
A round disc of moment of inertia ${I_2}$ about its axis perpendicular to its plane and passing through its center is placed over another disc of moment of inertia ${I_1}$rotating with an angular velocity $\omega $ about the same axis. The final angular velocity of the combination of disc is
A. $\dfrac{{{I_2}\omega }}{{{I_1} + {I_2}}}$
B. $\omega $
C. $\dfrac{{{I_1}\omega }}{{{I_1} + {I_2}}}$
D. $\dfrac{{\left( {{I_1} + {I_2}} \right)\omega }}{{{I_1}}}$
Answer
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Hint: The linear momentum depends on the mass and linear velocity. Since mass is constant when the linear velocity increases the linear momentum increases. In rotational motion, the moment of inertia is the analogue of mass. Similarly, the Angular momentum depends on the moment of inertia and the angular velocity. The radius of the body from the rotating axis and the angular velocity can affect the angular momentum.
Formula used:
Angular momentum
$L = I\omega $
Here, $I$ is the moment of inertia, and $\omega $ is the angular velocity.
Complete step by step solution:
Understand that, Angular velocity is the rate of change of position of the rotating body in its axis of rotation.
Angular momentum of a disc of the moment of inertia ${I_1}$ and rotating about its axis with angular velocity $\omega $ is given as,
${L_1} = {I_1}\omega $
When a round disc of the moment of inertia ${I_2}$ is placed over the first one and it rotates with angular velocity $\omega^{'}$, then the angular momentum is given as
${L_2} = \left( {{I_1} + {I_2}} \right)\omega^{'}$
Here ${I_1}$ is the moment of inertia of the first disc.
If the external torque is absent then the angular momentum is conserved.
${L_1} = {L_2}$
$
{I_1}\omega = \left( {{I_1} + {I_2}} \right)\omega^{'} \\
\\
$
Rearrange the above equation for $\omega^{'}$,
$\omega^{'} = \dfrac{{{I_1}\omega }}{{\left( {{I_1} + {I_2}} \right)}}$
$\therefore$ Hence, the option (C) is correct.
Additional information:
Angular momentum can be exchanged in a closed system. But it will be conserved. The torque cannot be exerted externally in the closed system. Also when the system is subjected to the collision the angular momentum will be conserved from all the directions. Angular momentum can give an account for the distribution of mass in an object. While integrating the mass of the particles of the body and their distance from the axis of rotation we can find the moment of inertia. When the rotation is constant, the system will have a constant angular momentum.
Note:
The angular velocity of the combined system will be $\dfrac{{{I_1}}}{{\left( {{I_1} + {I_2}} \right)}}$times the angular velocity $\omega $ of the disc having moment of inertia ${I_1}$. If the moment of inertia of the first disc is greater, then the angular velocity of the combined system can be a greater value. Hence the mass of the first disc is a depending factor. Gradually for greater mass, the angular velocity can be increased.
Formula used:
Angular momentum
$L = I\omega $
Here, $I$ is the moment of inertia, and $\omega $ is the angular velocity.
Complete step by step solution:
Understand that, Angular velocity is the rate of change of position of the rotating body in its axis of rotation.
Angular momentum of a disc of the moment of inertia ${I_1}$ and rotating about its axis with angular velocity $\omega $ is given as,
${L_1} = {I_1}\omega $
When a round disc of the moment of inertia ${I_2}$ is placed over the first one and it rotates with angular velocity $\omega^{'}$, then the angular momentum is given as
${L_2} = \left( {{I_1} + {I_2}} \right)\omega^{'}$
Here ${I_1}$ is the moment of inertia of the first disc.
If the external torque is absent then the angular momentum is conserved.
${L_1} = {L_2}$
$
{I_1}\omega = \left( {{I_1} + {I_2}} \right)\omega^{'} \\
\\
$
Rearrange the above equation for $\omega^{'}$,
$\omega^{'} = \dfrac{{{I_1}\omega }}{{\left( {{I_1} + {I_2}} \right)}}$
$\therefore$ Hence, the option (C) is correct.
Additional information:
Angular momentum can be exchanged in a closed system. But it will be conserved. The torque cannot be exerted externally in the closed system. Also when the system is subjected to the collision the angular momentum will be conserved from all the directions. Angular momentum can give an account for the distribution of mass in an object. While integrating the mass of the particles of the body and their distance from the axis of rotation we can find the moment of inertia. When the rotation is constant, the system will have a constant angular momentum.
Note:
The angular velocity of the combined system will be $\dfrac{{{I_1}}}{{\left( {{I_1} + {I_2}} \right)}}$times the angular velocity $\omega $ of the disc having moment of inertia ${I_1}$. If the moment of inertia of the first disc is greater, then the angular velocity of the combined system can be a greater value. Hence the mass of the first disc is a depending factor. Gradually for greater mass, the angular velocity can be increased.
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