PHYS103   Fall 2005    Homework 7

Giancoli Sixth Edition Chapter 5

Questions 5, 10, 13, 20

Problems 1, 3, 7, 8, 9, 18, 41, 49

 

 

 

Answers to Questions

 

 

5.     For the water to remain in the bucket, there must be a centripetal force forcing the water to move in a circle along with the bucket.  That centripetal force gets larger with the tangential velocity of the water, since .  The centripetal force at the top of the motion comes from a combination of the downward force of gravity and the downward normal force of the bucket on the water.  If  the bucket is moving faster than some minimum speed, the water will stay in the bucket.  If the bucket is moving too slow, there is insufficient force to keep the water moving in the circular path, and it spills out.

 

 

10.   She should let go of the string when the ball is at a position where the tangent line to the circle at the ball’s location, when extended, passes through the target’s position.  That tangent line indicates the direction of the velocity at that instant, and if the centripetal force is removed, then the ball will follow that line horizontally.  See the top-view diagram.

 

 

13.  The gravitational pull is the same in each case, by Newton’s 3rd law.  The magnitude of that pull is given by .  To find the acceleration of each body, the gravitational pulling force is divided by the mass of the body.  Since the Moon has the smaller mass, it will have the larger acceleration.

 

 

20.   A satellite remains in orbit due to the combination of gravitational force on the satellite directed towards the center of the orbit and the tangential speed of the satellite.  First, the proper tangential speed had to be established by some other force than the gravitational force.  Then, if the satellite has the proper combination of speed and radius such that the force required for circular motion is equal to the force of gravity on the satellite, then the satellite will maintain circular motion.

 

 

Solutions to Problems

 

1.         (a)        Find the centripetal acceleration from Eq. 5-1.

            (b)        The net horizontal force is causing the centripetal motion, and so will be the centripetal force.  

                                   

 

3.     The centripetal acceleration is .  The force (from Newton’s 2nd law) is .  The period is one year, converted into seconds. 

            The Sun exerts this force on the Earth.  It is a gravitational force.

 

 

7.     See the free-body diagram in the textbook.  Since the object is moving in a circle with a constant speed, the net force on the object at any point must point to the center of the circle.

(a)   Take positive to be downward.  Write Newton’s 2nd law in the downward direction. 

This is a downward force, as expected.

(b)            Take positive to be upward.  Write Newton’s 2nd law in the upward direction. 

This is an upward force, as expected.

 

 

8.     The centripetal force that the tension provides is given by .  Solve that for the speed.

 

9.     A free-body diagram for the car at one instant of time is shown.  In the diagram, the car is coming out of the paper at the reader, and the center of the circular path is to the right of the car, in the plane of the paper.  If the car has its maximum speed, it would be on the verge of slipping, and the force of static friction would be at its maximum value.  The vertical forces (gravity and normal force) are of the same magnitude, because the car is not accelerating vertically.  We assume that the force of friction is the force causing the circular motion.

Notice that the result is independent of the car’s mass.

 

 

18.  Consider the free-body diagram for a person in the “Rotor-ride”.   is the

normal force of contact between the rider and the wall, and  is the static frictional force between the back of the rider and the wall.  Write Newton’s 2nd law for the vertical forces, noting that there is no vertical acceleration. 

If we assume that the static friction force is a maximum, then

.

            But the normal force must be the force causing the centripetal motion – it is the

only force pointing to the center of rotation.  Thus .  Using , we have .  Equate the two expressions for the normal force and solve for the coefficient of friction.  Note that since there are 0.5 rev per sec, the period is 2.0 sec.

.

Any larger value of the coefficient of friction would mean that the normal force could be smaller to achieve the same frictional force, and so the period could be longer or the cylinder smaller.

 

There is no force pushing outward on the riders.  Rather, the wall pushes against the riders, so by Newton’s 3rd law the riders push against the wall.  This gives the sensation of being pressed into the wall.

 

 

41.        The expression for the acceleration due to gravity at the surface of a body is , where

 is the radius of the body.  For Mars, .  Thus

 

49.   The speed of an object in an orbit of radius r around a planet is given by , and is also given by , where T is the period of the object in orbit.  Equate the two expressions for the speed and solve for T.

      

For this problem, the inner orbit is at , and the outer orbit is at .  Use these values to calculate the periods.

Saturn’s rotation period (day) is 10 hr 39 min which is about .  Thus the inner ring will appear to move across the sky “faster” than the Sun (about twice per Saturn day), while the outer ring will appear to move across the sky “slower” than the Sun (about once every two Saturn days).