Question

A particle performs uniform circular motion with angular momentum 'L'. If the frequency of particles motion is halved and its KE is doubled then the angular momentum becomes –

• A. $\frac{L}{4}$
• B. $4L$
• C. $2L$
• D. $L/2$

Answer 1

Answer: (B)

$4L$

Answer 2

K.E. = $\frac{1}{2}$Iw² = $\frac{1}{2}$ (Iw)(w)

or K.E. = $\frac{1}{2}$Lw

or L = $\frac{2K.E.}{\omega}$

Now L' = $\frac{2(2K.E.)}{(\omega/2)}$= 4 L