Question

Angular velocity of a body of moment of inertia 100 $kg{-m}^2$ change from 2 rad/s to 4 rad/s in one minute. if constant torque acting on the body is 10 Nm, then the angle described by the body in one minute is:

• A. 180 rad
• B. 60 rad
• C. 600 rad
• D. 10 rad

Answer 1

Answer: (A)

180 rad

$\begin{array}{c}\text{Given, initial angular velocity}\left({\omega }_i\right)=2\text{rad/s.}\\ \text{and final angular velocity}\left({\omega }_f\right)=4\text{rad}/\text{s. and}\Delta t=1\text{min}=60\text{sec}\\ \text{Thus, angular acceleration}=\frac{{\omega }_f-{\omega }_i}{\Delta t}=\frac{2}{60}\text{rad}/s^2\\ =\frac{1}{30}{\text{rad/s}}^2\end{array}$

$\begin{array}{c}\text{Ther,}{\omega }_f^2={\omega }_i^2+2\alpha \Delta \theta \\ \therefore \Delta \theta =\frac{{\omega }_f^2-{\omega }_i^2}{2\alpha }=\frac{(4{)}^2-(2{)}^2}{2\times \frac{1}{30}}\text{rad.}=180\text{rad.}\\ \text{Thus, angle described by the body in}1min\text{is}180\text{rad.}\end{array}$