Minimum v0 for Tension in Inelastic Collision

[Jeremymu1195]

Homework Statement

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.

Homework Equations

∑F_r=mv²/r
T=1/2mv²
U=mgh

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),
letting m'=(m+M)∑F_r= T_b-m'g=m'v'²/L (*)

At the top,

∑F_r=T_t+m'g=m'v''²/L (**)

Using Energy,

1/2m'v'₂=1/2m'v''²+m'g(2L) (***)
v'²=v''²+4gL

v''²=v'²-4gL

I am going to pause here to see if there is anything I am missing.

 

[Doc Al]

You're on the right track. Find v'', then v', then v₀.