# Uniform circular motion help

• fsm
In summary, the conversation discusses the minimum speed needed to keep a 0.75kg tether ball at a radius of 1.2m when attached to a 1.6m massless string. The equation T*cos(theta)=(m*v^2)/r is suggested and the need for a free body diagram is mentioned. The tension in the y direction is found to affect the total tension and the equations T=(m*g)/sin(theta) and -sin(theta)=ma are suggested. The relationship between the vertical tension in the string and the velocity is discussed, with the conclusion that the vertical tension is related to the velocity through the acceleration due to gravity.
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.

fsm said:
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

can you use square root(rg)?

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.

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

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.

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.

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?

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

## 1. What is uniform circular motion?

Uniform circular motion is a type of motion in which an object moves in a circular path at a constant speed. This means that the object's velocity remains constant, but its direction changes continuously as it moves around the circle.

## 2. What causes an object to experience uniform circular motion?

An object experiences uniform circular motion when it is acted upon by a centripetal force, which is directed towards the center of the circle. This force is responsible for continuously changing the direction of the object's motion, keeping it in a circular path.

## 3. How is uniform circular motion different from non-uniform circular motion?

In uniform circular motion, the object's speed remains constant, while its direction changes. In non-uniform circular motion, both the speed and direction of the object's motion are changing.

## 4. What is the relationship between the radius of a circle and the speed of an object in uniform circular motion?

The speed of an object in uniform circular motion is directly proportional to the radius of the circle. This means that as the radius increases, the speed of the object also increases. Similarly, as the radius decreases, the speed of the object decreases.

## 5. How can we calculate the acceleration of an object in uniform circular motion?

The acceleration of an object in uniform circular motion can be calculated using the equation a = v^2/r, where a is the acceleration, v is the speed, and r is the radius of the circle. This equation shows that the acceleration is inversely proportional to the radius, meaning that as the radius increases, the acceleration decreases.

• Introductory Physics Homework Help
Replies
55
Views
612
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
17
Views
7K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
660
• Introductory Physics Homework Help
Replies
7
Views
3K
• Introductory Physics Homework Help
Replies
8
Views
2K