Question

A thin smooth rod of length l and mass M is rotating freely with angular speed ${{\text{ }\\!\\!\omega\\!\\!\text{ }}_{\text{0}}}$about an axis perpendicular to the rod and passing through the center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the bead reaches the ends of the rod, will be?
$\dfrac{M{{\omega }_{0}}}{M+3m}$
$\dfrac{M{{\omega }_{0}}}{M+m}$
$\dfrac{M{{\omega }_{0}}}{M+2m}$
$\dfrac{M{{\omega }_{0}}}{M+6m}$

Answer 1

Hint : We know that when the two masses reach the end of the rod there will be a change in mass distribution. Since no external torque acts on the rod system, the angular momentum should be conserved.

Complete step by step answer:

We have a rod of mass M and length l. So we know that the moment of inertia of a rod of length l and mass M with an axis perpendicular to the center of the rod is given by,
$\text{I}=\dfrac{\text{M}{{\text{l}}^{\text{2}}}}{12}$
So rotating rod at the beginning have two masses of negligible sizes and of mass m at its center. So the angular momentum associated with the rod is given by,
$\text{L}=\text{I }\\!\\!\omega\\!\\!\text{ }$
Where, L is the angular momentum. Let ${{\text{L}}_{\text{1}}}$ be the angular momentum at the beginning of the rotation and when the two masses are at the center.
${{\text{L}}_{1}}=\text{I}{{\text{ }\\!\\!\omega\\!\\!\text{ }}_{\text{0}}}=\left( \dfrac{M{{l}^{2}}}{12} \right){{\omega }_{0}}$ ……equation (1)
Where ${{\omega }_{0}}$ is the initial angular momentum.
When the two masses of mass m there is a change in mass distribution as a result there is a change in the moment of inertia. We shall denote the new moment of inertia by $\text{I }\\!\\!'\\!\\!\text{ }$, so the new moment of inertia will consist of the moment of inertia of the rod and the moment of inertia of the two beads at a length of L/2 from the center of the rod.
$I'=\left[ \left( \dfrac{M{{l}^{2}}}{12} \right)+m{{\left( \dfrac{l}{2} \right)}^{2}}+m{{\left( \dfrac{l}{2} \right)}^{2}} \right]$
And the angular frequency at this position be $\text{ }\\!\\!\omega\\!\\!\text{ }$. The angular momentum at this position is denoted by ${{\text{L}}_{\text{2}}}$ whose formula is given by,
${{L}_{2}}=I'\omega$
${{\text{L}}_{\text{2}}}=\left( \dfrac{\text{M}{{\text{l}}^{\text{2}}}}{12}+\dfrac{\text{m}{{\text{l}}^{\text{2}}}}{2} \right)\text{ }\\!\\!\omega\\!\\!\text{ }$ …… equation(2)
According to conservation of angular momentum, the angular of the system is a constant or conserved as long as there is no external force acting on the system.
$\therefore \text{ }{{\text{L}}_{1}}={{\text{L}}_{2}}$
$\left( \dfrac{\text{M}{{\text{l}}^{2}}}{12} \right){{\omega }_{0}}=\left( \dfrac{\text{M}{{\text{l}}^{\text{2}}}}{12}+\dfrac{\text{m}{{\text{l}}^{\text{2}}}}{2} \right)\text{ }\\!\\!\omega\\!\\!\text{ }$
$\left( \dfrac{\text{M}}{6} \right){{\omega }_{0}}=\left( \dfrac{\text{M}}{6}+\text{m} \right)\text{ }\\!\\!\omega\\!\\!\text{ }$
$\therefore \text{ }\omega =\dfrac{\text{M}{{\text{ }\\!\\!\omega\\!\\!\text{ }}_{\text{0}}}}{\text{M+6m}}$
Therefore , the answer to the question is option (D)- $\dfrac{\text{M}{{\text{ }\\!\\!\omega\\!\\!\text{ }}_{\text{0}}}}{\text{M+6m}}$

Note :
Moment of Inertia is defined as a quantity that determines how much torque is needed for angular acceleration. It depends on mass distribution and the distance of the mass from the axis.
If there was only one body of mass m present at any one end of the end, then option (A) will be the correct answer.
Angular Momentum is the rotational equivalent of linear momentum. They have the same form and are subject to the fundamental constraints of conservation laws, the conservation of momentum and the conservation of angular momentum.