Then end A of a rod slides down a smooth wall and its end B slides on a smooth floor. When AB makes angle `alpha` with the horizontal, A has speed v. The speed of B must be

A ladder of length L is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment, the speed of the end in contact with the horizontal floor is v and the ladder makes an angle theta=30^@ with horizontal. Then, the speed of the ladder's centre of mass must be

A ladder of length L is slipping with its ends against a vertical wall and a horizontal floor. At a certain moment, the speed of the end in contact with the horizontal floor is v and the ladder makes an angle theta=30^@ with horizontal. Then, the speed of the ladder's centre of mass must be

A uniform rod of length 2a is held with one end resulting on a smooth horizontal table makin an angle alpha with the vertical. When the rod is released,

A uniform rod is lying against a smooth wall and on a smooth floor. It is relased from rest when it makes an angle theta_(0) with the ground. Find the angle rod will make with the ground it just leaves the contact with the walls.

A conducting rod ab of length l, mass m and resistance R slides on a smooth, thick pair of metallic rails. The plane of the rails makes an angles theta with the horizontal. A magnetic field B acts along the perpendicular to the horizontal plane in upward direction. If the rod slides down on the rails at a constant speed v, then show that B=sqrt((mgRsintheta)/(l^(2)vcos^(2)theta)) .

A rod of length 10 meter has one end on smooth floor and the other end on smooth wall is released from the rest the position shown in figure. When the rod makes an angle of 37^(@) with the horizontal, answer the following: The velocity of centre of mass of rod is :

A thin rod MN free to rotation in the vertical plane about the fixed end N is held horizontal when the end M is released the speed of this end when the rod makes an angle alpha with the horizontal will be proportional to : (see figure)