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South Pasadena
A.P. Physics (Chapter 5)
Name ________________________________
Date ___/___/___ Period 1 2 3 4 5
CENTRIPETAL means “center seeking so centripetal acceleration is directed toward the center of a circle
in which an object is moving in uniform circular motion.
V
V0
Let’s see WHY it is directed towards the center.
An object is moving counterclockwise (CCW) in a circular path: It starts at
3:00 O’clock with initial velocity, v0. Its final velocity at 1:00 O’clock is V
The direction of the Acceleration can be found by finding the direction of the
change in velocity V (Remember that acceleration = V/t or (V –V0)/t
If we focus on just the v, we have (V –V0) This is the same as V + (–V0)

To the right, use the graphical tip to tail method to add these two vectors,
and then show the resultant:
(Start with vector V and then add vector V0)
Which direction does the resultant point? ____________________________
SO, THE DIRECTION OF CENTRIPETAL ACCELERATION IS ALWAYS
DIRECTED TO THE CENTER OF THE CIRCLE.

A 0.0135-kg stopper tied to a string is swung and makes 10 revolutions in 7.4 seconds.
a) Calculate the frequency of the motion. (Show your calculation)
b) If the stopper is swung with a radius of 0.85 m, what is its linear speed?
(Show your calculation)
c) What is the centripetal acceleration of the stopper’s motion? (Show work below)
d) What is the centripetal force acting on the stopper?

A ladybug sits halfway between the axis and the edge of a rotating turntable. What will
happen to the ladybug’s linear speed if:
a) the RPM rate is doubled?
_______________________
b) the ladybug moves to the edge of the turntable?
_______________________
c) both a and b occur?
_________________
d) What happens to her angular speed when she moves to the edge? ________________

If frequency is the inverse of Period, T and Period is the inverse of frequency, f, then
what would be the equations for linear speed and centripetal acceleration?
Linear Speed (for circular motion) = circumference of circle/Period = 2πr/T
Centripetal Acceleration, ac, (m/s2)= v2/r (radius)
Linear Speed (in terms of f) =
________________________
Centripetal Acceleration (in terms of π, r & f) =
________________________

A car of mass 1,000 kg travels at a constant speed around a flat curve
that has a radius of curvature of 100 m. The car is going as fast as it
can go without skidding. The car is on the left side of the circle and
the center of the circle is to the right. Draw a Free Body Diagram:

If the speed of the car were 20 m/s, what is the value of the coefficient of friction, ?

David swings his rock around his head to build momentum, when he faced Goliath. The
radius of the circle was 1.5 meters. If the Fc (centripetal force) acting on the rock was
22.8 N and the tangential velocity of the rock was 6.0 m/s, what was the mass of the
rock?

Sketch a diagram of the moving rock and indicate the direction of both
the centripetal acceleration and the centripetal force!

A 20 kg rover sits on a newly discovered Planet Z, which has twice the mass of Earth and
twice the diameter of Earth. What would be the gravitational force acting on the rover on
Planet Z? (Don’t use a calculator!)
___________

The Earth has a mass of 6.0 x 1024 kg. The Sun has a mass of 2.0 x 1030 kg. The Earth
orbits the sun in a circle of radius 1.5 x 1011 m. Without using your calculator, which of
the following choices is closest to the force of the Sun on the Earth?
a) 1012 N
b) 1022 N
c) 1032 N
d) 1042 N

Which graph below best represents the relationship between the magnitude of the
centripetal acceleration and the speed of an object moving in a circle of constant radius?
ac
ac
Speed
(a)
ac
Speed
(b)
ac
Speed
(c)
Speed
(d)

Remember that centripetal force can be a gravitational force,(e.g. 2 planets
exerting a gravitational force on each other), a tension force, (e.g. a string swinging a
rock in a circle), an applied force ,(e.g. a cup resting on a board moving in a vertical
circle) or even a frictional force ,(e.g. a box resting on a merry go round or a car going
around a curve).

A cup of water is being swung in a horizontal circle of radius 0.75 m at a constant speed.
a) What is the minimum speed of the cup of water to negotiate the top of the loop without
water spilling out of the cup? b) What is the frequency of this minimum speed?
HINTS: First draw a FBD of the water at the top. 2nd , if the cup rotates with a speed greater
than the minimum required to hold the water in the circle, then the bottom of the bucket exerts a
downward force. Below the minimum speed the water tends to fall out of the bucket. At the point
of minimum speed, the bottom of the cup exerts no force on the water when the cup + water are at
the top of the loop. The only force acting on the water at the top is the force of gravity (mg).
Therefore, net F = mg and net F = ma but a = ac = v2/r
So, mg = mv2/r mass cancels out calculate v
0.75 m
bucket or cup of water
Q. Why doesn’t the water fall out of the bucket when the bucket is at the top of the circle?
(For the same reason that passengers don’t fall out of a roller coaster when it is upside down.)
ANSWERS:
bullet 1:
v
− v0
a
bullet 2:
bullet 3:
a) 1.35 rev/sec or 1.35 Hz b) 7.2 m/s c) ac = 61 m/s2 d) Fc = 0.83 N
a) Doubles
b) Doubles
c) Quadruples
ac = 4 π2 r f2
d) Unchanged
bullet 4:
v = 2π r f
bullet 5:
see FBD to right
bullet 6:
 = 0.4
bullet 7:
mass of 950 grams (0.95 kilograms
bullet 8:
Both ac and Fc are directed towards the center of the circle.
bullet 9:
100 N
bullet 10:
b) 1022 N
bullet 11:
(b)
bullet 12:
FBD 1 force down mg a) v = 2.7 m/s b) v = 2π r / T so T = 1.7s
or f = 1/T = 0.57 rev/s or 34 rpm
FN
Ff
mg
actual answer =3.3 x 1022 N
ac is proportional to v2/r