Question

What torque is required to accelerate a wheel with inertia of 0.70kgm^2 from 1400 to 2100 rpm in 1 minute?

Answer 1

First, work out angular acceleration.

Answer 2

\[\alpha=\frac{\omega_f-\omega_i}{t}\]\[\frac{2100rev/min-1400rev/min}{min}\]\[\alpha=700rev/min^2\]

Answer 3

I guess next
\[\tau=I\alpha\]\[\tau=0.70kg\cdot m^2(700rev/min^2)\]\[\tau=490kg\cdot m\cdot rev/min^2\]Great... what am I supposed to do with that?

Answer 4

I guess I could
700rev/min * 2pi rad/rev * min^2/3600s^2
a=1.22 rad/s^2

Answer 5

Which gives me\[0.70kg\cdot m^2\times\frac{1.22rad}{s^2}=0.86\frac{kg\cdot m^2}{s^2}\cdot rad\]\[=0.86J\cdot rad\]Well that's no help.

Answer 6

All this is correct.
The unit in which torque is expressed is not J or J.rad but newton-metre (N.m), because it is the moment of a force, hence a force multiplied by a distance.

So your answer should read:

tau = 0.86 N.m