A uniform rod of length L and mass M is pivoted about a horizontal frictionless

Question

A uniform rod of length L and mass M is pivoted about a horizontal frictionless pin through one end. The rod is released from rest in the vertical position and given a slight nudge to the right. At the instant the rod has rotated 180 degrees and is agian vertical find.....
1) its angular speed
2) its angular acceleration
3) the speed of the center of mass of the rod
4) the acceleration of the center of mass of the rod ( both x and y components)
4) the components of the reaction for R on the rod at the pivot.

All answers are in terms of M, L, G

Thanks alot

Explanation / Answer

moment of inertia= ML^2/3 initial energy = MgL 1) its angular speed w final energy = Iw^2/2 = MgL so w = sqrt(2MgL/I) = sqrt(6g/L) 2) its angular acceleration alpha net torque = 0, so alpha = 0 3) the speed of the center of mass of the rod v v = w*(L/2) = sqrt(3gL/2) 4) the acceleration of the center of mass of the rod ( both x and y components) ax = (L/2)*alpha = 0 ay = v^2/(L/2) = 3g 5) the components of the reaction for R on the rod at the pivot. R - Mg = M*ay so R = Mg + M*3g = 4Mg

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