Uniform Circular Motion Tension

[Kyleman]

Homework Statement

A 100g bead is free to slide along an 80cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40 cm apart. When the pole is rotated about its axis, AB becomes horizontal. Find the tension in the string. Find the speed of the bead at B.

Homework Equations

a=v^2/r
F=ma

The Attempt at a Solution

So I figured out that the radius of the the circular motion the bead makes is 30cm. However, that's all I have. I'm not sure where to go from here because I don't know the velocity or acceleration.

 

[bel]

Draw a free body diagram of the bead and use the fact that the only forces are the force of gravity and the centripetal force. From the forces, calculate the acceleration, which will then, by $$a= v^2 / r$$ give you the velocity of the bead, since you already know the radius.

 

[Kyleman]

I thought that because the string is horizontal there is no gravity acting on it. How do I calculate the centripetal force too?

 

[Doc Al]

I thought that because the string is horizontal there is no gravity acting on it.

Why would you think that? Hint: Part of the string is not horizontal.

How do I calculate the centripetal force too?

bel gave you the equation for centripetal acceleration. Combine that with an analysis of the vertical and horizontal force components. Use Newton's 2nd law.

 

[Kyleman]

So would this be the correct formula?: Ft+mg=m v/r

 

[Kyleman]

sorry, Ft - force of the tension.

 

[Doc Al]

No, that is not correct.

Start by drawing a diagram identifying all the forces acting on the bead. Then consider the vertical and horizontal components separately.

 

[Kyleman]

okay, okay

is this correct? haha Fty=mg and Ftx=ma

Thanks for the help by the way.

 

[Doc Al]

Fty=mg and Ftx=ma

You are on the right track, but realize that there are two sections of string pulling the bead. (Draw yourself a picture.)

 

[KBark15]

So what's the answer?

 

[KBark15]

I got 980 N for the tension and 17.2m/s for the speed not sure if its right but at least its a real #

 

[Doc Al]

I got 980 N for the tension and 17.2m/s for the speed not sure if its right but at least its a real #

No, doesn't look right to me. But as you say, at least they are real numbers!

Show how you got those answers.

 

[tomotron3000]

Hello,
I am looking at the same problem.
I figured out the answers, but there is still something bothering me about the problem.
I understand why Fty=mg
but I don't understand why Ftx=mv^2/r.
Aren't both Ftx and mv^2/r going in the same direction, towards the pole?
Why is Ftx=mv^2/r?

 

[tomotron3000]

I have another question.
Given the question. How would you go about finding the lengths of each side. We know one side is 4, but how do you find the other sides? I just recognized a 3,4,5 side triangle. (maybe because I read that book Fermats Enigma) But had the number been something other then 4 I would not known how to do it.

 

[Neale.c]

I doing the same problem, and I think I understand what to do. Force in the y direction is mg, and Fx = mv^2/r, and then we can get the tension on BC by adding the two force vectors. The problem I'm running into is where to get velocity to find Fx?

 

[Delphi51]

Interesting problem! Does it look like this?

tomotron, your right triangle idea is brilliant! It need not be 3,4,5; it just has to fit the pythagorean equation.

 

[Doc Al]

I doing the same problem, and I think I understand what to do. Force in the y direction is mg, and Fx = mv^2/r, and then we can get the tension on BC by adding the two force vectors. The problem I'm running into is where to get velocity to find Fx?

Don't think of mv^2/r as a force, but as the result of apply Newton's 2nd law in the x direction. The acceleration in that direction is v^2/r. You'll end up solving for the velocity.

The only forces acting are the weight (mg) and the tension in the string. (Careful when considering the string tension.)