Question

A pulley $1m$ in diameter rotating at $600r.p.m.$ is brought to rest in $80s$. By constant force of friction on its shaft. How many revolutions does it moves?

• A. $200$
• B. $400$
• C. $600$
• D. $500$

Answer 1

The correct option is B $400$

Radius, $r=0.5$ $m$

Initial angular velocity, ${\omega }_o=600$ $rpm$

$⟹{\omega }_o=\frac{600}{60}=10$ $rpm$

$⟹{\omega }_o=10\times 2\pi$

$⟹{\omega }_o=20\pi$ $rad/s$

Time, $t=80$ $s$

Final angular velocity, $\omega =0$ $rad/s$

From kinematics:

$\omega -{\omega }_o=\alpha t$

$⟹\frac{-20\pi }{80}=\alpha$

$⟹\alpha =-0.25\pi$

Also, $\theta ={\omega }_{o}t+\frac{1}{2}\alpha t^2$

$⟹\theta =20\pi \times 80+\frac{1}{2}\times (-0.25\pi )\times 6400$

$⟹\theta =1600\pi -(0.25\times 3200\pi )$

$⟹\theta =1600\pi -800\pi$

$⟹\theta =800\pi$

$⟹\theta =400\times 2\pi$

$\therefore$ The pulley makes $400$ revolutions.