A 0.250 kg ball swung on a 1.3 m string in a vertical circle. If it's tangential speed at the bottom of the path is 4.2 m/s, what is the maximum speed the ball can have at the top and still move in a circle?

Answer 1

It is probably the minimum speed you mean.
The centripetal acceleration must be at least 9.81 m/s^2 on earth or the string will go slack and the ball will fall on your head.

Now you have given enough information already to find out.

By the way, any old mass m will do

v^2/r>/= g

ke at bottom =
= (1/2)m(4.2)^2

Ke at top = (1/2)mv^2
lost m g (2.6) potential on the way up
so
(1/2)m v^2 = (1/2)m(4.2)^2 - 2.6mg
v^2 = 4.2^2 - 5.2g
BUT
v^2/r >/= g

(4.2^2-5.2g)/1.3 >/= g
is that true on earth?
g = 9.81 m/s^2
(17.64 - 50 something) /1.3 is negative
so no way it is going fast enough at the bottom to even make it to the top