Question

A particle performs uniform circular motion with an angular momentum L. If the frequency of a particle's motion is doubled and its kinetic energy is halved, the angular momentum becomes:

• A. 2L
• B. 4L
• C. L/2
• D. L/4

Answer 1

Answer: (D)

L/4

Answer 2

Kinetic energy $E = \frac { 1 } { 2 } L \omega = \frac { 1 } { 2 } L \times 2 \pi n$

⇒$E \propto L \times n$

⇒ $\frac { L _ { 2 } } { L _ { 1 } } = \frac { E _ { 2 } } { E _ { 1 } } \times \frac { n _ { 1 } } { n _ { 2 } }$⇒$\frac { L _ { 2 } } { L _ { 1 } }$ $= \left[ \frac { E _ { 1 } / 2 } { E _ { 1 } } \right] \times \left[ \frac { n _ { 1 } } { 2 n _ { 1 } } \right]$ ⇒$L _ { 2 } = \frac { L _ { 1 } } { 4 } = \frac { L } { 4 }$