Question

A wheel is rotating about a fixed axis through its centre at 300 rpm. A constant torque starts acting on it opposes its motion. Before coming to rest it makes 25 complete rotations. If the moment of inertia of the wheel about the axis of rotation is $\left(\frac{10}{\pi }\right)Kg{m}^2$, the torque (in N-m) acting on it is

• A. 10
• B. 15
• C. 20
• D. 25

Answer 1

The correct option is A 10

$\omega =300rpm=10\pi rad{s}^{-1}$
${\omega }^2-{\omega }_0^2=2\alpha \theta 0-100{\pi }^2=2\times \alpha \times 25\times 2\pi \alpha =-\pi \tau =I\alpha =\frac{10}{\pi }\times \pi =10Nm$