Uniform circular motion - calculating force of tension

[janelle1905]

Homework Statement

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

Homework Equations

F=mv²/r

The Attempt at a Solution

a. Using sin(theta)=40/hyp and cos(90-theta)=40/H, and opp² + adj² = hyp2, the hypotenuse=50
The angles in the triangle are: Angle C=36.9^o and Angle B=53.1^o

But then to calculate the force of tension:
F_t=0.1v²/r

I'm not sure what to do, as there are two unknowns

 

[Redbelly98]

I can't see your figure yet (not approved), but here are two hints:

1. The centripetal force is not simply the rope tension. Did you draw a free-body diagram for the bead? All horizontal force components will contribute to the centripetal force.

2. r is not an unknown, you can figure it out from the dimensions of the triangle formed by the pole and string.

 

[janelle1905]

I can't see your figure yet (not approved), but here are two hints:

1. The centripetal force is not simply the rope tension. Did you draw a free-body diagram for the bead? All horizontal force components will contribute to the centripetal force.

2. r is not an unknown, you can figure it out from the dimensions of the triangle formed by the pole and string.

Regarding part 1, I did draw a FBD, and I know there is a force, F_c=mv2/r acting in the opposite direction as the Force of tension

2. I calculated the dimensions of the triangle, and r=30. But I am just wondering what do I set mv²/r equal to ?

Thanks for your help :)

 

[Redbelly98]

Since I still can't see the figure, can you verify something? From your description, I think that:

• A is the bottom end of the string
• B is where the bead is located on the string
• C is the top end of the string.

Am I right?

Regarding part 1, I did draw a FBD, and I know there is a force, F_c=mv2/r acting in the opposite direction as the Force of tension

Hmmm, not quite. mv²/r is the net force (sum of all forces) acting on the bead, and it acts horizontally towards the post to make the bead move in a circle.

2. I calculated the dimensions of the triangle, and r=30.

Yes, good.

But I am just wondering what do I set mv²/r equal to ?

Well, that is the key here. Can you list the forces, along with their directions, that act on the bead?

 

[janelle1905]

First off, yes you are correct about A being at the bottom, C at the top of the string etc...

Regarding the forces on the bead, this is what I think:

For the x-axis there are two forces, a positive force (F_T) and negative force acting in the opposite direction (F_c).
Therefore, F_T=F_C = mv²/r

 

[Redbelly98]

No, that's not quite right. There are no outward (away from the post) forces here, since there is nothing to provide that force.

There are 2 forces provided by the string, plus the force of gravity, for a total of three forces. I'll leave it to you to figure out their directions

So the usual procedure applies: consider both vertical and horizontal directions, and set

F_net = m a​

for the vertical as well as the horizontal case. Consider what the horizontal and vertical components are for each of the 3 forces I mentioned.

p.s. I'm logging off for the night, good luck ...

 

[rl.bhat]

First off, yes you are correct about A being at the bottom, C at the top of the string etc...

Regarding the forces on the bead, this is what I think:

For the x-axis there are two forces, a positive force (F_T) and negative force acting in the opposite direction (F_c).
Therefore, F_T=F_C = mv²/r

Since AB is horizontal, three forces must act on B.
Force F_T acts towards A, mg acts downwards and third force F_C must act along CB away from C.
F_Ccosθ balances the centripetal force m*v^2/r
F_Csinθ balances the weight mg.
That gives you velocity of the bead.

 

[janelle1905]

Hi - sorry for the delay in my response as I have been away for a few days.

Anyways - just to make sure I understand what you're saying:

F_net = ma = F_T-F_ccos53.1 - F_csin53.1
so:
ma = mv²/r - mv²/r(cos53.1) - mv²/rsin53.1
and then solve for v?

Thanks for your help :)

 

[PhaseShifter]

Have you attempted to solve this by calculating the vertical forces acting on the bead?

 

[Redbelly98]

Hi - sorry for the delay in my response as I have been away for a few days.

Anyways - just to make sure I understand what you're saying:

F_net = ma = F_T-F_ccos53.1 - F_csin53.1

Is that equation supposed to be for the horizontal or for the vertical force components?
You'll need one equation for each.

Also: Fc is really the string tension force due to the upper diagonal segment of string, in which case it is equal (in magnitude) to F_T, and acts in the direction towards C.

 

[janelle1905]

Okay...

I'm just wondering if the Force of tension for C-B and A-B are equal? And if not, how do you figure out what each is (with them being different?)

Thanks )

 

[Redbelly98]

Yes, they will be equal.