Maximize Your Swinging Ball Force with These Expert Tips

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 1K views
ritzcrackers
Messages
1
Reaction score
0
New poster is reminded to always show their work when starting schoolwork threads
Homework Statement
If you swing a ball on a string in a vertical circle, on the top of the circle can the force of tension ever be less than the weight force or will it always be greater? how about at the bottom of the circular path?
Relevant Equations
f=ma
idk
 
Physics news on Phys.org
Given that it does a perfect vertical circle in the top point you ll have $$\sum F=m\frac{v^2}{R} \Rightarrow T+B\geq 0 \Rightarrow T\geq -B$$ which simply means (since i take the positive direction pointing down and the force of tension points down in the upper point) that the force of tension can be anything and it depends on velocity. It will be zero if at the top point the mass has zero velocity, it will be infinite if it has infinite velocity or some in between value. It all depends on the velocity at the top point. In order to be greater than the weight it has to hold that $$m\frac{v^2}{R}-B\geq B\Rightarrow v\geq\sqrt{\frac{2BR}{m}}\geq \sqrt{2Rg}$$
 
Last edited:
Delta2 said:
Given that it does a perfect vertical circle in the top point you ll have $$\sum F=m\frac{v^2}{R} \rightarrow T+B\geq 0 \rightarrow T\geq -B$$ which simply means (since i take the positive direction pointing down and the force of tension points down in the upper point) that the force of tension can be anything and it depends on velocity. It will be zero if at the top point the mass has zero velocity, it will be infinite if it has infinite velocity or some in between value. It all depends on the velocity at the top point. In order to be greater than the weight it has to hold that $$m\frac{v^2}{R}-B\geq B\Rightarrow v\geq\sqrt{\frac{2BR}{m}}\geq 2Rg$$
You have not defined ##B##, assuming that it is the ball weight, but should be stated.

The last step should be an equality (although not technically wrong) and is missing a square root (physical dimensions don’t match).
 
Orodruin said:
You have not defined ##B##, assuming that it is the ball weight, but should be stated.

The last step should be an equality (although not technically wrong) and is missing a square root (physical dimensions don’t match).
Ok right I fixed the square root. Thanks for the correction.

Have to do another correction to myself, the force of tension is zero if the velocity is $$v=\sqrt{Rg}$$ (not if the velocity is zero).
 
Last edited: