Rotational Motion Question 100
Question: A cylinder of mass (M) and radius (R) is rotating with an angular velocity $(\omega)$. If the radius of the cylinder is halved keeping the mass same, what will be the new angular velocity?
Options:
A) $\omega$
B) $\frac{\omega}{2}$
C) $\frac{\omega}{4}$
D) $2\omega$
Show Answer
Answer:
Correct Answer: B
Solution:
The moment of inertia $(I)$ of a cylinder about its axis is given by $(\frac{1}{2}MR^2)$.
According to the conservation of angular momentum $(L = I\omega = \text{constant})$, if the radius is halved, the new moment of inertia $(I’)$ becomes
$(\frac{1}{2}M\left(\frac{R}{2}\right)^2$ = $\frac{1}{8}MR^2)$.
Therefore, the new angular velocity $(\omega’)$ must be such that
$(I’\omega’ = I\omega)$.
Therefore,
$\frac{1}{8}MR^2\omega’ = \frac{1}{2}MR^2\omega$
Solving for $(\omega’)$, we get $(\omega’ = \frac{\omega}{2})$.
