Rotational motion

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Question

One end of a straight uniform 1 m long bar is pivoted on horizontal table. It is released from rest when it makes an angle $30^{0}$ from the horizontal (see figure) . Its angular speed when it hits the table is given as $\sqrt{n} s^{- 1}$ , where n is an integer. The value of n is

Moderate

Solution

According to conservation of energy
$\begin{array}{l} m g (\frac{l}{2}) \sin θ = \frac{1}{2} I_{p} ω^{2} h e r e I_{P} = \frac{m l^{2}}{3} \\ \Rightarrow ω = \sqrt{\frac{3 g}{2 l}} = \sqrt{\frac{3 \times 10}{2 \times 1}} = \sqrt{15} r a d / s \end{array}$

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Similar Questions

The condition of equilibrium is satisfied when resultant of all forces acting on body vanishes and also no net torque acts on the body. However body can be different types of equilibrium as stable, unstable, neutral or may be having pure rotation when no resultant force acts on the body. Different conditions of equilibrium or rotational motion can be known by different types of energy it possesses. Column I mentions the state of body and column II mentions the type of energy it possesses. Match the correct options in two columns

COLUMN_I COLUMN_II

A) Body in stable equilibrium P) Rotational kinetic energy

B) Body in unstable equilibrium Q) Minimum potential

energy

C) Body in neutral equilibrium R)Maximum potential energy

D) Body in pure rotational motion S) Constant potential energy

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