A thin rod of length L and mass M is held vertically with one end on the floor and is allowed to fall. Find the velocity of the other end when it hits the floor, assuming that the end on the floor does not slip?

Explanation

Let M be the mass and L the length of the rod. When it is held vertically, its centre of mass is at a height (L2) from the floor, so that the potential energy of the stick is Mg (L/2). On releasing, the stick falls, i.e., it rotates about the end on the floor and 1 the potential energy is converted into rotational kinetic energy`1/2omega^2` , where I is the moment of inertia of the rod about the lower end and co the angular velocity when it hits the floor. Thus by conservation of mechanical energy,`Mg=L/2=1/2omega^2` or `Mg=L/2=1/2(ML^2)/3omega^2`[asI=1/3ML^2]` or `omega=sqrt(((3g)/L))` If v is the linear velocity of the end hitting the floor, thenv=romega=`sqrt(3gL)`.