Question

A grindstone has a moment of inertia of $10\, kgm ^{2}$. A constant torque is applied and the grindstone is found to have a speed of 200 rpm. 20 sec, after starting from rest, what will be the torque?

• A. $2.14 \,Ns$
• B. $6 \,Ns$
• C. $4 \,Ns$
• D. $3.33 \pi \,Ns$

Answer 1

Answer: (D)

$\omega_{0}=0 \omega=2 \pi n=2 \times \pi \times \frac{200}{60}$
$=\frac{20 \pi}{3} rad / sec$
angular acceleration $\alpha=\frac{\omega-\omega_{0}}{t}$
$=\frac{\frac{20 \pi}{3}-0}{20}$
$=\frac{20 \pi}{3 \times 20}$
$=\frac{\pi}{3} rad / s ^{2}$
$\therefore \tau=I \alpha=10 \times \frac{\pi}{3}$
$=3.33 \pi \,Ns$