JEEAdvanced
A thin uniform rod of mass M and length L is hinged at one end about…
Question
A thin uniform rod of mass M and length L is hinged at one end about a fixed frictionless horizontal axis and held horizontal, then released from rest. At the very instant of release, the hinge exerts a vertical reaction force of magnitude N on the rod. Express N as a percentage of the rod's weight Mg and give that percentage (i.e. report 100*N/(Mg)).
✓ Verified answer: 25checked by our engine — not a guess
Step-by-step solution
Torque about the hinge at release = Mg(L/2).
Moment of inertia about the end = M L^2/3.
So angular acceleration alpha = [Mg L/2] / [M L^2/3] = 3g/(2L).
The centre of mass is at L/2, so its downward linear acceleration is a_cm = alpha (L/2) = 3g/4.
Newton's second law vertically: Mg - N = M a_cm = M(3g/4), giving N = Mg - 3Mg/4 = Mg/4.
As a percentage, 100*N/(Mg) = 25.
Final answer25
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Physics Mechanics solutions
A wedge of mass M = 4 kg with a smooth inclined face making 45…JEE · PhysicsA block of mass 2 kg moving at 10 m/s slides onto the left end of a…JEE · PhysicsA particle is projected from the foot of a fixed inclined plane whose…JEE · PhysicsA small particle is placed at rest on the very top of a fixed smooth…JEE · PhysicsOn a frictionless horizontal line, a ball A of mass m moves with…JEE · PhysicsA solid uniform sphere (moment of inertia 2/5 M R^2 about its centre)…JEE · PhysicsA uniform flexible chain of total length 2 m lies on a frictionless…JEE · PhysicsA car travels on a curved road of radius 50 m that is banked at an…JEE · Physics