Question 417: Centre of Mass and Rotational Motion

Question

A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be _________ rad s$$^{-2}$$.

Accepted Answer

In this problem, the net force on the cylinder is the tension $$T$$ in the rope, which is equal to the force applied to the rope:
$$F=T=52.5~\mathrm{N}$$
The force causes the cylinder to accelerate with an angular acceleration $$\alpha$$, which is related to its linear acceleration $$a$$ and the radius of the cylinder $$R$$ by the equation:
$$\alpha=\frac{a}{R}$$
The linear acceleration $$a$$ of the cylinder can be found using the formula $$F=ma$$:
$$ma=F=52.5~\mathrm{N}$$
where $$m=5~\mathrm{kg}$$ is the mass of the cylinder. Solving for $$a$$, we get:
$$a=\frac{F}{m}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}}=10.5~\mathrm{m/s^2}$$
Substituting this value of $$a$$ into the equation for $$\alpha$$, we get:
$$\alpha=\frac{a}{R}=\frac{10.5~\mathrm{m/s^2}}{0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$$
Therefore, the angular acceleration of the cylinder is $$15~\mathrm{rad/s^2}$$.
Alternate Method:
Let's first draw a free body diagram of the cylinder. The force $$F$$ applied to the rope creates a tension in the rope, which in turn exerts a force on the cylinder in the opposite direction. This force is given by:
$$T=F$$
where $$T$$ is the tension in the rope. The cylinder also experiences a torque due to the tension in the rope, which causes it to rotate. The torque is given by:
$$	au=TR$$
where $$R$$ is the radius of the cylinder.
The net torque on the cylinder is equal to the product of the moment of inertia $$I$$ of the cylinder and its angular acceleration $$\alpha$$:
$$	au=I\alpha$$
The moment of inertia of a hollow cylinder about its geometrical axis which is parallel to its length is given by:
$$I=MR^2$$
where $$M$$ is the mass of the cylinder.
Substituting the given values, we get:
$$	au=TR=I\alpha=MR^2\alpha$$
Solving for $$\alpha$$, we get:
$$\alpha=\frac{T}{MR}=\frac{F}{MR}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}\cdot0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$$
Therefore, the angular acceleration of the cylinder is $$15~\mathrm{rad/s^2}$$.

Centre of Mass and Rotational Motion

A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be _________ rad s$^{-2}$.

Integer / Numeric Answer Type