# A straight rod of length L is released on a friction less horizontal floor in a vertical position .As it falls( also slips) , the distance of a point on the rod from the lower end, which follows a quarter circular locus.A) L/2   B) L/4  C) L/8  D) None.

54 Points
11 years ago

none as no point on the rod makes circular path.only the lower piont will shift L/2 distancefrom its initial position.

lj
115 Points
6 years ago
The disc tangents of point should be L/4 . It can be derived easily using instantaneous axis of rotation. They , if you still can not, ask in a different thread. Hope it helps.
Aviral Omar
9 Points
6 years ago
Draw the FBD at any general point of time wherein mg acts at the centre of the rod and Normal acts at the point where it touches the floor.Balancing forces, we get N=mg.The reason the rod slips is due to unbalanced torques. A point which moves with a circular locus will be one about which the rod can be assumed to be rotating. Such a point is the Instantaneous Axis of Rotation.Assume this point is at distance r from bottom of rod. Balancing torque about this point,mg(l/2-r)sin θ=Nrsin θ. Put N=mg and r comes out to be l/4.