# Rotational kinetic energy and power

warmfire540
A 30-cm diameter circular saw blade has a mass of 0.9 kg distributed uniformly in a disc.
(a) What is its rotational kinetic energy when it is operating at 4000 rpm?
(b) What average power must be applied to bring the blade from rest to its operating 4000 rpm?

Just making sure I'm doing this right...

a. Inertia=(1/2)mr^2
I=(1/2)(.9)(.15)^2
I=.0101
w=418.88
K=(1/2)Iw^2
K=(1/2)(.0101)(418.88)^2
K=886.08

b. P=average force
torque=r*F
torque=I*ang (ang being angular velocity)
so:
I*ang=r*f
F=I*ang/r

ergh...now i come to a block!
I don't know how to find acceleration without time..or how to find Force..am i on the right track here?

Homework Helper
torque=I*ang (ang being angular velocity)

No, torque = I*alpha where alpha is the angular acceleration. ("I*ang" is angular momentum)

warmfire540
No, torque = I*alpha where alpha is the angular acceleration. ("I*ang" is angular momentum)

okay..well i said the wrong letter, but i still don't know where to go from here..

rohanprabhu
firstly, you said that: P = average force. If by 'P' you mean Power, then this is totally wrong. Power is average energy per unit time [it actually is the rate of change of energy w.r.t time].

secondly, the second question is incomplete. Power talks about how much energy is supplied in unit time. A disc rotating at a constant angular velocity, 4000rpm has a specific amount of energy which does not change. Hence, to reach that energy, any amount of power is sufficient. The difference is just in how fast that rpm is achieved. More power means, the rpm will be achieved in smaller time and less power means that the rpm will be achieved in lesser time. A complete question would be something like:

(b) What average power must be applied to bring the blade from rest to its operating 4000 rpm in 200 seconds?

To do this kind of question, you just take the initial RKE and the final RKE to get the change in RKE. This change must equate the work done on it. And the source supplies this energy. Then divide this energy by the time it needs to be done in and you'll get the average power.

warmfire540
Okay, the professor said I just pick a reasonable time.. so I'll say t=15s
so now P=w/t
P=886.08/15
Power=59.072
done?

rohanprabhu
Okay, the professor said I just pick a reasonable time.. so I'll say t=15s
so now P=w/t
P=886.08/15
Power=59.072
done?

Yes. However, a 'reasonable time' is truly meaningless. In a car, you achieve 8000 rpm [it's a small, economy car] in less than 1 second in the first gear whereas it may take around 8-9 seconds in the 5th gear. But anyways, as long as you get the concept.. nothing else really matters 