
A cylinder of mass $ 250kg $ and radius $ 2.60m $ is rotating at $ 4.00{\text{ }}rad/s $ on a frictionless surface when two more identical non-rotating cylinders fall on top of the first. Because of friction between the cylinders, they will eventually all come to rotate at the same rate. What is this final angular velocity?
(A) $ 0.5\omega $
(B) $ 0.33\omega $
(C) $ 0.1\omega $
(D) $ \omega $
Answer
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Hint: We know that from the law of conservation of angular momentum when there is no external torque acting on it then the total angular momentum of the system is conserved and therefore it does not change. We use the conservation of angular momentum because we have a collision of cylinders that results in a change in rotation so therefore we conserve angular momentum.
The angular momentum of the rotating objects is given by,
$ L = I\omega $
Here $ I $ is the moment of inertia and $ \omega $ is the angular velocity of the rotating object.
Complete Step By Step Answer:
As we already said there will be no change in the angular momentum and therefore,
$ {L_f} = {L_i} $ ……. (1)
Here $ {L_f} $ is the final angular momentum and $ {L_i} $ is the initial angular momentum.
The angular momentum of the rotating objects is given by,
$ L = I\omega $ ……. (2)
Here $ I $ is the moment of inertia and $ \omega $ is the angular velocity of the rotating object.
Let the initial cylinder has a moment of inertia I. Given that two more identical cylinders fall on the top of the initial cylinder and have the same dimension as that of the initial cylinder and therefore we can say that the total moment of inertia of the three cylinders increases and it will be equal to $ 3I $ .
If we write equation (1) in terms of equation (2) we get,
$ 3I{\omega _f} = I{\omega _i} $ …… (2)
$ \Rightarrow {\omega _f} = \dfrac{{{\omega _i}}}{3} $
$ \Rightarrow {\omega _f} = 0.33\omega $
Correct Answer: Therefore the correct option is B.
Note:
Before we proceed to apply the conservation of angular momentum law, we must properly analyse if there are any net external torques or not. Because in the presence of even a single torque, this law will become invalid. This is because the angular moment of the system is bound to change. This law of conservation of angular momentum can be used in various problems based on rotational motion.
The angular momentum of the rotating objects is given by,
$ L = I\omega $
Here $ I $ is the moment of inertia and $ \omega $ is the angular velocity of the rotating object.
Complete Step By Step Answer:
As we already said there will be no change in the angular momentum and therefore,
$ {L_f} = {L_i} $ ……. (1)
Here $ {L_f} $ is the final angular momentum and $ {L_i} $ is the initial angular momentum.
The angular momentum of the rotating objects is given by,
$ L = I\omega $ ……. (2)
Here $ I $ is the moment of inertia and $ \omega $ is the angular velocity of the rotating object.
Let the initial cylinder has a moment of inertia I. Given that two more identical cylinders fall on the top of the initial cylinder and have the same dimension as that of the initial cylinder and therefore we can say that the total moment of inertia of the three cylinders increases and it will be equal to $ 3I $ .
If we write equation (1) in terms of equation (2) we get,
$ 3I{\omega _f} = I{\omega _i} $ …… (2)
$ \Rightarrow {\omega _f} = \dfrac{{{\omega _i}}}{3} $
$ \Rightarrow {\omega _f} = 0.33\omega $
Correct Answer: Therefore the correct option is B.
Note:
Before we proceed to apply the conservation of angular momentum law, we must properly analyse if there are any net external torques or not. Because in the presence of even a single torque, this law will become invalid. This is because the angular moment of the system is bound to change. This law of conservation of angular momentum can be used in various problems based on rotational motion.
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