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

The system comprises a mass attached to a string, executing circular motion. The centripetal force formula is applicable to determine the string's tension.The tension generates the necessary centripetal force for circular motion, calculated as:\[F_c = M \cdot \omega^2 \cdot R,\]where:- \( M \) denotes the mass,- \( \omega \) represents the angular velocity,- \( R \) is the radius of the circular path.Angular velocity \( \omega \) relates to rotations per second \( n \) (given as \( \frac{3}{\pi} \)) by:\[\omega = 2\pi n.\]Substituting \( n \):\[\omega = 2\pi \cdot \frac{3}{\pi} = 6.\]Consequently, the string tension is:\[T = M \cdot \omega^2 \cdot L = M \cdot 6^2 \cdot L = M \cdot L^2.\] Final Answer: \( M \cdot L^2 \).