Circular motion

  • Thread starter PhysKid24
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  • #1
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a ball is whirled around with astring in the vertical plane. at the top of th ewhirl, what is the minimum speed which the ball can have before leaving the string? do you do sum of the forces and just set v = vmin? Then it says the maximum tension the string can have at the bottom is Tmax, what is vmax, the maximum speed of the ball without breaing the string, is it again sum of the forces? thanks
 

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  • #2
Gokul43201
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So basically, you are given the tension at the bottom of the swing, and asked for the velocity there.

Yes, simply use [tex]F(net) = ma [/tex]
to find v.
 
  • #3
Doc Al
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centripetal acceleration

PhysKid24 said:
a ball is whirled around with astring in the vertical plane. at the top of th ewhirl, what is the minimum speed which the ball can have before leaving the string?
I don't know what you mean by "leaving" the string. I suspect the question is "what's the minimum speed at the top of the motion that will keep the string taut". (Or do you mean "what's the minimum speed at the top of the motion that will break the string"?)
do you do sum of the forces and just set v = vmin?
You apply Newton's 2nd law, realizing that the ball is centripetally accelerated.
Then it says the maximum tension the string can have at the bottom is Tmax, what is vmax, the maximum speed of the ball without breaing the string, is it again sum of the forces?
Again, as Gokul43201 explains, apply Newton's 2nd law. But realize that the ball is centripetally accelerated.

Assuming I understand the problem: At the top of the motion, you want the minimum speed. Set the tension to zero and solve for the speed. At the bottom, you want the maximum speed; set the tension to Tmax and solve for the speed.
 

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