In This post you will get answer of Q.
A light rope is wound around a hollow cylinder of mass $4, kg$ and radius $40, cm.$ If the rope is pulled with a force of $40, N.$ it’s angular acceleration is
Solution:
Torque, $tau=I alpha$
$F times r =M r^{2} alpha$
$40 times 0.4=4 times(0.4)^{2} alpha$
$16=0.64 alpha$
$alpha=frac{16}{0.64}$
$=frac{16}{64} times 100$
$=25, rad / s ^{2}$
Solution:
Torque, $tau=I alpha$
$F times r =M r^{2} alpha$
$40 times 0.4=4 times(0.4)^{2} alpha$
$16=0.64 alpha$
$alpha=frac{16}{0.64}$
$=frac{16}{64} times 100$
$=25, rad / s ^{2}$