Answer to Question #139000 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

$L=mvR=m\omega R^2=2\cdot\frac{1}{2}m R^2\omega=2I\omega$ $\Rightarrow$ $\omega=\frac{L}{2I}$

$<M>=R<F>=\frac{R}{2}(F_1+F_2)=\frac{R}{2}(m\omega_1^2R+m\omega_2^2R)=\frac{mR^2}{2}(\omega_1^2+\omega_2^2)=I(\omega_1^2+\omega_2^2)=\frac{L_1^2+L_2^2}{4I}$

$<M>=\frac{3^2+2^2}{4\cdot0.125}=26$ N•m

$\varphi=2\pi n=\frac{\omega_1+\omega_2}{2}t=\frac{t}{2I}(L_1+L_2)$

$n=\frac{t}{4\pi I}(L_1+L_2)$

$n=\frac{1.5}{4\cdot3.14\cdot0.125}\cdot(3+2)\approx4.78$