Question

A wheel is rotating about a fixed axis through its center 300 rpm. A constant torque starts acting on it, opposing 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 $⟮\frac{10}{\pi }⟯kg{m}^2$, 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

The initial angular velocity is ${\omega }_0=300\times ⟮\frac{\pi }{30}⟯$

$=10\pi rad{s}^{-1}$

We have,

${\omega }^2={\omega }_0^2+2\alpha \theta$

$0=⟮10\pi {⟯}^2-2\alpha ⟮25\times 2\pi ⟯$

$\Rightarrow \alpha =\pi$

Also,

$\tau =I\alpha$

$\tau =⟮\frac{10}{\pi }⟯\times \pi$

= 10 Nm