Question

A wheel of radius 0.4 m can rotate freely about its axis as shown in the figure. A string is wrapped over its rim and a mass of 4 kg is hung. An angular acceleration of 8 rad $-s^{-2}$ is produced in it due to the torque. Then, moment of inertia of the wheel is $(g=10m{s}^{-2})$ :

• A. $2kg-m^2$
• B. $8kg-m^2$
• C. $4kg-m^2$
• D. $1.36kg-m^2$

Answer 1

The correct option is D $1.36kg-m^2$

$\begin{array}{c}\text{Given, value of}R=0.4m\text{.}\\ \text{Let, mass goes down with acceleration,}a.\\ \text{Thus,}40-T=4a---\left(1\right)\\ \text{Let, moment of inertia of wheel = I.}\end{array}$

$\begin{array}{c}\text{Thus,}\tau =I\alpha [\tau =\text{Torque},\alpha =\text{angular acceleration}\\ \Rightarrow \tau =I\times 8\end{array}$

$\text{In}e{q}^n-\left(i\right)\to a=R\alpha \Rightarrow 40-T=4\times 0.4\times 8\Rightarrow T=27.2N$

$\begin{array}{cc}\text{we can write,}\tau =T\times R& =27.2\times 0.4N-m\\ & =10.88Nm.\end{array}$

$So,10.88=I\times 8\Rightarrow I=1.36kg{m}^2$