Question

A small mass attached to a string, rotates on a frictionless table top as shown. It 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
A. Increase by factor of 4
B. Increase by factor 8
C. Remain constant
D. Increase by a factor 2

Answer 1

Since a small mass attached to a string rotates on a frictionless table top. As we know that the tension in the string increases by pulling the string, causing the radius of the circular motion to decrease by a factor of 2 then we use a formula of kinetic energy.

Complete answer:
As we know that a small mass attached to a string rotates on a frictionless table. So, there will be a kinetic energy and angular momentum will be produced. So, the kinetic energy in terms of angular momentum is given by
$k.E.=\dfrac{{{I}^{2}}}{2I}$
And $2k.E.=\dfrac{{{L}^{2}}}{2I}\text{ }\left( 1 \right)$
From the angular momentum conservation about centre L is constant.
$\therefore L=\text{constant}$
And I is the moment of inertia
$\therefore I=m{{r}^{2}}$
From a rod
Now,
$kE'=\dfrac{{{L}^{2}}}{2{{\left( m\dfrac{r}{2} \right)}^{2}}}=\dfrac{{{L}^{2}}}{2\times \dfrac{m{{r}^{2}}}{4}}=\dfrac{2L}{m{{r}^{2}}}$
$=\dfrac{2{{L}^{2}}}{m{{r}^{2}}}=2.\dfrac{{{L}^{2}}}{I}$
Using equation 1 we get

& K.E’=2 \times 2KE \\\ & \implies KE'=4KE \\\ \end{aligned}$$ KE is increased by a factors of 4 **So, the correct answer is “Option A”.** **Note:** Here, Be careful while calculating the $$K.E. = \dfrac{{{L}^{2}}}{2I}$$ and where $$I=m{{r}^{2}}$$ For a circular disc and for a rod $$I={{\left( m\dfrac{r}{2} \right)}^{2}}=\dfrac{m{{r}^{2}}}{4}$$ Put the value at the exact time where it is needed to solve the problem for kinetic energy.