Question

A constant torque acting on a uniform circular wheel of moment of inertia ${}^{\prime }$ , so that its angularmomentum changes from $A_0$ to 4$A_0$ in 4 sec. The average power due to torque on the wheel in4 sec is

• A. $\frac{A_0^2}{2I}$
• B. $\frac{3A_0^2}{4I}$
• C. $\frac{15A_0^2}{8I}$
• D. $\frac{A_0^2}{8I}$

Answer 1

The correct option is C $\frac{15A_0^2}{8I}$

$\begin{array}{c}\text{Let the angular momentum be givenas}\\ [L=A_0+\frac{3A_0}{4}t]\text{where}t=\text{time.}\\ \Rightarrow \text{Lorque}\tau =\frac{dL}{dt}=\frac{3A_0}{4}\\ \text{and angulare velocity (W)}\\ \omega =\frac{1}{I}[A_0+\frac{3A_0}{4}t]\end{array}$

$\begin{array}{cc}& \Rightarrow \text{Power}P=[\cdot \omega =\frac{3A_0^2}{4I}[1+\frac{3t}{4}]\\ & \Rightarrow \text{average Power}{P}_{\text{on}}=\frac{{\int }_0^{4}P\cdot dt}{4\sec }\\ & P_{ON}={\int }_0^4\frac{\frac{3A_0^2}{4I}\cdot (1+\frac{3t}{4})\cdot dt}{4}\\ & P_{\text{ow}}=\frac{15A_0^2}{8I}\end{array}$