Mr. Curtin's Physics

Advanced Placement Test Chapter 11

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This is a test from a previous year covering the Chapter 11 material.  Although all tests are different, working through these problems should help you prepare for this year's test.  The answers are available at the end of the test.

Short Answer.  Answer briefly in the space provided.

Three uniform rigid bodies --- a solid sphere, a solid cylinder, and a hollow cylinder --- are placed at the top of the incline. If they are released from rest at the same time at the same elevation and roll without slipping, which reaches the bottom first? Which reaches the bottom last? Which will have more total kinetic energy at the bottom?

Problems. Show your work clearly and explain your steps. Try to spend no more than 15 minutes on each problem.

A large sphere rolls without slipping across a horizontal surface.  The sphere has a constant translational velocity of 10 m/s, a mass of 25 kg, and a radius of 0.2 meters.   The moment of inertia of a sphere is 2/5 MR^2.  The sphere approaches a 25 degree incline of height 3 meters as shown below and rolls up without slipping.

a.  Calculate the translational kinetic energy of the sphere as it rolls along the surface.

b.  Calculate the rotational kinetic energy of the sphere as it rolls along the surface.

c.  Calculate the magnitude of the sphere's velocity as it leaves the top of the incline.

d.  Specifiy the direction of the sphere's angular velocity as it leaves the surface.

e.  Neglecting air resistance, calculate the horizontal distance from the top of the incline to the point at which the sphere lands back on the surface.

f.  If the sphere was to slide without friction (instead of rolling), would your answer to part e be more or less?  You need not justify this answer with a calculation.

Problem 2

A billiard ball has a mass M, radius R, and moment of inertia about its center of mass of 2/5 MR2. The ball is struck by a cue stick along a horizontal line through the ball's center of mass so that the ball initially slides with a velocity vo as shown above. As the ball moves across the rough table with a coefficient of friction of m, its motion gradually changes from pure slipping or linear motion to pure rolling without slipping.

1. Assuming constant acceleration, develop an expression for the linear velocity v of the ball while it is slipping and starting to roll.
2. Assuming constant angular acceleration, develop an expression for the angular velocity w of the ball while it is slipping and starting to roll. (Remember v doesn't equal Rw until it is purely rolling.)
3. Determine the time at which the ball begins to roll without slipping.
4. When the ball is struck it acquires an angular momentum about the fixed point P on the surface of the table. During the subsequent motion, its angular momentum about that point doesn't change despite the constant unbalanced frictional force. Explain why this is so.

A 1 kg ball is moving horizontally with a velocity of 10 m/s, as shown below, when it makes a glancing collision with lower end of a bar that was hanging vertically at rest before the collision. For the system consisting of the ball and the bar, linear momentum is not conserved in this collision (because there is an unbalanced force at the pivot), but kinetic energy is conserved. The bar, which has a length, L, of 1.2 meters and a mass, M, of 3 kg, is pivoted about the upper end. Immediately after the collision the ball moves with a speed, v, at an angle, q, with its original direction. The bar swings freely, and after the collision reaches a maximum angle of 90 degrees with respect to the vertical. The moment of inertia of the rod is 1/3 ML2. Ignore all friction.

1. Determine the angular velocity of the bar immediately after the collision.
2. Determine the speed, v, of the 1 kg ball immediately after the collision.
3. Determine the magnitude of the angular momentum of the object about the pivot just before the collision.
4. Determine the angle q.

Answers:

Short Answer.

Ball reaches first, Hollow cylinder reaches last. All three have the same total kinetic energy.

Problem 1

1. KEtrans = ½ m v2 = ½ (25 kg) (10 m/s)2= 1250 J

2. KErot = ½ I w2 = ½ (2/5 MR2)(v/R)2 = ½ (2/5)(25 kg)(10 m/s)2
3. = 500 J

4. Conservation of Energy

KE total init = GPE final + KE trans final + KE rot final

1250 J + 500 J = mgh + ½ mv2 + ½ Iw2 = mgh + ½ mv2 + ½ (2/5 mR2)(v/R)2

1750 J = mgh + (7/10) mv2

= (25 kg)(9.8 m/s2)(3 m) + (7/10)(25 kg)v2

v = 7.61 m/s

d.  Angular velocity is into the page.

e.

f.      The distance would be LESS.  The rolling ball has more total energy.

Problem 2

(2/5)R  a   =   mg

a   =  (5/2) (mg/R)            (Positive since angular velocity is increasing)

w  =  wo   +  a t   =   0  +  (5/2) (mg/R)t  =    (5/2) (mg/R)t

Problem 3