Q. Two particles of mass m are connected by a light rope on a sphere shown in figure.It can move within the guide after a quarter circular shape .Find the speed of the masses when the upper mass just leaves the guid.
For the top ball to leave the sphere, it has to overcome its weight, m*g. To do that, the necessary force would come from the balls spinning. That force would be:
Fc = mass * acceleration = m * (v^2 /R)
But, then, with these balls being connected by a string a quarter of a sphere apart, it means the balls will always be 45 degrees apart. Or, the Fc would be acting in a 45 degree angle. Thus:
Fc = mass * acceleration = m * (v^2 /R) * cos 45
Fc = centripetal force
v = linear speed
So, now, this needs to be equal to the weight of the ball:
Fc = m*g
m*g = m * (v^2 /R) cos 45
The masses cancel out. We can multiply by r/cos 45.
g*R/cos 45 = v^2
Then, take the square root:
sqrt (g*R/cos 45) = v
So, that would be the velocity of the balls just before the top ball comes off the sphere.
Good luck, 8235. I hope this helps.