A 12.00 meter ramp is inclined at 30 degrees and a ball is held at the top of the ramp and then released. The ball rolls without slipping. The mass of the ball is 0.500 kg and its radius is 0.050 meters. The ball reaches the bottom of the ramp with a speed of 5.00 m/s.

A.) Calculate the rotational inertia of the ball

B.) Calculate the rotational kinetic energy of the ball.

## Answers (3)

This question can be done with Forces, or with Energies. i perfer to do it with forces simply because its more interesting, but it is very long, so ill do out both.

I'm not gonna do the question for you (thats cheating, and no fun), ill just give you some guidelines:

A)Energy:

Ek bottom = Ep top

1/2 I w^2 = mgh

w is omega

Mass, linear velocity, are given. w is simply velocity/radius(given), for h (height), use simple trig. h = 12sin30

simple.

-----------------------OR----------------------

A) Forces:

much more interesting, and makes you use your brain, but longer :(

first find the accelleration (linear)

initial velocity = 0

final vel. = 5

displacement = 12

obviously use v^2 = u^2 + 2aS, to find a.

then find the equivalent force. F=ma. draw a force diagram, showing this gravational force (NB. remember, grvity is a force directly downwards - Newton and his apple)and the force of friction (as there is no slipping!!). then resolve the grav. force into perpendicular and parallel compontents. the i. direction (or x-axis if you wish) is Fsin30, and the j. direction (or y-axis) is Fcos30.

now get the resultant force. ie. Gravity - Friction. Friction is uR, where u is the coefficent of friction (we shall leave it in terms of u for now if it's not given in the question.

Next, you need the rotational accelleration. L=Iw. L is sum of moments (rotational forces) I is moment of inertia (rotational inertia) and w is angular Accelleration. The moment in question is due to the force of friction, Friction=uR, and moment = Force by Distance so, L = uR by Radius(given)

now, we know L = I by uR by Radius by w. just rearrange to get w = ......

now, we have 2 eqasions in u, R, a, I and w. thats 5 variables. woah!

but!, I is what we need to find, u is in both equasions and we can cancel off, R is calculable, and a and w are related through a simple formula. not so difficult after all.

lets start with R... this is the reaction force in the j. direction (perpendicular to the inclined plane) but there is no movement in this plane, so forces are balanced, so R is equal to the j. component of the gravitational force (already calculated)

leave I, and u alone for the time being.

as for a, and w:

a = wr

convert w to a, and we have already calculated a !!

now you should have 2 equations in u, and I

rearrange them and use simultaneus equations to solve for I (it may have to be in terms of u if the u's dont cancel off. just assume this is what they are looking for as they dont explicitly give the coeffecient of friction)

Now part B - easy

B) once you have I, the rotational kenetic energy is simple to get:

just use:

Ek = (1/2) (I) (w^2). NB w here, represents angular VELOCITY, not acc/n

a very similar formula to the 1/2 m v^2 for linear kenetic energy!

hope this helps!!

A.) I = 2mr^2/5 (for solid sphere)

I = 2*(0.500 kg)(0.050 m)^2/5

I = 5.00e-4 kg-m^2

B.) K(rotational) = Iω^2/2

The speed of the ball at the bottom of the ramp would be the same as the speed of a point on the tip of the ball with respect to its center, so:

ω = v/r

K(rotational) = I(v/r)^2/2

K(rotational) = (5.00e-4 kg-m^2)(5.00 m/s / 0.050 m)^2 / 2

K(rotational) = 2.5 J

A.) Calculate the rotational inertia of the ball

Use the following to solve:

1/2v^2 x ( mass + (inertia/radius^2)) = mgh

B.) Calculate the rotational kinetic energy of the ball.

Use the following to solve:

rotational KE = 1/2 x inertia x omega(rad/s)^2

omega = angular speed in rad/s = 5m/s / radius = 100 rad/s