A thin smooth rod of length L and mass M is r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.The initial moment of inertia of the rod is $I_{rod} =

Vedclass · Accepted Answer

$\frac{M\omega_0}{M + 6m}$

Vedclass · Answer

(D) Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.The initial moment of inertia of the rod is $I_{rod} = \frac{ML^2}{12}$. The beads are at the centre,so their initial distance from the axis is $0$. Thus,$I_i = \frac{ML^2}{12}$.The initial angular momentum is $L_i = I_i \omega_0 = \left( \frac{ML^2}{12} \right) \omega_0$.When the beads reach the ends,their distance from the axis is $L/2$. The final moment of inertia is $I_f = \frac{ML^2}{12} + 2 \left( m \left( \frac{L}{2} \right)^2 \right) = \frac{ML^2}{12} + \frac{2mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{L^2(M + 6m)}{12}$.Using $L_i = L_f$: $\left( \frac{ML^2}{12} \right) \omega_0 = \left( \frac{L^2(M + 6m)}{12} \right) \omega_f$.Solving for $\omega_f$: $\omega_f = \frac{M\omega_0}{M + 6m}$.

Similar Questions

$A$ block of ice is placed at the center of a circular table. The table rotates with an angular velocity $\omega$ about a vertical axis passing through its center. If the ice melts without evaporation,what will happen to the rotational speed of the system?

State the law of conservation of angular momentum and explain it with the example of a girl sitting on a swivel chair.

The angular momentum of a system of particles is not conserved:

$A$ swimmer curls their body before entering the water so that...

$A$ mouse of mass $m$ jumps on the outside edge of a rotating ceiling fan of moment of inertia $I$ and radius $R$. The fractional loss of angular velocity of the fan as a result is

Vedclass Test Series

Exam Paper Generator

Online Exam Module