Question

A uniform ring is rotating about the vertical axis with angular velocity \[\theta \] initially. A point insect (S) having the same mass as that of the ring starts walking from the lowest point \[{P_1}\] and finally reaches the point \[{P_2}\] (as shown in figure). If the final angular velocity of the ring is \[\dfrac{\theta}{x}\], find the value of \[x\].

Answer 1

Hint: Use the formula for the angular momentum of an object, moment of inertia of the ring about an axis passing through its centre and perpendicular to plane of the ring. Also use the formula for moment of inertia of an object. Determine the initial and final angular momentum of the system and use the law of conservation of angular momentum to calculate final velocity of the ring.

Formulae used:
The angular momentum \[L\] of an object is
\[L = I\omega \] …… (1)
Here, \[I\] is the moment of inertia of the object and \[\omega \] is the angular velocity of the object.
The moment of inertia \[I\] of the ring is about an axis perpendicular to the plane of ring and passing through its centre is
\[I = \dfrac{{m{R^2}}}{2}\] …… (2)
Here, \[m\] is the mass of the ring and \[R\] is the radius of the ring.
The expression for moment of inertia \[I\] of an object is
\[I = m{R^2}\] …… (3)
Here, \[m\] is the mass of the object and \[R\] is the distance from the axis of rotation.

Complete step by step answer:
We have given that initially the ring is rotating with angular velocity \[\theta \].
\[{\omega _i} = \theta \]
We have to calculate the final velocity of the ring.
Let \[m\] be the mass of the ring which is the same as the mass of the insect S and \[R\] be the radius of the ring.The insect walks on the ring from the lowest point \[{P_1}\] to the point \[{P_2}\].According to equation (1), the initial angular momentum \[{L_i}\] of the ring-insect system is
\[{L_i} = I\theta \]
Here, \[I\] is a moment of inertia of the ring about an axis passing through its centre and perpendicular to the plane of the ring.

According to equation (1), the final angular momentum \[{L_f}\] of the ring-insect system is
\[{L_f} = I\omega + I'\omega \]
Here, \[I'\] is the moment of inertia of the insect at point \[{P_2}\] about the axis of rotation and \[\omega \] is the final angular velocity of the ring-insect system.
According to the law of conservation of angular momentum, the initial angular momentum of the ring-insect system is equal to the final angular momentum of the ring-insect system.
\[{L_i} = {L_f}\]
\[ \Rightarrow I\theta = I\omega + I'\omega \]
\[ \Rightarrow \left( {\dfrac{{m{R^2}}}{2}} \right)\theta = \left( {\dfrac{{m{R^2}}}{2}} \right)\omega + m{R^2}\omega \]
\[ \Rightarrow \dfrac{\theta }{2} = \dfrac{\omega }{2} + \omega \]
\[ \Rightarrow \dfrac{\theta }{2} = \dfrac{{3\omega }}{2}\]
\[ \therefore \omega = \dfrac{\theta }{3}\]

Hence, the value of x is 3.

Note: The students should not forget to take initial angular momentum of the insect zero as initially it is on the same axis about which the moment of inertia is to be measured which makes the moment of inertia of the insect zero. Also the students should keep in mind that the final angular velocity of the ring and insect is the same as the insect is walking on the ring.