NLM O30

1. Force Analysis on a Massless Rod

The rod is described as “light,” meaning it is massless ($I_{rod} = 0$). The only mass in the rotating system is the bead. Let’s consider the torque equation about the pivot O. Since the rod is massless, the net torque required to rotate the rod itself is zero. The torque is generated solely by the bead’s weight and the rod transmits this interaction.

However, a more rigorous approach is to check the normal reaction force $N$ between the rod and the bead. The equation of motion for the bead in the tangential direction (perpendicular to the rod) involves gravity and the normal force.

$$ mg \cos\theta + N = m(r \alpha) $$

Simultaneously, the angular acceleration $\alpha$ is determined by the torque $\tau$ exerted by the bead on the pivot system.

$$ \tau = I \alpha \implies (mg \cos\theta) r = (mr^2) \alpha $$ $$ \alpha = \frac{g \cos\theta}{r} $$

Substituting this $\alpha$ back into the force equation:

$$ mg \cos\theta + N = m r \left( \frac{g \cos\theta}{r} \right) $$ $$ mg \cos\theta + N = mg \cos\theta \implies N = 0 $$

2. Implication of Zero Normal Force

Since the Normal Reaction $N$ is zero, the frictional force $f$, which is defined as $f \le \mu N$, must also be zero regardless of the coefficient of friction $\mu$.

Conclusion: The bead experiences no forces from the rod (neither perpendicular support nor radial friction). It is effectively in free fall, subject only to gravity.

3. Kinematics of the Bead

Since the bead falls freely starting from rest at a horizontal distance $l$ from the pivot:

• Horizontal position: $x(t) = l$ (Constant)
• Vertical position: $y(t) = \frac{1}{2}gt^2$ (Downwards)

The rod simply rotates to track the bead’s position. The angle $\theta$ made with the horizontal is given by simple trigonometry in the triangle formed by the pivot, the initial position, and the current bead position:

$$ \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{y(t)}{x(t)} $$ $$ \tan\theta = \frac{\frac{1}{2}gt^2}{l} = \frac{gt^2}{2l} $$ $$ \theta = \tan^{-1}\left( \frac{gt^2}{2l} \right) $$

4. Analyzing the Options

Since $N=0 \implies f=0$, the motion is independent of $\mu$. The bead behaves as if $\mu$ is negligible in all cases.

• (a) If $\mu$ is negligible: Formula holds.
• (b) If $\mu$ is finite: Since $N=0$, friction is still 0, so the formula still holds.