
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
426.1k+ views
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.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

Current Loop as Magnetic Dipole and Its Derivation for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

Waves Class 11 Notes: CBSE Physics Chapter 14
