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Question:

A small mass attached to a string rotates on a frictionless table top as shown. If the tension of the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will:

Remain constant.

Decrease by a factor of 2.

Increase by a factor of 2.

Increase by a factor of 4.

Solution:

K.E.=L²/2I
From angular momentum conservation about centre.
L→constant
I=mr²
K.E.'=L²/2(mr'²)
r'=r/2
Therefore, K.E.' = L²/2(m(r/2)²)=L²/2(mr²/4) = 4(L²/2mr²) = 4K.E.