What Is the Minimum Speed for Tether Ball Stability?

[fsm]

A 0.75kg tether ball is attached to a 1.6m massless string. What is the minimum speed needed to keep the ball at a radius of 1.2m? If the radius was replaced with a string of the same length what would be the tension of each string if the ball's speed is 5.5m/s?

For part 1 I think the equation to use is T*cos(theta)=(m*v^2)/r. Now I'm stuck because I have two unknowns being the tension and the speed. I'm not sure but is T=(m*g)/sin(theta)?

For Part 2 I think it has something to do with adding the lengths of the rope.
Other than that I'm lost.

 

[Dorothy Weglend]

A 0.75kg tether ball is attached to a 1.6m massless string. What is the minimum speed needed to keep the ball at a radius of 1.2m? If the radius was replaced with a string of the same length what would be the tension of each string if the ball's speed is 5.5m/s?

For part 1 I think the equation to use is T*cos(theta)=(m*v^2)/r. Now I'm stuck because I have two unknowns being the tension and the speed. I'm not sure but is T=(m*g)/sin(theta)?

For Part 2 I think it has something to do with adding the lengths of the rope.
Other than that I'm lost.

I would suggest drawing a free body diagram that shows the forces on the ball. That's what I find most helpful. Once you've done that, you'll see what forces need to balance, and you will end up being able to write the equations (you should have two, with two unknowns), that you can solve.

Dorothy

 

[Stevedye56]

can you use square root(rg)?

 

[fsm]

Motion diagrams do help. I have one set up. I know the weight of the ball is points down to the ground. I know tension has a x and y component. With uniform circular motion the tension in the y direction doesn't affect the total tension. There is also no acceleration in the y direction either. I'm just not seeing it.

 

[Dorothy Weglend]

The cord will need to balance the downward pull of gravity. That's how you can find the tension in the cord.

Are you using the angle the cord makes with the ball?

Dorothy

 

[Dorothy Weglend]

Whoops. Sorry. Hit submit too soon.

If you are using the angle that cord makes with the ball, then it looks to me like you are basically done.

 

[EthanB]

The tension in the y direction absolutely affects the total tension. It will, though, need to be equal to the weight of the ball.

The tension in the x direction will have components coming from both ropes and will be the amount required to keep the ball moving uniformally at the given radius.

 

[fsm]

When the ball spinning at a minimum speed that the tension in the x direction is only from the 1.6m string. As the ball spins it is only enlongating the bottom string to 1.2m and is not kepping it taut. Therefore the bottom string has no tension at the minimum speed and is the radius of the balls path. Someone explained to me why the tension in the y direction doesn't affect the minimum speed (this is what I meant to say in post #4)but I forgot why. So is the tension -sin(theta)=ma for the y diection?

 

[Dorothy Weglend]

You were closer in your original message. I believe the equation for the tension in the Y direction was right there, I'm not sure why you changed it.

If you solve these equations for v, don't you end up with a g in the result? Keeping the radius and length of string constant, the only thing that determines the velocity then is g. So it seems to me that the vertical tension in the string is definitely related to v.

Dorothy