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Chapter 5: Circular Motion;
Gravitation
Curves
• Forces act in straight lines
• An object will move in a straight line if there is
NO force acting on it or if the net force acts in
the direction of the motion.
• BUT, what if the net force acts at an angle to
the motion?
Come Full Circle
• From the last example, what would happen if
the magnitude of the force and its angle to
the motion remained constant?
Uniform Circular Motion
• An object that moves in a circle at constant
speed is said to experience uniform circular
motion.
Kinematics of UCM
• aR = v 2 / r
• Remember:
frequency (f) is the number of times the
object completes a circle per second.
period (T) is the amount of time it takes to
complete a circle, or T = 1/f
• Also remember v = d/t. In a circle, distance
means circumference(2πr) and time means
period. So v = 2r
T
Example
• A 150g ball at the end of a string is revolving
uniformly in a horizontal circle of radius
0.600m. The ball makes 2.00 revolutions in a
second. What is its centripetal acceleration?
Answer
• v = (2πr)/T = (2*3.14*0.600m)/(0.500s) =
7.54m/s
• aR = v2/r = (7.54m/s)2 / (0.600m) = 94.8m/s2
Dynamics of UCM
• From chapter 4 we learned that acceleration
requires a force.
• So what causes the centripetal acceleration
that causes UCM?
• That would be the centripetal force, ΣFR
• We start with ΣF = ma and through the
wonders of substitution we get ΣFR = maR or
• ΣFR = mv2 / r
Beware
• Centripetal force is NOT a new kind of force.
• The term “centripetal” simply tells you that
the force is “center seeking”.
• Example: Picture swinging a child around by
the arms. In order to keep the child moving in
a circle you have to pull in with your arms. You
pulling with your arms is no different from you
pulling a box.
Example
• Estimate the force a person must exert on a
string attached to a 0.150kg ball to make the
ball revolve in a horizontal circle of radius
0.600m. Also note the ball makes 2.00
revolutions per second.
Solution
• ΣFR = mv2 / r = m(2πr)2 / r =
(0.150kg)(7.54m/s)2 / (0.600m) ≈ 14N
Example
• A 0.150kg ball on the end of a 1.10m long
cord(of negligible mass) is swung in a vertical
circle.
• A.) determine the minimum speed the ball
must have at the top of its arc so that it
continues moving in a circle.
• B.) Calculate the tension in the cord at the
bottom of the arc assuming the ball is moving
at twice the speed from part A.
You Going Into a Curve
• When you are in a car that enters a curve you
get pulled toward the door.
• This is because of Newton’s First Law; you
want to keep going straight.
• So what makes you move toward the center of
the curve?
The Car Going Into the Curve
• The car also wants to go straight when it
enters a curve.
• What makes the car move in a curve?
• What happens if this force is not enough?
Example
• A 1000kg car rounds a curve on a flat road of
radius 50m at a speed of 14m/s. Will the car
make the turn or will it skid off if…
a. the pavement is dry and the coefficient of
static friction is μs = 0.6
b. the pavement is icy and μs = 0.25
Solution
• The only force acting in the horizontal
direction is friction.
• We need to see if this force provides enough
centripetal force to maintain the curve.
• FR = maR = mv2/r = (1000kg * (14m/s)2)/50m =
3900N this is how much force is needed.
• Ffr = μsFN = μsmg =
(0.60)(1000kg)(9.8m/s2)=5900N this is the
max friction by the car, so we are fine.
Solution
• On the icy road the Ffr becomes,
Ffr = μsFN = μsmg = (0.25)(1000kg)(9.8m/s2) =
2500N
• This is too low. The car would skid off the
road.
Slamming on the Brakes
• The situation gets even worse if your wheels
lock up. (granted this doesn’t happen much
anymore thanks to the wonders of anti-lock
brakes)
• When the wheels are rolling there is a spot on
the tire that is stationary to the ground, so the
friction is static.
• When the tire locks and slides as a whole unit,
the friction is kinetic, which is lower.
Handling the Curves
• There are a few ways we can make curves
safer.
– Reduce v (slow down when entering a curve)
– Increase r (make the curve more gradual)
– Increase friction (new asphalt technology)
– Make the curve an incline (called a banked curve)
Banked Curves
• By banking the curve we create a normal force
that is at an angle, which creates 2
components.
• The x component of FN (FN sinθ) adds to the FR.
• We can set the horizontal normal force, Fnsinθ
equal FR, mv2/r to find the correct angle.
Example
• What banking angle is need for an express
way off-ramp curve of radius 50m at a design
speed of 50km/hr (or 14m/s)
Solution
• FN sinθ = mv2/r
• We need to find FN
• We know the vertical forces are balanced
(because the car is not floating)
So FN cosθ = mg
or FN = mg/cosθ
• We can plug this into FN sinθ = mv2/r
Solution
•
mg
cos 
•
•
•
•
mg tanθ = mv2/r
tanθ = v2 /rg
tanθ = (14m/s)2 / (50m)(9.8m/s2) = 0.40
Using tan-1 we get θ = 22o
sinθ = mv2/r
Falling Bodies
• At this point Newton had already constructed
his 3 laws when he was pondering falling
objects.
• He already concluded that falling objects must
have a force exerted on them, but what was
doing the exerting?
High Minded
• The story goes that Newton watched an apple
fall to the ground and was struck by an
inspiration: if gravity can reach tree tops and
even mountain tops can it reach the moon?
• It was this simple question (along with a lot of
math) that set in motion a chain of thinking
that connects Isaac Newton to Neil Armstrong
to Curiosity and beyond.
Haters gonna hate
• Newton ran into a great deal of resistance to
his theory of gravity.
• What do you think was/were the issue(s)?
Pro-active
• Newton did more than just bring up good
questions; he set out to find the answers.
• We already know that the acceleration due to
gravity on the surface of the Earth is 9.8m/s2.
• Let’s do an example that will cast some light
on the nature of gravity.
That’s no space station
• The Moon is 3.84x108 m from the Earth and
travels in a near perfect circle. The period of
the moon is 27.3days = 2.36x106 s. Find the
centripetal acceleration, aR, of the Moon.
Analysis
• The distance from the center of the Earth to
the Moon is 60 times the distance from the
center of the Earth to its surface.
• 602 = 3600
• Divide 9.8m/s2 by 3600 and compare it to your
answer from the last problem.
• What do you conclude?
Mass Effect
• We already stated in chapter 4 that objects
with more mass need more force to reach the
same acceleration as a less massive object.
• We also know from Newton’s 3rd that if the
Earth is pulling on the Moon, then Moon is
pulling on the Earth.
• Therefore, Newton concluded that the force of
gravity is proportional to both the masses.
Put it all together
• So far we got the force is related to 1/r2 and
also related to m1 x m2
• Together they make NEWTON’S LAW OF
UNIVERSAL GRAVITATION:
m1m2
F G 2
r
Near Earth Gravity
m1m2
• When we use F  G r 2 on the surface of
the earth, m1 becomes the mass of the Earth
(5.98E24kg), m2 becomes the mass of the
object on the surface, and r becomes the
radius of the Earth (6.38E6m).
Weighty Subject Matter
• As we have already stated in chapter 4, we call
the force of gravity the weight of the object,
mg.
mmE
m1m2
• So we can rewrite F  G 2 as mg  G 2
r
• The mass cancels out and we get
rE
mE
g G 2
rE
Example
• Estimate g on the top of Mt. Everest, 8848m
above the Earth’s surface.
Solution
• We need to add the distance from the surface
to the summit to the radius of the Earth.
• 6.38E6m + 8.8E3m = 6.389E6m
mE
• We plug this value in as “r” in g  G 2
r




11 Nm 2
24
 6.67 x10
2  5.98 x10 kg
kg 

2
g

9
.
77
m
/
s
2
6
6.389 x10 m


Geosynchronous Orbit
• An object in geosynchronous orbit is an object
that hovers over the same spot above the
Earth. It revolves at the same rate that the
Earth rotates.
• Why is this important?
• How is this possible?
Ready for Launch
• What is the necessary height above the Earth
a satellite must reach to obtain
geosynchronous orbit?
• What is that satellite’s speed? (That is how
fast does the Earth rotate?)
Solution
• Start at FR = maR,
• Substituting (GmsatmE)/r2 in for FR
and v2/r for aR gives us:
• (GmsatmE)/r2 = msatv2/r
• Substituting (2πr)/T for v gives us
• (GmE)/r2 = (2πr)2/rT2 (note that T = 1day =
86,400s)
• We need to solve this for r.
Solution Cont.
• Algebra gives us



2
2
2
24
Gm
T
6
.
67
x
10
Nm
/
kg
5
.
98
x
10
kg 86,400s 
3
E
r 

2
4
4 2
• Which equals 7.54 x 1022 m3
• Taking the cube root gives us r = 4.23 x 107m.
• Finally we are ready to find v
2
As The World Turns
• We can solve (GmmE)/r2 = mv2/r for v and get,
GmE
v

r
6.67 E  11Nm

/ kg 2 5.98 E 24kg 
 3070m / s
4.23E 7 m
2
Meet Johannes Kepler
• Over 50 years before Newton published his 3
laws of motion and the law of gravitation,
German astronomer Johannes Kepler(15711630) published many works including his 3
laws of planetary motion. (Kepler btw based
his work off the earlier work of Tycho
Brahe(1546-1601)).
Kepler’s 3 Laws
• 1: The path of each planet about the Sun is an
ellipse with the Sun at one focus.
• 2: Each planet moves so that an imaginary line
drawn from the Sun to the planet sweeps out
equal areas in equal periods of time.
• 3: The ratio of the squares of the periods of
any two planets revolving about the Sun is
equal to the ratio of the cubes of their mean
distances from the Sun.
Our Focus is on 3
• The last law can be represented
mathematically as follows:
 T1

 T2
2

 r1 
   

 r2 
3
• We can rearrange them to get
 r13
 2
 T1
  r23
   2
  T2



Newton over Kepler
• Where Newton really shined was in his ability
to show that all 3 of Kepler’s Laws can be
derived mathematically using the Law of
Universal Gravitation.
Imperfections Reveal the True Beauty
• Since Kepler’s time we have gotten somewhat
better at measuring the orbits the planets and
we have seen that they do not exactly follow
Kepler’s perfect ellipses.
• Newton fixes this as well. He stated that
everything pulls on everything else. Other
planets tug on a planet and cause
disturbances in its orbit.
Something is Wrong with Uranus
• These disturbances actually lead to the
discovery of 2 planets!
• Astronomers could not account for the
weirdness in the orbit of Uranus.
• The only thing that could fix the math would
be if there was more mass beyond Uranus.
• This lead us to look for and find Neptune and
Pluto.