
A particle performs uniform circular motion with an angular momentum L. If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes:
A) $4L$
B) $2L$
C) $\dfrac{L}{2}$
D) $\dfrac{L}{4}$
Answer
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Hint: Angular momentum is defined as the measure of the rotational momentum of the rotating body which is equal to the product of the angular velocity of the system and the moment of the inertia to the axis. Angular momentum is a vector quantity.
Complete step by step solution:
Given data:
Initial angular momentum = L
Initial angular frequency = $\omega $
Initial Kinetic energy = k
Final angular frequency, ${\omega ‘} = 2\omega $
Final Kinetic energy, ${k’} = \dfrac{k}{2}$
Final angular momentum =?
We know that angular momentum is given by the formula, $L = mvr$
And also we know that, $v = r\omega $
Thus L can be written as, $L = m\omega {r^2}$
The kinetic energy is given by the formula, $k = \dfrac{1}{2}m{v^2}$
We can also write angular momentum in terms of Inertia and angular frequency as, $L = I\omega $
Thus kinetic energy becomes, $k = \dfrac{1}{2}I{\omega ^2}$
$\therefore $Final Inertia, ${I’} = \dfrac{1}{8}I$
Thus substituting the value of ${I’}, {\omega '}$, we get,
$\Rightarrow {L'} = \dfrac{1}{8}I \times 2\omega = \dfrac{1}{4}I\omega $ $\Rightarrow \left( {\because {L'} = {I'}{\omega '}} \right)$
$\Rightarrow {L'} = \dfrac{L}{4}$ $\left( {\because L = I\omega } \right)$
Thus the final angular momentum becomes, ${L'} = \dfrac{L}{4}$.
Hence the correct option is D.
Additional Information:
1. Kinetic energy is defined as the energy possessed by a body by its motion. It is the energy of motion.
2. Angular frequency is also called a circular frequency or radial frequency and is defined as the measurement of the angular displacement per unit time.
Note: 1. Jean Buridan who is the discoverer of momentum also discovered angular momentum.
2. As the mass increases, the angular momentum also increases. Hence we can say that the mass will be directly proportional to the angular momentum.
3. Angular momentum also depends on the rotational velocity and the rotational inertia. Whenever the object changes its shape, the angular velocity changes, and thus the angular momentum also changes.
Complete step by step solution:
Given data:
Initial angular momentum = L
Initial angular frequency = $\omega $
Initial Kinetic energy = k
Final angular frequency, ${\omega ‘} = 2\omega $
Final Kinetic energy, ${k’} = \dfrac{k}{2}$
Final angular momentum =?
We know that angular momentum is given by the formula, $L = mvr$
And also we know that, $v = r\omega $
Thus L can be written as, $L = m\omega {r^2}$
The kinetic energy is given by the formula, $k = \dfrac{1}{2}m{v^2}$
We can also write angular momentum in terms of Inertia and angular frequency as, $L = I\omega $
Thus kinetic energy becomes, $k = \dfrac{1}{2}I{\omega ^2}$
$\therefore $Final Inertia, ${I’} = \dfrac{1}{8}I$
Thus substituting the value of ${I’}, {\omega '}$, we get,
$\Rightarrow {L'} = \dfrac{1}{8}I \times 2\omega = \dfrac{1}{4}I\omega $ $\Rightarrow \left( {\because {L'} = {I'}{\omega '}} \right)$
$\Rightarrow {L'} = \dfrac{L}{4}$ $\left( {\because L = I\omega } \right)$
Thus the final angular momentum becomes, ${L'} = \dfrac{L}{4}$.
Hence the correct option is D.
Additional Information:
1. Kinetic energy is defined as the energy possessed by a body by its motion. It is the energy of motion.
2. Angular frequency is also called a circular frequency or radial frequency and is defined as the measurement of the angular displacement per unit time.
Note: 1. Jean Buridan who is the discoverer of momentum also discovered angular momentum.
2. As the mass increases, the angular momentum also increases. Hence we can say that the mass will be directly proportional to the angular momentum.
3. Angular momentum also depends on the rotational velocity and the rotational inertia. Whenever the object changes its shape, the angular velocity changes, and thus the angular momentum also changes.
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