Answer to Question #141807 in Mechanics | Relativity for Ojugbele Daniel

Question

An angular momentum of a fly wheel having a momentum of inertia of 0.125kgm^2 decreases form 3.0kgm^2/s to 2.0kgm^2/s in a period of 1.5secs.
(i) what is the average torque acting on the fly wheel during this period.
(ii) assuming a uniform angular acceleration, through how many revolution will the fly wheel have turned?

Answer 1

Average torque$=I\times\frac{\Delta w}{\Delta t}=0.125\times\frac{2-3}{1.5}=0.083$

$\tau=I\alpha$

$0.083=0.125\times\alpha$

$\alpha=0.67$