Help: A ball attached to two strings (circular mtoion)

[Cate]

Homework Statement

A ball of mass 1.21 kg is attached to a rotating shaft by two strings, putting the ball into circular motion. Tension on the upper string is 35N. (The length of the strings is 1.6m)

a) draw a free body diagram for forces acting on the ball.
b) determine the tension in the lower string
c) determine the net force acting on the ball
d) determine the speed of the ball

Homework Equations

ac= v^2/r

Fc=mac

The Attempt at a Solution

My problem is with a) I always have a hard time doing free body diagrams and breaking down the x and y components. Once I get that I know how to solve for c) and d). Some direction would be much appreciated!

 

[CaptainZappo]

If the ball is moving in uniform circular motion, is there any net force in the vertical direction? Perhaps you can use this to your advantage.

 

[Cate]

yes i assumed that the fnet in the y direction is zero. but what about the two strings?

 

[Cate]

could someone please point me in the right direction as to find b)?

 

[Doc Al]

I assume the ball is traveling in a horizontal circle? A diagram would be helpful.

The strings apply their tension forces to the ball. Consider that the ball is centripetally accelerating and that the radial component of the net tension on the ball provides the centripetal force.

 

[Dick]

The sum of the vertical components of the tensions must equal mg to keep the ball suspended. The sum of the horizontal components must equal mv^2/r. But I don't see any way to actually get a number for anything unless you know the angle between the strings or the radius of rotation r.

 

[Cate]

I assume the radius of rotation would be 1.6m

 

[Doc Al]

I assume the radius of rotation would be 1.6m

Isn't that the length of the strings? (Total length? Or length of each?) Are the strings horizontal?

 

[Cate]

yes, 1.6 m is the length og the strings (individual) but it's also the disatnce between the two strings on the rod

 

[Cate]

The diagram looks like #14 at this web page

http://teachers.oregon.k12.wi.us/fishwild/APPhysics/ForceMotion2.pdf

 

[Doc Al]

yes, 1.6 m is the length og the strings (individual) but it's also the disatnce between the two strings on the rod

Apply a little triangle geometry and you can figure out the radius of the ball's circular path.

 

[Cate]

what do you mean trianle geometry? why isn't the radius 1.6m? how will this help me solve for the lower tension

 

[Doc Al]

what do you mean trianle geometry? why isn't the radius 1.6m? how will this help me solve for the lower tension

If the string is 1.6m and it is at some angle with the horizontal, how can the radius also be 1.6m?

Draw yourself a diagram. The rod and the two strings form an equilateral triangle.

 

[Cate]

o.k so if it's an equaliteral trangle then all three sides are 1.6m so I'm trying to find the height of the triangle?

 

[Doc Al]

That's right. Or just make a right triangle by drawing in the radius. (What are the angles of an equilateral triangle?)

 

[Cate]

I got the radius is about 1.39 m

 

[Doc Al]

I got the radius is about 1.39 m

Looks good.

 

[Cate]

thanks for your help, but could you give me a hint has to how to solve for the lower tension?

 

[Doc Al]

What's the net horizontal force on the ball? (Call the lower tension T.) That force is the centripetal force.