Question: The angular momentum of a flywheel having a rotational inertia of $\mathbf{0}\mathbf{.}\mathbf{140}\mathit{k}\mathit{g}{\mathit{m}}^{\mathbf{2}}$about its central axis decreases from${\mathbf{\text{3.00to0.800kgm}}}^{\mathbf{\text{2}}}\mathbf{\text{/sin1.50s.}}$(a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant 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?

1. The rotational inertia of the flywheel is$I=0.140kg{m}^2$
2. The initial angular momentum of the wheel is$L_i=3.00kg\frac{m^2}{s}$
3. The final angular momentum of the wheel is$L_f=0.800kg\frac{m^2}{s}$

The period for the applied torque is$\Delta t=1.50s$ .

Use the concept of Newton’s second law in angular form and use the expression of angular momentum in terms of rotational inertia and angular velocity. Use the concept of the work done on the wheel as product of torque and its rotational angle and rate of the work done is the average power of the flywheel.

Formula:

$$\begin{gathered} \mathit{\tau }\mathbf{=}\frac{\mathbf{\Delta }\mathbf{L}}{\mathbf{\Delta }\mathbf{t}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{f}}\mathbf{-}{\mathbf{L}}_{\mathbf{i}}}{\mathbf{\Delta }\mathbf{t}} \\ \mathit{L}\mathbf{=}\mathit{I}\mathit{\omega } \\ \mathit{\theta }\mathbf{=}{\mathit{\omega }}_{\mathbf{i}}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{\alpha }{\mathit{t}}^{\mathbf{2}} \\ \mathit{W}\mathbf{=}\mathit{\tau }\mathit{\theta } \\ {\mathit{P}}_{\mathbf{a}\mathbf{v}\mathbf{g}}\mathbf{=}\frac{\mathbf{W}}{\mathbf{\Delta }\mathbf{t}} \end{gathered}$$

(a)

According to Newton’s second law in angular form, the sum of all torques acting on a particle is equal to the time rate of the change of the angular momentum of that particle.

$$\begin{gathered} \frac{d\overrightarrow{L}}{dt}={\overrightarrow{\tau }}_{net} \\ \tau =\frac{\Delta L}{\Delta t}=\frac{L_f-L_i}{\Delta t} \\ \Rightarrow \tau =\frac{\left(0.800kg.\frac{m^2}{s}\right)-\left(3.00kg.\frac{m^2}{s}\right)}{1.50s} \\ \Rightarrow \tau =-1.467N.m \end{gathered}$$

The negative sign indicates that the average torque is going along negative z axis and wheel rotates in clockwise sense and slows down.

In magnitude, the average torque acting on the wheel is,

$\tau =1.467N.m$

(b)

The relation between the angular momentum, rotational inertia and angular velocity is

$$\begin{gathered} L=I\omega \\ \omega =\frac{L}{I} \end{gathered}$$

The angular acceleration is constant, hence

$$\begin{gathered} \tau =I\alpha \\ \Rightarrow \alpha =\frac{\tau }{I} \end{gathered}$$

According to the rotational kinematical equation as

$$\begin{gathered} \theta ={\omega }_{i}t+\frac{1}{2}\alpha t^2 \\ \Rightarrow \theta =\frac{L_i}{I}t+\frac{1}{2}\frac{\tau }{I}{t}^2 \\ \Rightarrow \theta =\frac{\left(3.00kg.\frac{m^2}{s}\right)\times \left(1.50s\right)}{0.140kg.m^2}+\frac{1}{2}\frac{\left(-1.467N.m\right)\times {\left(1.50s\right)}^2}{0.140kg.m^2} \\ \Rightarrow \theta =20.4rad \end{gathered}$$

(c)

The work done on the wheel is

$$\begin{gathered} W=\tau \theta \\ W=\left(-1.467N.m\right)\times \left(20.4rad\right) \\ W=-29.9J \end{gathered}$$

(d)

The average power of the wheel is the rate of work done.

$$\begin{gathered} P_{avg}=-\frac{W}{\Delta t} \\ \Rightarrow P_{avg}=-\frac{-29.9J}{1.50s} \\ \Rightarrow P_{avg}=19.9W \end{gathered}$$