Question 753: Centre of Mass and Rotational Motion

Question

A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s
-1
about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,
are gently placed symmetrically on the disc in such a manner that they are touching each other along the
axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest
relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s
-1
) of the system is

Accepted Answer

Consider the disc and the rings together as a system. Since the external torque $\vec{\Gamma}_{	ext {ext }}$ is zero, the angular momentum of the system about the vertical axis through $O$ is conserved, which can be expressed as:
$$ I_i \omega_i = I_f \omega_f $$
Initially, the moment of inertia is due to the disc alone, and its value is:
$$ I_i = \frac{1}{2} M R^2 = \frac{1}{2} 	imes 50 	imes 0.4^2 = 4 \mathrm{~kg \cdot m}^2 $$
The moment of inertia of a ring with mass $m$ and radius $r$ about an axis perpendicular to its center is $m r^2$. Using the theorem of parallel axes, the ring's moment of inertia about the axis of rotation is given by:
$$ I_{	ext{ring}} = m r^2 + m d^2 = 2 m r^2 $$
where $d = r$ is the distance between the parallel axes. Thus, the moment of inertia of the complete system in the final configuration is:
$$  \begin{aligned}  I_f &= \frac{1}{2} M R^2 + 2 m r^2 + 2 m r^2 \  &= \frac{1}{2} (50)(0.4)^2 + 2 \left(2(6.25)(0.2)^2ight) = 5 \mathrm{~kg \cdot m}^2  \end{aligned}  $$
With no external torque acting on the system, the angular momentum is conserved:
$$  \begin{aligned}  I_i \omega_i &= I_f \omega_f \  4 	imes 10 &= 5 \omega_f \  \omega_f &= 8 \mathrm{~rad/s} \end{aligned}  $$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type