A thin and uniform rod of mass and length is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact point with the floor without slipping. Which of the following statement(s) is/are cor

Question

A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle $60 °$ with vertical [2019]

[ $g$ is the acceleration due to gravity]

Smart Achievers · Accepted Answer

(1, 2, 3)

The rod is released from rest so that it falls by rotating about its contact point with the floor without slipping.

$Gain in kinetic energy = loss in potential energy$

$\frac{1}{2} I ω^{2} = m g \frac{l}{2} (1 - \cos 60 °)$

$∴$ $\frac{m l^{2}}{3} ω^{2} = m g \frac{l}{2} \Rightarrow ω = \sqrt{\frac{3 g}{2 l}}$

$Now, τ = I α$

$∴$ $m g \times \frac{l}{2} \sin 60 ° = \frac{1}{3} m l^{2} α \Rightarrow α = \frac{3 \sqrt{3} g}{4 l}$

$Further, a_{t} = \frac{l}{2} α = \frac{3 \sqrt{3} g}{8}$

$Also a_{r} = ω^{2} \frac{l}{2} = \frac{3 g}{2 l} \times \frac{l}{2} = \frac{3 g}{4}$

For vertical motion of centre of mass

$m g - N = m (a_{r} \cos 60 ° + a_{t} \cos 30 °)$

$∴$ $m g - N = m [\frac{3 g}{4} \times \frac{1}{2} + \frac{3 \sqrt{3} g}{8} \times \frac{\sqrt{3}}{2}]$

$∴$ $N = \frac{M g}{16}$

Solution

1 The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$

2 The radial acceleration of the rod's center of mass will be $\frac{3 g}{4}$

3 The normal reaction force from the floor on the rod will be $\frac{M g}{16}$

4 The angular acceleration of the rod will be $\frac{2 g}{L}$

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Solution

1 The angular speed of the rod will be 3g2L

2 The radial acceleration of the rod's center of mass will be 3g4

3 The normal reaction force from the floor on the rod will be Mg16

4 The angular acceleration of the rod will be 2gL

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1 The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$

2 The radial acceleration of the rod's center of mass will be $\frac{3 g}{4}$

3 The normal reaction force from the floor on the rod will be $\frac{M g}{16}$

4 The angular acceleration of the rod will be $\frac{2 g}{L}$