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

decrease by a factor of 2

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remain constant

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increase by a factor of 2

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increase by a factor of 4

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Solution

The correct option is D increase by a factor of 4
From the law of conservation of angular momentum
$m v r = m v^{'} \frac{r}{2}$
$v^{'} = 2 v$
so $\frac{K E}{K E_{2}} = \frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m v^{' 2}} = \frac{1}{4}$

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

The correct option is D increase by a factor of 4From the law of conservation of angular momentummvr=mv′r2v′=2vso KEKE2=12mv212mv′2=14

The correct option is D increase by a factor of 4
From the law of conservation of angular momentum
$m v r = m v^{'} \frac{r}{2}$
$v^{'} = 2 v$
so $\frac{K E}{K E_{2}} = \frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m v^{' 2}} = \frac{1}{4}$