Question

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

• A. decrease by a factor of 2
• B. remain constant
• C. increase by a factor of 2
• D. increase by a factor of 4

Answer 1

The Correct Option is D

To solve this problem, we need to consider the physics of rotational motion and the relationship between the tension, radius, and kinetic energy of the rotating mass.

First, let's understand the mechanics involved:

• The kinetic energy K of an object in circular motion is given by the formula: K = \frac{1}{2} m v^2, where m is the mass and v is the tangential velocity.
• In such a scenario, the centripetal force required to keep the object moving in a circular path is provided by the tension in the string. This force is given by: F = \frac{m v^2}{r}, where r is the radius of the circular path.
• If the radius of rotation decreases, but the angular momentum remains constant (as no external torque acts), the velocity must adjust accordingly. The angular momentum L is given by: L = m v r.

According to the principle of conservation of angular momentum:

m v_1 r_1 = m v_2 r_2

Here, when the radius r is reduced to half, r_2 = \frac{r_1}{2}, we have:

v_2 = \frac{v_1 r_1}{r_2} = \frac{v_1 r_1}{\frac{r_1}{2}} = 2v_1

This means that the new tangential velocity becomes twice the original velocity.

Substituting this into the kinetic energy formula gives us:

K_2 = \frac{1}{2} m (2v_1)^2

K_2 = 2^2 \times \frac{1}{2} m v_1^2 = 4 K_1

This results in the kinetic energy increasing by a factor of 4.

Thus, the correct answer is: increase by a factor of 4.