A rope is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30N? What is the linear acceleration of the rope?

Solution

We have,
Mass of cylinder, m=3 kg
Radius of the cylinder, r = 40 cm = 0.4 m

Moment of inertia, I = mr² = 3 (0.4)² = 0.48 kg m²
Since the rope leaves the cylinder tangential, therefore torque by tension in string is,
$straight tau space equals space rF space equals space 0.4 space straight x space 30 space equals space 12 space Nm$
Also,
$straight tau space equals thin space straight I space straight alpha$
Therefore,
$straight alpha space equals space straight tau over straight I space equals space fraction numerator 12 over denominator 0.48 end fraction space equals space 25 space rad divided by sec squared$
Linear acceleration of rope is,
$straight a space equals space straight r space straight alpha space equals space 0.4 space straight x space 25 space equals space 10 space straight m divided by straight s squared$