The compound pendulum shown below consists of a 50-kg uniform disk rigidly attac
ID: 1850978 • Letter: T
Question
The compound pendulum shown below consists of a 50-kg uniform disk rigidly attached to a massless rod. The pendulum rotates freely in a vertical plane around a horizontal axis at point O. The diameter of the disk is 0.8 m and the center of the disk is 0.8 m from O. At the instant shown the pendulum is rotating counterclockwise around O at angular velocity ? = rad/s CCW. At the instant shown, find the angular acceleration of the pendulum and the magnitude of the reaction force at O.
The answers are ?=9.4397 rad/s^2 counterclockwise for the angular acceleration and 607.1 N for the reaction force at O. Thanks in advance!
http://i50.tinypic.com/2hqxo1w.jpg
Explanation / Answer
Torque about O, T = mg*0.8Sin60 = 50*9.81 *0.8*Sin60 = 340 N-m
MOI about O, I = MOI about G + m*0.8^2 = 1/2*mr^2 + m*0.8^2 = 1/2*50*(0.8/2)^2 + 50*0.8^2 = 36 kg-m^2
= T/I = 340/36 = 9.44 rad/s^2
Centrifugal force = mL^2 = 50*0.8*3^2 = 360 N
Tangential force = m(L) = 50*0.8*9.44 = 377.6 N
Component of weight mg in centrifugal direction = mgCos60 = 50*9.81*Cos60 = 245.25 N
Component of weight mg in tangential direction = mgSin60 = 50*9.81*Sin60 = 424.8 N
Total force in centrifugal direction = 360 + 245.25 = 605.25 N
Total force in tangential direciton = 377.6 - 424.8 = - 47.2 N
Net reaction force = ((-47.2)^2 + 605.25^2) = 607.1 N
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