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.
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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.
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