Question

A stone is tied to a string of length L is whirled in a vertical circle with the other
end of the string at the centre. At a certain instant of time, the stone is at its
lowest position and has a speed U. The magnitude of the change in its velocity as
it reaches a position where the string is horizontal is

Answer 1

Given: A stone is tied to a string of length L is whirled in a vertical circle with the other  end of the string at the centre. At a certain instant of time, the stone is at its  lowest position and has a speed u.

To find: The magnitude of the change in its velocity as  it reaches a position where the string is horizontal ?

Solution:

• Now we have given stone is at its  lowest position and has a speed u and height is L.

• Kinetic Energy of body when it is at bottom = 1/2 mu²

• At quarter position, work done against gravity is same.

• So by energy conservation, we have:

                 1/2 mu² = 1/2 mv² + mgh

                 1/2 mu² - 1/2 mv² = mgl

                 1/2 (u² - v²) =  gl

                 v² - u² = -2gl

• Adding 2u² on both sides, we get:

                  v² - u² + 2u² = 2u² - 2gl

                  v² + u² = 2 (u² - gl)

                 √(  v² + u²) = √(2 (u² - gl) )

Answer:

         So the change in magnitude of velocity will be √(2 (u² - gl) )

Answer 2

Answer:

Loss in KE = gain in PE

or, 1/2 mu² - 1/2mv² = mgl

or, 1/2m (u² - V²) = mgl

or, u²-V² = 2gl

or, V²=u² - 2gl

Change in velocity = √(u + V)

=√(u² +u² - 2gl)

= √{2 (u² – gl)}