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A uniform rigid rod of length 2 m is pivoted at one end about a…
Question
A uniform rigid rod of length 2 m is pivoted at one end about a horizontal frictionless axis and held horizontally. It is released from rest. At the instant of release, find the magnitude of the linear acceleration (in m/s^2) of the free end of the rod. Take g = 10 m/s^2 and moment of inertia of a rod about one end as ML^2/3.
✓ Verified answer: 15checked by our engine — not a guess
Step-by-step solution
Gravity acts at the rod's centre, a distance L/2 from the pivot, giving torque tau = M g (L/2).
With I = M L^2/3, angular acceleration alpha = tau/I = [M g (L/2)]/[M L^2/3] = 3g/(2L) = 3*10/(2*2) = 7.5 rad/s^2.
The free end is at distance L from the pivot, so its linear (tangential) acceleration = alpha*L = 7.5*2 = 15 m/s^2.
Final answer15
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