# Power of rotational motion

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1. May 8, 2016

### RoboNerd

1. The problem statement, all variables and given/known data
An object of moment of inertia I is initially at rest. a net torque T accelerates the object to angular velocity omega in time t.

The power with which the object is accelerated is?

The right answer apparently is [ I * omega^2 ] / [2 * t].

Could anyone please explain why this approach is right?

Here's what I did:

2. Relevant equations
torque = I * angular acceleration
power = torque * omega
3. The attempt at a solution
power = torque * omega = I * angular acceleration * omega = I * (final omega/delta t) * final omega [I was finding the maximum power here]

However, the solutions have everything I got, with an additional 2 in the denominator.
Is the 2 supposed to be used for calculating the average omega in the rotational motion?

If so, the problem just says "the power," not "average power," so is it OK to make such an assumption?

2. May 8, 2016

Check that.

3. May 8, 2016

4. May 8, 2016

### RoboNerd

I accounted for the angular acceleration in my work, though.

5. May 8, 2016

### CWatters

Try again :-)

6. May 8, 2016

### CWatters

Sorry we cross posted...

Power = Torque * angular velocity

7. May 8, 2016

### RoboNerd

Ohh OK. that is exactly what I meant.

8. May 8, 2016

### CWatters

Oops looks like I miss read your OP. I see omega _is_ the angular velocity.

9. May 8, 2016

### RoboNerd

OK, so is the reason why they have a 2 in the denominator for the accepted answer the fact that they are accounting for average omega?

10. May 8, 2016

### CWatters

The equation for the KE in a moving object is 0.5mV2.. What's the same for a flywheel ?

11. May 8, 2016

### RoboNerd

I do not believe they are asking for kinetic energy of rotation in this problem.

12. May 8, 2016

### CWatters

Not explicitly but remember

Power = energy/time

(or rather the change in energy/change in time)

13. May 8, 2016

### RoboNerd

well I could have done that approach. The question is not what approach I could have done, but rather, if they were doing for average power.

OK, going your way I would have (0.5) * I * omega^2 /2, which is the "right" answer.

14. May 8, 2016

### CWatters

15. May 8, 2016

### RoboNerd

I did not mean "average power", I was talking about taking the average omega with it being = (wi + wf)/2 with the expression turning into wf/2 because wi is equal to zero.

16. May 8, 2016

### CWatters

I can see where you are coming from but I get there a slightly different way...

The work done by Torque T is given by
Work = Tθ
where θ = Angular Displacement

Now
T = Iα
where
I = Moment of inertia
α = angular acceleration

so
Work = Tθ = Iαθ

α = ω/t
and
θ = ωt/2

so the work done is

Work = I * ω/t * ωt/2 = 0.5Iω2

and the power is that divided by time..

Power = 0.5Iω2/t

17. May 9, 2016

### lychette

18. May 9, 2016

### lychette

your first post is absolutely correct.
Power = T x ω
The object accelerates from zero up to ω and therefore the power increase from zero up to Tω after t seconds
If the quoted answer is 0.5Tω then this is the average power supplied during the acceleration.

19. May 9, 2016

### RoboNerd

OK, thanks so much for the input!!

20. May 9, 2016

### lychette

The power is not constant !
It increases from zero to a max of Tω where ω is the final angular velocity
This is how the value of 0.5Tω comes about for the average power (assuming uniform accelerarion)
Mistakes like this cause great confusion !!