Question

A wheel of moment of inertia $5\times {10}^{-3}kg{m}^2$ is making $20rev/s$. The torque required to stop it in $10s$ is :

• A. $2\pi \times {10}^{-2}Nm$
• B. $2\pi \times {10}^{2}Nm$
• C. $4\pi \times {10}^{-2}Nm$
• D. $4\pi \times {10}^{2}Nm$

Answer 1

Answer: (A)

$2\pi \times {10}^{-2}Nm$

$\omega =20\times 2\pi$
$\tau =I\omega$

So, $\tau \Delta t=5\times {10}^{-3}\omega$

$\tau =\frac{5\times {10}^{-3}\times 20\times 2\pi }{10}=2\pi \times {10}^{-2}Nm$