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$^{-2}$.
Answer (integer)
15
Solution
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:
<br/><br/>
$F=T=52.5~\mathrm{N}$
<br/><br/>
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:
<br/><br/>
$\alpha=\frac{a}{R}$
<br/><br/>
The linear acceleration $a$ of the cylinder can be found using the formula $F=ma$:
<br/><br/>
$ma=F=52.5~\mathrm{N}$
<br/><br/>
where $m=5~\mathrm{kg}$ is the mass of the cylinder. Solving for $a$, we get:
<br/><br/>
$a=\frac{F}{m}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}}=10.5~\mathrm{m/s^2}$
<br/><br/>
Substituting this value of $a$ into the equation for $\alpha$, we get:
<br/><br/>
$$\alpha=\frac{a}{R}=\frac{10.5~\mathrm{m/s^2}}{0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$$
<br/><br/>
Therefore, the angular acceleration of the cylinder is $15~\mathrm{rad/s^2}$.<br/><br/>
<b>Alternate Method:</b><br/><br/>
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:
<br/><br/>
$T=F$
<br/><br/>
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:
<br/><br/>
$\tau=TR$
<br/><br/>
where $R$ is the radius of the cylinder.
<br/><br/>
The net torque on the cylinder is equal to the product of the moment of inertia $I$ of the cylinder and its angular acceleration $\alpha$:
<br/><br/>
$\tau=I\alpha$
<br/><br/>
The moment of inertia of a hollow cylinder about its geometrical axis which is parallel to its length is given by:
<br/><br/>
$I=MR^2$
<br/><br/>
where $M$ is the mass of the cylinder.
<br/><br/>
Substituting the given values, we get:
<br/><br/>
$\tau=TR=I\alpha=MR^2\alpha$
<br/><br/>
Solving for $\alpha$, we get:
<br/><br/>
$$\alpha=\frac{T}{MR}=\frac{F}{MR}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}\cdot0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}$$
<br/><br/>
Therefore, the angular acceleration of the cylinder is $15~\mathrm{rad/s^2}$.
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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