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Physics Question on System of Particles & Rotational Motion

A constant torque of 3.14 Nm3.14\,Nm is exerted on a pivoted wheel. If the angular acceleration of the wheel is 4π rad/s2,4\pi \,rad/s^{2}, then the moment of Inertia of the wheel is

A

0.25 kg−m20.25\,kg-m^{2}

B

2.5 kg−m22.5\,kg-m^{2}

C

4.5 kg−m24.5\,kg-m^{2}

D

25 kg−m225\,kg-m^{2}

Answer

0.25 kg−m20.25\,kg-m^{2}

Explanation

Solution

Torque, τ=Iω\tau =I\omega
Moment of inertia, I=τωI=\frac{\tau }{\omega }
=3.144×π=\frac{3.14}{4\times \pi }
=3.144×3.14=14=\frac{3.14}{4\times 3.14}=\frac{1}{4}
=0.25 kg−m2=0.25\,kg-m^{2}