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JEE Advanced · Physics · 6. Work Power Energy

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]

  1. A The radial acceleration of the rod's center of mass will be 3g4
  2. B The angular acceleration of the rod will be 2gL
  3. C The angular speed of the rod will be 3g2L
  4. D The normal reaction force from the floor on the rod will be Mg16
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation


Using conservation of energy
K+U=0
\(\frac{1}{2} I_0 \omega^2=-\Delta U\left(I_0=\right.\) moment of inertia about Hinge \()\)
12ml23ω2=--mgl4
ω=3g2lC is correct
aradial=ω2.l2=3g2ll2=3g4A is correct
Now, τ=I.α
mg.l2sin60=ml23.αα=33g4l
B is incorrect
Acceleration in vertical direction av=αl2sin60o+ω2l2cos60o
=av=33g832+3g8
av=9g16+6g16=15 g16
Now, using NLM \(\Rightarrow m g-N=m a_v \Rightarrow N=m g\)\(-m a_v=m g-\frac{15}{16} m g\)
N=mg16D is correct
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