Question

define the velocity of chain while leaving the smooth table some part hanging​

Answer 1

Explanation:

Mass of hanging chain part is 3M

and its CM is 6l

below table.

Hence, initial potential energy is U 1 =− 3M g 6l

= 18Mgl

When whole chain slips off, then hanging mass is M and the CM of hanging part is 2l

below table.

Hence , final potential energy U 2 =−Mg 2l

Since total energy is conserved, hence-U 2 +K 2

=U 1 +K 1

⟹ 21

Mv 2 − 2Mgl =− 18Mgl +0

⟹v 2 = 98gl

⟹v= 32 2gl