Answer to Question #142043 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

(i) torque change in angular momentum / time required

$\frac{3-2}{1.5} = 0.66 \;kgm^2/s^2$

(ii) $I = 0.125\;kgm^2$

$L = 1 \;km^2/s$

$w = \frac{L}{I} = \frac{1}{0.125} = 8 \;rads^{-1}$