Circular Motion / Universal Gravitation concept WS (honors)
1)
A car travels around a curve with constant speed.
a. Does the velocity of the car change in this process? Explain.
Yes; the direction the car is traveling changes at each instant so the velocity is changing.
b. Is the car accelerated? If so what is the direction of that acceleration?
Yes, since its velocity is changing. The direction of the acceleration is TOWARD THE CENTER OF
THE CURVE.
2)
Two cars travel with constant speed around the same curve, car A at twice the speed of car B. Which car
experiences the larger acceleration? How much larger?
Since vA = 2 vB and ac  v², then car A experiences FOUR TIMES THE ACCELERATION of car B.
3)
A car travels at the same constant speed around two curves, curve 1 having twice the radius of curve 2. For
which of these curves is the car's acceleration larger? How much larger?
Since r1 = 2 r2 and ac  1/r, then the acceleration through curve 2 is TWICE THAT of curve 1
4)
A car travels around a flat curve with constant speed. What is the direction of the net force acting upon the
car?
The net force is acting TOWARD THE CENTER OF THE CIRCLE (and is called the centripetal force).
5)
Back to item 4: Is there a maximum speed at which the car will be able to negotiate the curve? If so, what
factors determine this maximum speed? Explain.
The only force able to apply centripetal force to the car in this situation is the FRICTIONAL FORCE
between the tires and the road. Therefore, the maximum speed of the car will be that in which the
amount of centripetal force needed (mv²/r) is equal to the amount of friction available (µF N).
Fc = F f
mv²/r = µ FN = µ (mg)
v=
gr
The factors that affect this maximum speed are the coefficient of friction and the radius of the curve
(assuming ‘g’ is constant).
6)
If the curve is banked, is it possible for a car to negotiate the curve even when there is no frictional force at all
(i.e. icy road surface)? Explain.
Yes, it is. The normal force of the road on the car will apply the needed centripetal force (actually, not
the entire FN, just the amount pointing in the horizontal direction – toward the center of the curve). As
long as the speed is low enough so Fc = FN (horizontal), the car will negotiate the curve successfully
without friction.
7)
You are riding a vertical circle ride at an amusement park (upside-down at top; right side-up at bottom). At
which point along the ride is the force exerted by the seat on your butt the least? Explain.
At the TOP of the circle. At that point the seat is only pushing on you to help the gravitational force
pull you in a circular path (Fc = Fg + FN). At the side, the normal force is supplying all of the needed
centripetal force (Fc = FN); at the bottom the normal force has to supply all of the needed centripetal
force PLUS cancel out the gravitational force, which is trying to pull you away from the center of the
circle. ( Fc = FN – Fg )
8)
Assuming no safety devices the ride in #7, is there a minimum speed which must be maintained by the ride in
order that passengers not leave the correct path of motion? Explain.
Yes; you have to travel fast enough so that the amount of gravitational force that is acting on you (mg)
is exactly the amount that is needed to make you go in that circle (mv²/r); therefore, F g = Fc . That
minimum speed would be v =
gr . If you go any slower, Fg will pull you into a smaller-radius
circle, and you will leave the track.
9)
(Referring to the ride in #7) is there a maximum speed which must not be surpassed in order that the
passengers not leave the correct path of motion? Explain.
No – assuming the track is of infinite strength. No matter how fast you go, the track will always be
able to provide a normal force to help the gravitational force pull you into a circle.
10) You can swing a bucket of water around in a vertical circle and the water will not come out of the bucket.
What keeps the water in the bucket? Explain.
Since the water is traveling tangent to the circle at this point, there is no reason for it to come straight
down out of the bucket; it is actually trying to travel horizontally. The real question is: Why doesn’t it
continue its straight-line path at that point? ANSWER: The bottom of the bucket pushes the water
down, apply the Fc needed to make it go in a circle.
11) If the bottom of the water bucket fell out when the bucket was located halfway along the downward side of the
circular path, what would be the resulting motion of the water?
At this point the water is moving straight downward (tangent to the circle). Therefore, it will go
straight down to the ground.
12) Two objects are attracting each other gravitationally. If the mass of one object doubles and the mass of the
other object triples, what happens to the gravitational force between them?
Since Fg  m1 m2, then the gravitational force will INCREASE BY A FACTOR OF 6.
13) Two objects, which are 1000 km apart, are attracting each other gravitationally. If they are moved apart an
additional 2000 km, what happens to the gravitational force between them?
r1 = 1000 km and r2 = 3000 km, so r2 = 3 r1. Since Fg  1/r², then the gravitational force will DECREASE
BY A FACTOR OF 9.
14) If you were to travel to the moon, you would cross a certain spot at which the net gravitational force on you
(from the earth and the moon anyway) would be zero. Where would this spot be located; closer to earth,
closer to the moon, or equal distance from both? Explain.
Since the earth is much more massive than the moon (  81 times), the gravitational force exerted on
you by the earth is greater than that applied to you by the moon when you are equidistant from both.
Moving closer to the moon – and, therefore, further from the earth – will cause those two gravitational
forces to become closer to the same value. Thus, the spot at which F g (from earth) = Fg (from moon) would be
much closer to the moon.
How much closer? Figure it out – you can do it!