Question

A flywheel of radius $R$ and of moment of inertia $I$ can be rotated in vertical plane without friction about its natural axis. A light string is wound over the wheel and a body of weight $W$ is suspended at the other end of string. The angular acceleration of wheel when suspended body is released is:

• A. $\frac{WR}{I}$
• B. less than $\frac{WR}{I}$
• C. greater than $\frac{WR}{I}$
• D. $\ge$$\frac{WR}{I}$

Answer 1

The correct option is B less than $\frac{WR}{I}$

Using the force balance we get
$-T+W=\frac{W}{g}a$
Using the torque balance for fly wheel we have
$I\alpha =TR$
For no slip condition, $\alpha =\frac{a}{R}$
Thus
$I\frac{a}{R^2}=T$
Putting this value of $T$ in the first equation we get
$-\frac{Ia}{R^2}+\frac{W}{g}a=W$
$(\frac{I}{R^2}+\frac{W}{g})a=W$
$a=\left(\frac{W}{-\frac{I}{R^2}+\frac{W}{g}}\right)$
$\alpha =\frac{a}{R}=\frac{WR}{\frac{W}{g}{R}^2-I}$
Thus we get option B as the correct answer.