A rod with length L and mass M is pivoted about a horizontal, frictionless pin t
ID: 1468396 • Letter: A
Question
A rod with length L and mass M is pivoted about a horizontal, frictionless pin through one end. It is released from rest with its center just to the right of directly above the pivot point, so the rod rotates clockwise about the pin. Consider the time interval between the release of the rod and the instant that the rod is horizontal as shown. Does the moment of inertia of the rod about the pivot point increase, decrease, or remain the same in this time interval? Explain. Does the magnitude of the torque on the rod about the pivot point increase, decrease, or remain the same in this time interval? Explain. Does the magnitude of the angular acceleration of the rod about the pivot point increase, decrease, or remain the same in this time interval? Explain. Does the magnitude of the angular momentum of the rod about the pivot point increase, decrease, or remain the same in this time interval? Explain. Does the kinetic energy of the rod increase, decrease, or remain the same in this time interval? Explain. Does the gravitational potential energy of the rod increase, decrease, or remain the same in this time interval? Explain.Explanation / Answer
Here ,
1) moment of inertia of rod is given as
I = m*l^2/3
hence , the moment of inertia will be same.
2)
as torque is given as
T = mg * L *sin(theta)/2
theta is the angle of rod with vertical
the torque on the rod will increase
3)
as angular acceleration = torque/moment iof inertia
hence ,
the angular acceleration will incerase
4)
as the angular speed of the rod will increase
the magnitude of angular momentum will increase
5)
as the angular speed of rod increase
the kinetic energy of rod will increase.
6)
as the centre of mass is coming down ,
the gravtitational potential energy of the rod will decrease
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