Answer to Question #123777 in Mechanics | Relativity for Ojugbele Daniel

Question

The angular momentum of a flywheel having a Moment of inertia 0.125kgm^3 decreases from 3.0 to 2.0kgm^2/s^2 in a period of 1.5s
(i) Assuming a uniform angular acceleration, through how many revolutions will the flywheel have tuned?
(ii) how much work is done?

Answer 1

Solution.

$I=0.125kgm^2;$

$L1=3.0kgm^2/s;$

$L_2=2.0kgm^2/s;$

$t=1.5s;$

i)$M=\dfrac{\Delta L}{\Delta t};$

$M=\dfrac{1kgm^2/s}{1.5s}=0.7kgm^2/s;$

$M=I\epsilon\implies \epsilon=\dfrac{M}{I};$

$\epsilon=\dfrac{0.7kgm^2/s}{0.125kgm^2}=5.6rad/s^2;$

$\epsilon=\dfrac{\Delta\omega}{\Delta t};$ $\Delta \omega=\dfrac{\Delta\phi}{\Delta t};$

$\Delta\phi=\epsilon(\Delta t)^2;$

$\Delta\phi=5.6rad/s^2\sdot(1.5s)^2=12.6rad;$

$n=\dfrac{12.6rad}{2\pi rad}=8;$

ii )$A=\dfrac{I\omega_1^2}{2}-\dfrac{Iw_2^2}{2};$

$L_1=I\omega_1;$

$\omega_1=\dfrac{3.0kgm^2/s}{0.125kgm^2}=24rad/s;$

$\omega_2=\dfrac{2.0kgm^2/s}{0.125kgm^2}=16rad/s;$

$A=\dfrac{0.125kgm^2\sdot(24rad/s)^2}{2}-\dfrac{0.125kgm^2\sdot(16rad/s)^2}{2}=20J;$

Answer: $i)n=8;$

$ii)A=20J.$