Question

small ring of mass M is attached at one end of a light string of length L equal to 0.6 metre the other end of the string is tied to a small block of mass bsmall ring of mass M is attached at one end of a light string of length L equal to 0.6 metre the other end of the string is tied to a small block of B of mass 2m the ring is free to move on a smooth horizontal rod the block B is released from rest from position shown the velocity of the ring when a string becomes vertical Eyes ​

Answer 1

Given:  Length L equal to 0.6 metre

To find: The velocity of the ring when a string becomes vertical?

Solution:

• Now, we have given that the ring can move freely on the string until its not made vertical.

• Now, after the string is vertical, lets consider the velocity of the ring as v.

• Now, we know the energy conservation theorem as energy can neither be created nor destroyed but can be converted from one form of energy to the another.

• This means that a system always has the same amount of energy, unless it's added from the outside.

• So applying the energy conservation, we get:

                1/2 x mv² = m x g x l + 1/3 x m x g x l

                1/2mv² = 4/3 x m x g x l

                v² = 8 x g x l  /3

                v = √8gl / 3

• Putting the value of l, we get:

                v = √8 x 9.8 x 0.6 / 3

                v = √47.04/3

                v = 15.68 m/s

Answer:

    The velocity of the ring when a string becomes vertical is 15.68 m/s.