Someone help me to solve this sum from physics quckly( plz explain in step and full calculations)?

Question

A particle of mass m suspended from a string of l is given an velocity of root4gl horzontally.when the particle is at lowest point so that it moves in a vertical plane.

Find the height above the lowest point when the string becomes slacked.

In the same setup find out the max height of part above the lowest point.

Answer 1

For the maximum height above the lowest point, the problem is most easily done through energy considerations.  Put the gravitational potential energy reference at the lowest point.  In that case, all of the mechanical energy is in kinetic energy:

T = (1/2)m v^2 = (1/2) m (4gl) = 2mgl

At the highest point, all of the mechanical energy will have been transferred to gravitational potential energy, so

U = mgh = 2mgl

Masses and g cancel:  h = 2l

This, of course, neglects air resistance and other frictional losses.

This result indicates that you've pushed it hard enough that it'll describe a half-circle from the lowest point, coming to rest directly above where it started.  In that case, the particle will then fall straight down.

As for where the string will go slack, that will happen as soon as the string is no longer providing any of the centripetal force necessary to hold the particle in its "orbit".  Clearly, this will not happen as long as the particle's height above the lowest point is less than l.  As soon as the string is pointing horizontally (i.e. at height l), gravity and the string will be perpendicular.  From that point on, a portion of the weight will be acting as centripetal force, and as that happens the tension in the string will begin to drop off.  So, we need to find the point at which the component of gravity that acts towards the center equals the necessary centripetal acceleration.

I'm afraid that's as far as I'm going to take this problem.  It's late here, and I'm sleepy.  Make a sketch, though, and decompose gravity into centripetal and tangential parts.  It should be possible to get a relation between the centripetal component of gravity and the height of the particle above the midline.  Relate that to centripetal acceleration (a = (v^2)/r).  Since speed is constantly being reduced by the tangential acceleration, you'll want to approach this from energy considerations again (unless you know calculus, actually).  You should end up with a fairly simple equation to solve.

Hope this helps.