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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
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Verified Answer
The correct answer is:
increase by a factor of 4
From the law of conservation of angular momentum
So,
$$
\begin{aligned}
m v r & =m v^{\prime} \frac{r}{2} \\
y^{\prime} & =2 y \\
\frac{k_1}{k_2} & =\frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2} \\
\frac{k_1}{k_2} & =\frac{y^2}{v^{\prime 2}} \\
& =\frac{y^2}{(2 v)^2} \\
\frac{k_1}{k_2} & =\frac{1}{4} \\
k_1 & =\frac{1}{4} k_2 \\
k_2 & =4 k_1
\end{aligned}
$$
So,
$$
\begin{aligned}
m v r & =m v^{\prime} \frac{r}{2} \\
y^{\prime} & =2 y \\
\frac{k_1}{k_2} & =\frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2} \\
\frac{k_1}{k_2} & =\frac{y^2}{v^{\prime 2}} \\
& =\frac{y^2}{(2 v)^2} \\
\frac{k_1}{k_2} & =\frac{1}{4} \\
k_1 & =\frac{1}{4} k_2 \\
k_2 & =4 k_1
\end{aligned}
$$
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