A small mass attached to a string rotates on a frictionless table top as shown. If the tension on 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 [image:7_image8.emf]

Question

Dr. Abubacker · Accepted Answer

(a) $K.E.\  = \frac{L^{2}}{2I}$
$\because$ From angular momentum conservation about centre
$L \rightarrow$ constant
$I = mr^{2}$
$K.E.’ = \frac{L^{2}}{2(mr^{'2})}\ \ r^{'} = \frac{r}{2}$
K.E’. $= 4$ K.E.
K.E. is increased by a factor of 4

Q. A small mass attached to a string rotates on a frictionless table top as shown. If the tension on 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

Explanation

Reviewed by Dr. Abubacker

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