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]
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=3√3g4L⇒av=(αL2)sin60o+ω2L2cos600 where av is the net vertical accelration .
⇒av=3√3g8√32+3g8⇒av=9g16+6g16=15mg16
Let N be the normal reaction at the point of contact of rod.
Net vertical forces = mav
⇒mg−N=mav⇒N=mg16
Hence option B is also correct.