Question

The angular momentum of a flywheel having a rotational inertia of 0.200 kg · m2 about its central axis decreases from 3.80 to 0.600 kg · m2/s in 1.30 s
(b) Assuming a uniform angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the average power of the flywheel?

Answer 1

Answer:

14.3065 rad

-36.777 J

-28.29 W

Explanation:

$L_f$ = Final angular momentum = 0.6 kgm²/s

$L_i$ = Initial angular momentum = 3 kgm²/s

I = Moment of inertia = 0.2 kgm²

Torque is given by

$\tau=\dfrac{L_f-L_i}{t}\\\Rightarrow \tau=\frac{0.6-3.8}{1.3}\\\Rightarrow \tau=-2.46\ Nm$

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2$

Initial angular speed is given by

$\omega_i=\dfrac{L_i}{I}$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}$

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=\dfrac{L_it+\dfrac{1}{2}\tau t^2}{I}\\\Rightarrow \theta=\dfrac{3.8\times 1.3+\dfrac{1}{2}\times -2.46\times 1.3^2}{0.2}\\\Rightarrow \theta=14.3065\ rad$

The angle is 14.3065 rad

Work done is given by

$W=\tau \theta\\\Rightarrow W=-2.46\times 14.95\\\Rightarrow W=-36.777\ J$

The work done on the wheel is -36.777 J

Power is given by

$P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{-36.777}{1.3}\\\Rightarrow P=-28.29\ W$

The power is -28.29 W