Two identical, uniform rods are configured as shown above, and the assemble is released at rest (in the position shown). Find, immediately after release:
bc (angular acceleration) (magnitude and direction)
ab (magnitude and direction)
Inertia = I = 1/3 * m*l^2 = 1.8*0.840^2/3 = 0.42 kg m^2
m = mass = 1.8 kg, l = length of the rod = 0.84
Torque = I* alpha
alpha = angular acceleration
Torque = mgx
g = acceleration due to gravity = 9.8 ms^-2
x = perpendicular distance between the line of action and mass
x= l/2 cos (45)= 0.84/2 cos (45) = 0.29 m
Torque = mgx=1.8*9.8*0.29 = 5.11 Nm
alphaBC=angular acceleration of BC = torque/Inertia = 5.11/0.42 = 12.2 rad/s^2 (in the direction perpendicular to the plane of the paper using right hand thumb rule)
aB = linear acceleration = 0 (because the point is not moving)
Two identical, uniform rods are configured as shown above, and the assemble is released at rest (...
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