A 2.50 kg ball is attached to a 3.00 m bar and swung in a vertical circle. If the ball does not leave the circular loop, what minimum speed must it have at the top of the arc?

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Answer 1 · 2017-06-05T00:42:28

#v=sqrt(29.4)~~5.42217668469m/s#

#sumF=ma#

Since the ball is moving in a circle, we can say that #a=v^2/r#

There are two forces acting on the ball. We have the force of gravity and the normal force and they both point downwards at the top of the arc:

#mg+N=m(v^2/r)#

We are finding the minimum speed that the ball must have. That means that there is no normal force:

#mg+0=m(v^2/r)#

Now plug in the values we know and solve for #v#. #m=2.5kg#, #g=9.8m/s^2#, #r=3m#:

#2.5(9.8)=2.5(v^2/3)#

Solve for #v#:

#9.8=v^2/3#

#3(9.8)=v^2#

#v=sqrt(29.4)~~5.42217668469m/s#