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Chapter 5: Problem 546

A particle performing uniform circular motion has angular momentum \(L\)., its angular frequency is doubled and its \(K . E\). halved, then the new angular momentum is \(\\{\mathrm{A}\\} 1 / 2\) \\{B \(\\} 1 / 4\) \(\\{\mathrm{C}\\} 2 \mathrm{~L}\) \(\\{\mathrm{D}\\} 4 \mathrm{~L}\)

Short Answer

Expert verified
The new angular momentum is \(L' = \frac{1}{2} L\), which corresponds to option A.

Step by step solution

01

Define given variables

Initially, Angular momentum L Angular frequency ω Kinetic energy K.E After the changes, New angular frequency = 2ω New kinetic energy = (1/2) K.E
02

Introduce the formulas for angular momentum and kinetic energy

The angular momentum formula is: \(L = Iω\), where I is the moment of inertia and ω is the angular frequency. The kinetic energy formula for rotational motion is: \(K.E = \frac{1}{2} Iω^2\)
03

Express the moment of inertia in terms of angular momentum and angular frequency

Using the angular momentum formula, we can express moment of inertia as: \(I = \frac{L}{ω}\)
04

Express the initial kinetic energy in terms of the moment of inertia and angular frequency

Using the moment of inertia expression from step 3 and the kinetic energy formula: \(K.E = \frac{1}{2}(\frac{L}{ω})ω^2 = \frac{1}{2}Lω\)
05

Calculate the new angular momentum

The new kinetic energy is half of the initial, so: \(\frac{1}{2}K.E = \frac{1}{2}(\frac{1}{2}Lω) = \frac{1}{4}Lω\) Since the new angular frequency is 2ω: \(\frac{1}{2}K.E = \frac{1}{4}L(2ω) = \frac{1}{2}L(ω)\) New angular momentum \(L' = \frac{1}{2} L\) The new angular momentum is \(L' = \frac{1}{2} L\), which corresponds to option A.

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Most popular questions from this chapter

A particle of mass \(\mathrm{m}\) slides down on inclined plane and reaches the bottom with linear velocity \(\mathrm{V}\). If the same mass is in the form of ring and rolls without slipping down the same inclined plane. Its velocity will be \(\\{\mathrm{A}\\} \mathrm{V}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~V})}\) \(\\{\mathrm{C}\\}(\mathrm{V} / \sqrt{2})\) \(\\{\mathrm{D}\\} 2 \mathrm{~V}\)

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