A particle performing uniform circular motion has angular momentum L. What will be the new angular momentum, if its angular frequency is doubled and its kinetic energy is halved?

The angular momentum (L) of a particle performing uniform circular motion is given by the formula:

L = mvr

where m is the mass of the particle, v is the velocity, and r is the radius of the circular path.

The kinetic energy (K) of the same particle is given by the formula:

K = 1/2 mv^2

Given that the angular frequency is doubled and the kinetic energy is halved, we need to find the new angular momentum.

First, let's express the velocity v in terms of the angular frequency ω and the radius r:

v = ωr

Substitute this into the formula for the angular momentum:

L = m(ωr)r = mωr^2

If the angular frequency ω is doubled, the new angular momentum L' is:

L' = m(2ω)r^2 = 2mωr^2 = 2L

Next, let's consider the change in kinetic energy. If the kinetic energy is halved, the new kinetic energy K' is:

K' = 1/2 K = 1/4 mv^2

Substitute the expression for the velocity v = ωr into this formula:

K' = 1/4 m(ωr)^2 = 1/4 mω^2r^2

Since the kinetic energy is proportional to the square of the velocity (or the square of the angular frequency), halving the kinetic energy implies that the new angular frequency ω' is:

ω' = sqrt(2)ω

Substitute this into the formula for the new angular momentum L' = 2mωr^2:

L' = 2m(sqrt(2)ω)r^2 = 2sqrt(2)mωr^2

So, the new angular momentum is 2sqrt(2) times the original angular momentum.