Rotation, Conservation of Energy, Power

[jtw2e]

Homework Statement

A rotating uniform cylindrical platform that forms the base of carnival ride has a mass of 248 kg and radius of 4.8 m. The platform slows down from 4.6 rev/s to rest in 14.9 s when the motor is disconnected.

Determine the power output required to maintain a steady angular speed of 4.6 rev/s. (hp)

Homework Equations

I have KEi = Eloss

The Attempt at a Solution

All I've been able to do so far is find the initial KE(rotational) = 41286.86J.
This would equal the Energy lost. I know it shouldn't take this amount just to keep it going though. It should take a smaller input of energy to keep it rotating at 4.6rev/s (or 28.9 rad/s).

Just a bit stuck on this one :(

 

[Fewmet]

In the absence of friction or other dissipative forces, it will take no energy to maintain the constant rotational velocity.

Do you see how to use the given information to find the forces opposing the rotation?

 

[jtw2e]

In the absence of friction or other dissipative forces, it will take no energy to maintain the constant rotational velocity.

Do you see how to use the given information to find the forces opposing the rotation?

It seems to me that whatever opposes rotation is the F_friction. I guess it's a constant retarding force over the full 14.9s. But I don't really see how to tie it together. Would the retarding force be a torque?

 

[Fewmet]

It seems to me that whatever opposes rotation is the F_friction. I guess it's a constant retarding force over the full 14.9s. But I don't really see how to tie it together. Would the retarding force be a torque?

Yes: the friction acts to bring the disk to rest over those 14.9 seconds. You could view this as torque, but that's tricky: torque depends on distance from the axis of rotation, and the friction presumable acts all along the radius of the disk.

I often find it is helpful to consider what you can find, even if it does not look like it'll lead to the answer. What can you figure out given that the rotating disk goes from an energy of 41286.86 J to rest in 14.9 seconds?

 

[jtw2e]

Yes: the friction acts to bring the disk to rest over those 14.9 seconds. You could view this as torque, but that's tricky: torque depends on distance from the axis of rotation, and the friction presumable acts all along the radius of the disk.

I often find it is helpful to consider what you can find, even if it does not look like it'll lead to the answer. What can you figure out given that the rotating disk goes from an energy of 41286.86 J to rest in 14.9 seconds?

... I can find that the angular acceleration \alpha = \omega/time = 1.9398 rad/s².

But I think that's an instantaneous acceleration.

The only other thing I can think of is:
F_friction = I\alpha

 

[Fewmet]

... I can find that the angular acceleration \alpha = \omega/time = 1.9398 rad/s².

But I think that's an instantaneous acceleration.

The only other thing I can think of is:
F_friction = I\alpha

All reasonable thoughts. (A caution: to use \alpha = \omega/time you need\Delta\omega and \Deltat. That gives the acceleration that caused that change in rotational velocity).

(Maybe it is just my monitor, but the Greek letters are all looking superscripted.)

The power generated by the force of friction to bring the disk from 41286.86 J to rest in 14.9 seconds is 41286.86 J /14.9 seconds. Do you then see how to use that to get the answer?

 

[jtw2e]

...The power generated by the force of friction to bring the disk from 41286.86 J to rest in 14.9 seconds is 41286.86 J /14.9 seconds. Do you then see how to use that to get the answer?

How does that work? Because Power is Work / Time ... not KE/t right?

I feel dumber every time I try. :/

 

[Fewmet]

No reason to feel dumb. Note that when I asked for what you could find out you came up with stuff that I had not been clever enough to think of.

You should, however, take this opportunity to fill a gap in your understanding. It looks like you are a little fuzzy about he relationship between work and energy.

Yes: power = work/time. I always emphasize with my students that work done to a body changes its energy by the same amount. If you do 200 J of work to lift a mass, you have increased its energy by 200 J. If I stop a baseball that was moving with a kinetic energy of 15 J, I did 15 J of work to it.

Back to the original problem: it takes no work (and therefore requires the expenditure of no power) to keep a body rotating in the absence of friction. It is a body in motion remaining in uniform motion. But friction is acting on the spinning disk in the problem, so you need to do work if you want to keep the disk rotating uniformly. What you have worked out is that friction does 41286.86 J of work to the disk in 14.9 s, which is a rate of 2770.930201 Watts. If we assume friction is a constant. You will always need 2770.930201 W to keep the disk going.

Make sense? Do you find the reasoning compelling? If not, I can say more. It is after midnight in Easter US, though, and I have a conference to attend in the morning. I am going to sleep. I'll check back here in before I leave.