mi/h)
Motion of projectile
without gravity
If the tangential speed
is high enough the
projectile will fall
“around the earth”
Circular Motion Dynamics
Satellites and Weightlessness
How can someone in the space shuttle be experiencing weightlessness when the
spacecraft is experiencing nearly as much gravitational acceleration (g) as someone
on the surface?
An airborne athlete and an astronaut both experience weightlessness
Circular Motion Dynamics
Satellites and Weightlessness
Our experience of weight is really the normal force of the surface we are in contact
with pushing back on our body
You are made aware of this idea when traveling on a high speed elevator or riding on
an amusement park ride.
The normal force has a small magnitude at the top of the loop (where the rider
often feels weightless) and a large magnitude at the bottom of the loop (where
the rider often feels heavy).
Circular Motion Dynamics
Apparent Weightlessness
Consider a roller coaster track that has a series of hills and dips as shown below. The
black arrows show that the centripetal acceleration is directed towards the center of the
circular shaped arcs as the car moves along the track.
The forces acting on the car at positions A, B and D are shown below
Circular Motion Dynamics
Apparent Weightlessness
Consider a roller coaster moving over a series of hills as shown on the previous slide.
At the bottom of a hill the track has do do two things:
i) It has to support the car’s weight which means that it has to provide
a normal force (FN) upwards
ii) It also has to change the motion of the car because the path that the
car follows is curved (not straight!). Because the direction of motion
changes constantly the track has tp provide a centripetal force
towards the center of the turn.
Therefore…… FN = Fg + FC = mg + mv2 / r
= m (g + v2 / r )
Circular Motion Dynamics
Apparent Weightlessness
FN
At the top of the hill the track can push upwards or pull downwards
depending on the car’s speed.
i) If the car travels slowly over the top FN is directed upwards but is a
little bit smaller than Fg (down) because the car starts to lift just a little bit
off the track because it wants to continue in a straight line path as it
goes over the top
ii) If the car travels fast over the top FN is directed downwards because
the car actually lifts a lot off the track but a second set of wheels below
the track stop the car from moving upwards so the track applies a force
back down (3rd Law) on the car (Fapp)
Notice that both the top and bottom of the hill there is an unbalanced
force (FNET) towards the center of the turn. This is the centripetal force
(FC) that allows the car to follow the curved track
Also note that at some speed between fast and slow the passengers in
the car (and the car itself) will not receive a normal force and so will feel
weightless.
Circular Motion Dynamics
Torque
The ability of a force to rotate an object about some axis is
measured by quantity called torque, .
Torque like force is a vector quantity. Its magnitude is:
 = Frperp
where rperp is the lever arm and is the perpendicular distance
from the axis of rotation to a line drawn along the direction of
force.
If there are several forces acting on the object then the net
torque is obtained by summing the torques produced by
each of these forces., thus
net =   = 1 + 2 +…..= F1r1perp + F2r2perp +…..
F


r
rperp
A student pushes on a door furthest from the
hinges (1.5m) with a force of 15N at an angle of 350
to the plane of the door. Determine the torque on
the door.
 = Frperp = F r sin
 = 13 Nm
= (15N) (1.5m) sin350
Note: Positive torque is assigned
counterclockwise and negative torque
clockwise
Circular Motion Dynamics
Rotational Equilibrium
The first condition for equilibrium that we have discussed is namely, the sum of all the external forces is zero.
With no net forces acting on an object it obeys Newton’s first law, I.e. no accelerations and thus no changes in
motion. (F= 0) The object could still rotate and change its rate of rotation if the torques don’t add to zero.
The second condition for equilibrium, therefore, is that there are no changes in rotation. This happens when the
sum of the external torques adds to zero, so we have
 = 0
Position of the Axis of Rotation
If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque; the
location of the axis is completely arbitrary.
The Center of Gravity
The center of gravity is that point in or near an object where all the torques due to the weight of the object add to zero
no matter how the object is orientated. It is what we usually call the balance point. The x component is found from
the equation
xcg = m1x1 + m2x2 +… = i mi+xi
m1 + m2 +…
M
The y and z components of the center of gravity are found in a completely analogous manner.
The center of gravity of a symmetrical body that is homogenous must lie on the axes of symmetry.
Circular Motion Dynamics
Rotational Equilibrium
A hungry 700 N bear walks out on a beam in an attempt to retrieve some “goodies”
hanging at the end. The beam is uniform, weighs 200 N, and is 6.00 m long; the goodies
weigh 80.0 N. a) Draw a FBD for the beam. b) When the bear is at x=1.00 m, find the
tension in the wire and the components of the reaction force at the hinge. c) If the wire
can withstand a maximum tension of 900 N, what is the maximum distance the bear can
walk before the wire breaks?
a)
Ty = T sin600 = 0.866T
700 N
X
Tx = T cos600 = 0.5 T
H
3m
3m
V
80 N
200 N
b) If x = 1 m then
left end = (-700 N)(1m) - (200 N)(3 m) - (80 N)(6 m) + (0.866T)(6 m)
Equating this to zero gives: T = 342 N
From Fx = 0,
H = 0.5 T = 171 N
From Fy = 0,
V = 980 N - 0.866T = 683 N
c) If T = 900 N
left end = (-700 N)(X) - (200 N)(3 m) - (80 N)(6 m) + (779.4 N)(6 m)
Equating this to zero and solving for x gives:
x = 5.13 m