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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 60o with vertical? [g is the acceleration due to gravity]


A

The angular acceleration of the rod will be 2gL

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B

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

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C

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

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D

The angular speed of the rod will be 3g2L

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Solution

The correct option is D

The angular speed of the rod will be 3g2L


When the rod has rotated by 60o, the centre of mass gets shifted vertically by l/4
Using conservation of energy,
ΔK+ΔU=0
12I0ω2=ΔU
12mL23ω2=(mgL4)
ω=3g2L
Hence option D is correct.
aradial=ω2L2=3g2LL2=3g4Hence option C is correctSince, τ=I0α (About hinge)
α=mgL2sin60omL23=33g4Lav=(αL2)sin60o+ω2L2cos600 where av is the net vertical accelration .
av=33g832+3g8av=9g16+6g16=15mg16
Let N be the normal reaction at the point of contact of rod.
Net vertical forces = mav
mgN=mavN=mg16
Hence option B is also correct.


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