A mass m with velocity v0 makes an inelastic collision with second mass M that is suspended by a string of length L. The velocity v0 is perpendicular to the vertical string. After the collision the combined masses, m+M, rotate in a vertical plane around the point of suspension of the string. Find the minimum value v0 such that the string remains always under tension for a complete rotation.
The Attempt at a Solution
The condition for the an inflexible string to remain under tension is that it never goes slack, i.e. length=L for the entire rotation.
Since the collision is inelastic, momentum is conserved while kinetic energy is not. So,
mv0=(m+M)v' where v' is the magnitude of the x-directed velocity of the combined masses (m+M) after the collision. so v'=m/(m+M)v0.
At the bottom of the vertical circle (In Polar Coordinates),
∑Fr= Tb-m'g=m'v'2/L (*)
At the top,
I am going to pause here to see if there is anything I am missing.