Question

A string of length \( L \) is fixed at one end and carries a mass of \( M \) at the other end. The mass makes \( \frac{3}{\pi} \) rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is \_\_\_\_ ML.

Hint:

For rotational motion, the tension in the string acting on a rotating mass is related to the square of the angular velocity and the length of the string. The formula \( T = M \cdot \omega^2 \cdot L \) helps relate the tension to the rotational speed.

Answer 1

In this problem, the system consists of a rotating mass at the end of a string. The mass is undergoing circular motion, so we can apply the formula for centripetal force to find the tension in the string. The tension in the string provides the centripetal force required for the circular motion. The centripetal force \( F_c \) is given by: \[ F_c = M \cdot \omega^2 \cdot R, \] where: - \( M \) is the mass, - \( \omega \) is the angular velocity, - \( R \) is the radius of the circular path. The angular velocity \( \omega \) can be related to the number of rotations per second \( n \) (which is \( \frac{3}{\pi} \) rotations per second) by: \[ \omega = 2\pi n. \] Substituting the given value for \( n \), we get: \[ \omega = 2\pi \cdot \frac{3}{\pi} = 6. \] Thus, the tension in the string is: \[ T = M \cdot \omega^2 \cdot L = M \cdot 6^2 \cdot L = M \cdot L^2. \] Final Answer: \( M \cdot L^2 \).