Question

A wheel starting with angular velocity of $10radian/sec$ acquires angular velocity of $100radian/sec$ in 15 seconds. If moment of inertia is $10kg-m^2$, then applied torque (in newton-metre) is

• A. $900$
• B. $100$
• C. $90$
• D. $60$

Answer 1

The correct option is D $60$

$\begin{array}{c}\text{Let, initial angular velocity}\left({\omega }_0\right)\text{and final angular velocity}(\omega {)}_t.\\ \text{Given,}{\omega }_0=10\text{rad/sec}\\ \text{and,}{\omega }_t=100\text{rad}/sec\text{.}\\ \text{Also,}\Delta t=15sec\text{.}\end{array}$

$\text{Thus, angular acceleration}\left(\alpha \right)=\frac{{\omega }_t{\omega }_0}{\Delta t}=\frac{90}{15}rad/s^2=6rad/s^2$

$\begin{array}{c}\text{we are given that, moment of inertia of wheel}\left(I\right)=10kg-m^2\text{.}\\ \text{we know,}\text{Torque}\left(\tau \right)=I\alpha \\ \text{Thus, applied torque}=(10\times 6)N-m=60\times Nm\end{array}$