Question

A ball of mass M at one end of a string of length L rotates in a vertical circle just fast enough to prevent the string from going loose the speed of the ball at the bottom of the circle is:a)sqrt2gl b)sqrt3gl c)sqrt4gl d)sqrt5gl e)sqrt7gl

Answer 1

Answer:

First of all, as a piece of information always remember that the minimum velocity at the bottom of the vertical circle is √5gl,

where "g" = gravity and "l" is the radius of the vertical circle.

This is the minimum velocity to complete the vertical circle without slagging the string at any intermediate point.

This can be proved using "conservation of energy theorem."

We start the proof with the following statement:

KE1 +PE1 = KE2 + PE2

REFER TO THE ATTACHED PHOTO TO UNDERSTAND BETTER.

Another information you should keep in mind , the minimum velocity at the top point should be (√gl)