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

Step 1: Understand the Given Information

The problem involves a rotating mass \( M \) attached to a string of length \( L \), making \( \frac{3}{\pi} \) rotations per second. The question asks for the tension \( T \) in the string. The mass describes a circular path with a radius \( R \), and the string makes an angle \( \theta \) with the vertical axis.

Step 2: Angular Velocity

The number of rotations per second (frequency) is given as \( \frac{3}{\pi} \) rotations per second. The angular velocity \( \omega \) is related to the frequency by:

\( \omega = 2\pi \times \text{frequency} = 2\pi \times \frac{3}{\pi} = 6 \, \text{rad/s} \)

Step 3: Centripetal Force

The mass \( M \) is undergoing circular motion with a radius \( R \). The centripetal force required to keep the mass in its circular path is given by:

\( F_{\text{centripetal}} = M \omega^2 R \)

Substituting the value of \( \omega \) (which is 6 rad/s), we get:

\( F_{\text{centripetal}} = M \times 6^2 \times R = 36 M R \)

Step 4: Tension and Vertical Component

The tension \( T \) in the string has both vertical and horizontal components. The vertical component of the tension balances the gravitational force acting on the mass:

\( T \cos \theta = Mg \)

The horizontal component provides the centripetal force:

\( T \sin \theta = M \omega^2 R = 36 M R \)

Step 5: Finding Tension \( T \)

Using the relationship between the vertical and horizontal components of tension:

\( \frac{T \sin \theta}{T \cos \theta} = \frac{36 M R}{Mg} = \frac{36 R}{g} \)

This simplifies to:

\( \tan \theta = \frac{36 R}{g} \)

Step 6: Final Expression for Tension

Finally, the tension \( T \) in the string can be expressed as:

\( T = 36 M L \)

Conclusion

The tension in the string is \( \mathbf{36 M L} \).