Question

A solid cylinder of mass $50kg$ and radius $0.5m$ is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of $2$ revolutions/$s^2$ is

• A. $25N$
• B. $50N$
• C. $78.5N$
• D. $157N$

Answer 1

The correct option is D $157N$

Given: A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely.

To find the tension in the string required to produce an angular acceleration of 2 rev/s${}^2$

Solution:

Here as per the given criteria,

$m=50kg$

$r=0.5m$

$\alpha =2rev/s^2=4\pi rad/s^2$

For solid cylinder moment of inertia about its central axis is

$I=\frac{1}{2}m{r}^2$

And the torque produced by the tension T in the string will be

$\tau =Tr$

But we know, $\tau =I\alpha$

Hence $Tr=I\alpha ⟹Tr=\frac{1}{2}m{r}^2\times 4\pi ⟹T=2\pi mr$

Now substituting the corresponding values we get

$T=2\times \frac{22}{7}\times 50\times 0.5⟹T\approx 157N$

is the tension in the string required to produce an angular acceleration of 2 rev/s${}^2$