# Finding the change in Rotational Kinetic Energy

## Homework Statement

A sphere with a 3 kg * m^2 moment of inertia about an axis through its center has its angular velocity changed by 4 rad/s owing to a 2 N*m torque applied about that axis for 6s. If its initial angular velocity is 10 rad/s, the change in its rotational kinetic energy in J is...
The answer according to the book is 144 J.

## Homework Equations

1/2Iiwi^2 - 1/2Ifwf^2 = change in Ke
T=I(a)

## The Attempt at a Solution

The wording of the question is kind of throwing me off because i don't understand the part about the Torque.

Thanks

dynamicsolo
Homework Helper
It will be helpful to you to know that there are analogies in rotational mechanics to the definitions and equations you learned in linear mechanics.

Back in linear mechanics, you have the definition of work:

W = F · delta_x

and the work-kinetic energy theorem:

W = KE_f - KE_i .

In rotational mechanics, the rotational work will now be

W_rot = (torque) · (delta_theta) ,

with delta_theta being the angle the object has turned through while the torque is applied.

The work-kinetic energy theorem still applies,
now with

KE_rot = (1/2) · I · (w^2) .

Ok so the initial KE_rot = 1/2 * I * (w^2)
= 1/2 * 3 * (10^2) = 150
I see what you are saying that W_rot = torque * delta theta
but how would that apply I am not given an angle?
and am I finding the W_rot and subtracting it from 150?

tiny-tim
Homework Helper
A sphere with a 3 kg * m^2 moment of inertia about an axis through its center has its angular velocity changed by 4 rad/s … If its initial angular velocity is 10 rad/s, the change in its rotational kinetic energy in J is...
The answer according to the book is 144 J.

… I am not given an angle?
and am I finding the W_rot and subtracting it from 150?

Hi crazyog!

You don't need an angle or a torque or anything else for this part of the question …

You know it goes from 10 rad/s to 14 rad/s … so what is the increase in rotational KE?

Oh ok so when they say "changed by 4 rad/s" it means it is now 14 m/s
so that would be

1/2(3)(14^2) = 294 J
294 - 150 = 144 J

Thank you!

dynamicsolo
Homework Helper
You don't need an angle or a torque or anything else for this part of the question …

Oh, yes, quite so! Since they gave that information, I thought we'd need it somewhere. (Presumably, there are other parts to this problem...)

You know it goes from 10 rad/s to 14 rad/s … so what is the increase in rotational KE?

I wonder how they expected the student to know that the angular speed was being increased. The problem states that the "angular velocity (is) changed by 4 rad/s". You get a different magnitude for the rotational KE change if the angular speed goes to 6 rad/sec...

tiny-tim
Homework Helper
I wonder how they expected the student to know that the angular speed was being increased. The problem states that the "angular velocity (is) changed by 4 rad/s". You get a different magnitude for the rotational KE change if the angular speed goes to 6 rad/sec...

yes … you're right! …

Now I'm wondering why I assumed it was increasing!

dynamicsolo
Homework Helper
And the change in angular momentum (rotational impulse) is

delta_L = (torque)·(delta_t) .

So the change in angular momentum is

delta_L = +/- (2 N·m)·(6 sec) = +/-12 kg·(m^2)/sec ,

so the new angular momentum would be either

L' = (3 kg·m^2)·(10 rad/sec) + 12 kg·(m^2)/sec
= 42 kg·(m^2)/sec

or

L' = (3 kg·m^2)·(10 rad/sec) - 12 kg·(m^2)/sec
= 18 kg·(m^2)/sec ,

which are consistent with final angular speeds of 14 or 6 rad/sec .

So the other information in the problem doesn't provide a clue either...

So a second valid answer would be -96 J for the change in rotational KE. (I think the solver for the textbook must have assumed the speed was increased.)

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