Question

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

A.Remain constant.
B.Increase by a factor of 2.
C.Increase by a factor of 4.
D.Decrease by a factor of 2.

Answer 1

Hint: We will find out the initial kinetic energy and then the angular momentum (both initial and final). Then we will equate both the angular momentums in order to find out the value of the final velocity. At last, we will divide the initial kinetic energy by the final kinetic energy in order to get our answer.

Formula used: $K.E = \dfrac{1}{2}m{v^2}$, $\vec L = \vec r \times m\vec v$
The angular velocity of the mass given in the question will be $\omega r$.
The length of the string/the radius of the circle is r.

Complete step-by-step answer:
As we know the formula for kinetic energy in terms of mass and velocity is $K.E = \dfrac{1}{2}m{v^2}$.
Putting the given value of the velocity in the above formula of kinetic energy, we get the initial kinetic energy as-
$\Rightarrow K.{E_i} = \dfrac{1}{2}m{v^2} \\\ \\\ \Rightarrow K.{E_i} = \dfrac{1}{2}m{\omega ^2}{v^2} \\\$
When we are increasing the tension in the string by pulling, the radius is decreased. The direction of the tension will be along the thread towards the centre. The torque of the point O in the figure will be zero as the formula for torque is $\vec \tau = \vec r \times \vec F$ which means the difference between the point of application and the axis. The axis is vertical in nature and passes through point O through which the line of action of tension was passing hence the value of r will be zero making the value of torque as zero.
Since the value of torque is zero, the angular momentum is conserved. This means that the angular momentum before the decreasing of the radius will be equal to the angular momentum after the decreasing of the radius.
If we equate the magnitude of the initial and the final angular momentum, we get-
Initial angular momentum- $mvr$.
$\Rightarrow {{\vec L}_i} = \vec r \times m\vec v \\\ \\\ \Rightarrow {L_i} = \sin 90mvr \\\$
As we can see from the figure that the angle between the velocity and the radius is 90 degrees.
$\Rightarrow {L_i} = mvr$
Let the changed velocity in the final angular momentum be v’.
The radius was decreased by half. Hence, $r = \dfrac{r}{2}$.
From above we get the final angular momentum-
$\Rightarrow {L_f} = m \times v'\dfrac{r}{2}$
Equating both the momentums we get-
$\Rightarrow mvr = mv'\dfrac{r}{2} \\\ \\\ \Rightarrow v = \dfrac{{v'}}{2} \\\ \\\ \Rightarrow v' = 2v \\\$
The final kinetic energy by putting these values will be-
$\Rightarrow K.{E_f} = \dfrac{1}{2}mv{'^2} \\\ \\\ \Rightarrow K.{E_f} = \dfrac{1}{2}m{\left( {2v} \right)^2} \\\ \\\ \Rightarrow K.{E_f} = \dfrac{1}{2}m4{v^2} \\\$
Dividing initial kinetic energy by the final kinetic energy, we get-
$\Rightarrow \dfrac{{K.{E_i}}}{{K.{E_f}}} = \dfrac{{\dfrac{1}{2}m{v^2}}}{{\dfrac{1}{2}m4{v^2}}} \\\ \\\ \Rightarrow \dfrac{{K.{E_i}}}{{K.{E_f}}} = \dfrac{1}{4} \\\ \\\ \Rightarrow K.{E_f} = 4K.{E_i} \\\$

Hence, option C is the correct option.

Note: Kinetic energy, energy in which the action of an entity or particle has. When research that passes energy to an entity is performed using a net power, the body accelerates and absorbs cinematic strength. Kinetic energy is an attribute of a moving body or a molecule that not only relies on acceleration but also time. The movement of the force may include the transfer (or force along the way) of the axis, the acceleration or some mixture of movements.