A small mass attached to a string rotates on frictionless table top as shown. If the tension is the string…
A small mass attached to a string rotates on frictionless table top as shown. If the tension is 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
increase by a factor of 2
increase by a factor of 4
decrease by a factor of 2
Solution
From the law of conservation of angular momentum
So,
$\begin{aligned}
m v r & =m v^{\prime} \frac{r}{2} \\
v^{\prime} & =2 v \\
\frac{k_1}{k_2} & =\frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2} \\
\frac{k_1}{k_2} & =\frac{v^2}{v^{\prime 2}} \\
& =\frac{v^2}{(2 v)^2} \\
\frac{k_1}{k_2} & =\frac{1}{4} \\
k_1 & =\frac{1}{4} k_2 \\
k_2 & =4 k_1
\end{aligned}$
