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Question: A bar length l carrying a small mass m at one of its ends rotates with a uniform angular speed ω in ...

A bar length l carrying a small mass m at one of its ends rotates with a uniform angular speed ω in a vertical plane about the midpoint of the bar. During the rotation, at some instant of time when the bar is horizontal, the mass is detached from the bar but the bar continues to rotate with the same ω. The mass moves vertically up, comes back and reaches the bar at the same point. At that place, the acceleration due to gravity is g.
A) This is possible if the quantity ω2l2πg\dfrac{{{\omega ^2}l}}{{2\pi g}}is an integer
B) The total time of flight of the mass is proportional to ω2{\omega ^2}
C) The total distance travelled by the mass in air is proportional to ω2{\omega ^2}
D) The total distance travelled by the mass in air and its total time of flight are both independent on its mass

Explanation

Solution

The combination of translational motion and the rotational motion of a rigid body is called rolling motion. According to the law of conservation of angular momentum, when there is no external couple acting, total angular momentum of a rigid body or system of particles is considered to be conserved.

Complete step by step answer:
We know that in rotatory motion, the rotational kinetic energy (K.E.) of a bod
is expressed as:
K.E=12Iω2K.E = \dfrac{1}{2}I{\omega ^2}
Length of bar = l (Given)
Mass = m (Given)
Angular speed = ω\omega (Given)
Acceleration due to gravity = g (Given)
v=12ωl T=2vg=ωlg or n×2πω=ωlg  v = \dfrac{1}{2}\omega l \\\ T = \dfrac{{2v}}{g} = \dfrac{{\omega l}}{g} \\\ or \\\ n \times \dfrac{{2\pi }}{\omega } = \dfrac{{\omega l}}{g} \\\
As we know, the bar of length l completes n rotations within the time period T.
∴n=lω22πg Distan⁡ce travelled=2h=2v22g=l2ω24g  \therefore n = \dfrac{{l{\omega ^2}}}{{2\pi g}} \\\ Dis\tan ce{\text{ }}travelled = 2h = \dfrac{{2{v^2}}}{{2g}} = \dfrac{{{l^2}{\omega ^2}}}{{4g}} \\\
Hence, the present question has multiple answers. The statements given in the Options A, C and D are correct.

Note: If the moment of inertia of body changes from I 1 to I 2 , owing to the change in distribution of the mass of body, then the angular momentum of body changes from ω1{\omega _1}to ω2{\omega_2}such that:
I1ω1=I2ω2{I_1}{\omega _1} = {I_2}{\omega _2}