Question
A thin uniform rod (length = 1.3 m, mass = 4.1 kg) is pivoted about a horizontal frictionless pin through one of its ends. The moment of inertia of the rod through this axis is
1/3 m L2.
The rod is released when it is 55° below the horizontal. What is the angular acceleration of the rod at the instant it is released?
1/3 m L2.
The rod is released when it is 55° below the horizontal. What is the angular acceleration of the rod at the instant it is released?
Answers
angular acceleration
= (Torque)/(Moment of Inertia)
= (M*g*cos55*L/2)/[(1/3)*M*L^2)
= (3/2)*cos55*g/L = 6.5 s^-2
= (Torque)/(Moment of Inertia)
= (M*g*cos55*L/2)/[(1/3)*M*L^2)
= (3/2)*cos55*g/L = 6.5 s^-2