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Question: A uniform stick of length L is pivoted at one end on a horizontal table. The stick is held forming a...

A uniform stick of length L is pivoted at one end on a horizontal table. The stick is held forming an angle θ0\theta_0 with the table. A small block of mass mm is placed at the other end of the stick and it remains at rest. The system is released from rest.

A

The stick will hit the table before the block if cosθ0≥23cos\theta_0 \ge \sqrt{\frac{2}{3}}.

B

The contact force between the block and the stick immediately before the system is released if θ0=cos−1(23)\theta_0 = cos^{-1}(\sqrt{\frac{2}{3}}) is mgmg.

C

The contact force between the block and the stick immediately before the system is released if θ0=cos−1(23)\theta_0 = cos^{-1}(\sqrt{\frac{2}{3}}) is 00.

D

The contact force between the block and the stick immediately before the system is released if θ0=cos−1(23)\theta_0 = cos^{-1}(\sqrt{\frac{2}{3}}) is 23mg\sqrt{\frac{2}{3}}mg.

Answer

The stick will hit the table before the block if cosθ0≥23cos\theta_0 \ge \sqrt{\frac{2}{3}}. The contact force between the block and the stick immediately before the system is released (when θ0=cos−1(23)\theta_0 = cos^{-1}(\sqrt{\frac{2}{3}})) is 0.

Explanation

Solution

Here's the breakdown:

  1. Forces and Acceleration: When the stick is released, it rotates about the pivot. For the block to remain at rest relative to the stick, the stick must exert a force on the block that cancels the block's weight. Applying Newton's law in the vertical direction gives:

    mLα⋅cosθ0=mg  ⟹  α=gL⋅cosθ0mL\alpha \cdot cos\theta_0 = mg \implies \alpha = \frac{g}{L \cdot cos\theta_0}

    where α\alpha is the angular acceleration.

  2. Torque Analysis: Analyzing the torque on the stick (considering its distributed mass and the reaction force from the block) leads to a consistency condition:

    cosθ0=23cos\theta_0 = \sqrt{\frac{2}{3}}

    This condition determines when the contact force between the block and the stick vanishes.

  3. Critical Angle: If cosθ0≥23cos\theta_0 \ge \sqrt{\frac{2}{3}}, the stick rotates fast enough to hit the table before the block loses contact. When θ0=cos−1(23)\theta_0 = cos^{-1}(\sqrt{\frac{2}{3}}), the contact force needed to keep the block at rest is zero.