Question

A thin, uniform metal rod of mass \( M \) and length \( L \) is swinging about a horizontal axis passing through its end. Its maximum angular velocity is \( \omega \). Its centre of mass rises to a maximum height of \( \frac{L^2 \omega^2}{6g} \). (where \( g \) is the acceleration due to gravity)

• A. \( \frac{L^2 \omega^2}{6g} \)
• B. \( \frac{L^2 \omega^2}{g} \)
• C. \( \frac{L^2 \omega^2}{2g} \)
• D. \( \frac{L^2 \omega^2}{3g} \)

Hint:

In rotational motion problems, energy conservation is often the most effective method for solving for quantities like height or velocity. Always check for the total energy conversion between kinetic and potential forms.

Answer 1

The Correct Option is A

Step 1: Understanding the problem.
We are dealing with the motion of a thin metal rod swinging about a fixed point. The maximum height attained by the center of mass is related to the maximum angular velocity \( \omega \). This is a problem involving the principles of rotational motion and energy conservation.
Step 2: Energy Conservation Approach.
At the lowest point, the rod has rotational kinetic energy due to its angular velocity \( \omega \). At the highest point, all of this kinetic energy has been converted into potential energy as the center of mass rises by a height \( h \). Thus, applying conservation of energy: \[ \frac{1}{2} I \omega^2 = Mgh \] Where \( I \) is the moment of inertia of the rod about the pivot point. For a rod rotating about one end, the moment of inertia is given by: \[ I = \frac{1}{3} ML^2 \] Substituting into the energy equation: \[ \frac{1}{2} \times \frac{1}{3} ML^2 \omega^2 = Mgh \] Step 3: Solving for the height \( h \).
Simplifying the equation: \[ \frac{1}{6} ML^2 \omega^2 = Mgh \] Dividing both sides by \( Mg \): \[ h = \frac{L^2 \omega^2}{6g} \] Step 4: Conclusion.
Thus, the maximum height attained by the center of mass is \( \frac{L^2 \omega^2}{6g} \), which corresponds to option (A).