A uniform thin rod of length L and mass M is free to rotate on a frictionless pi

Question

A uniform thin rod of length L and mass M is free to rotate on a frictionless pin passing through one end (see figure). The rod is released from rest in the horizontal position.

(a) What is the speed of its center of gravity when the rod reaches its lowest position? (Use any variable or symbol stated above along with the following as necessary: g.)

vcg =

(b) What is the tangential speed of the lowest point on the rod when it is in the vertical position? (Use the following as necessary: L, M, g.)

vlower end =

A uniform thin rod of length L and mass M is free to rotate on a frictionless pin passing through one end (see figure). The rod is released from rest in the horizontal position. (a) What is the speed of its center of gravity when the rod reaches its lowest position? (Use any variable or symbol stated above along with the following as necessary: g.) vcg = (b) What is the tangential speed of the lowest point on the rod when it is in the vertical position? (Use the following as necessary: L, M, g.) vlower end =

Explanation / Answer

The angular acceleration is given by

T = I*a, where T = torque and I = moment of inertia

The (initial) torque due to gravity is M*g*L/2 (the center of gravity is at the middle of the rod).

The moment of inertia is M*L^2 / 3

a = T/I = M*g*L/2 / M*L^2/3

a = (3/2)*g/L

The translational acceleration is a*L = (3/2)*g

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