Centripetal force and tension in a string

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  • #1
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Hi everyone. I've seen this question posted before so I apologize, but I have a hard time understanding the responses and some posts in different threads seem to go different directions (I don't know what to follow!). I've given the problem a try and would greatly appreciate knowing if I'm on the right track! Thanks in advance!:smile:

Homework Statement



A 100 g bead is free to slide along an 80 cm long piece
of string ABC. The ends of the string are attached to a
vertical pole at A (bottom) and C (top), which are 40 cm 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


a2+b2=c2
FR= mv2/r

The Attempt at a Solution


Using pythagoras I figured out that the radius (AB) is .3m
Using sinθ (.4/.5) the angle at B is 53.1 degrees

Using a FBD, it appears that the bead has 2 forces acting on it in a vertical direction: mg and a force upwards ("Fc") from the BC string segment. The vertical component of this force must equal mg because there is no acceleration in the y direction (or, at least that's my logic).
So: Fc sin 53.1 = mg
Fc vertical = (.1)(9.8)/sin 53.1 = 1.23

This is where I come to problems. For the horizontal component, it seems like Fnet would equal the centripital force pulling inwards towards A plus the horizontal component of the force from BC string. Would the equation be Fc(cosθ) = mv2/r? How would I solve this without knowing Fc horizontal or velocity? I don't even know if I am on the right track, I feel so confused :frown: Any advice?
 

Answers and Replies

  • #2
Doc Al
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Using a FBD, it appears that the bead has 2 forces acting on it in a vertical direction: mg and a force upwards ("Fc") from the BC string segment. The vertical component of this force must equal mg because there is no acceleration in the y direction (or, at least that's my logic).
So: Fc sin 53.1 = mg
Fc vertical = (.1)(9.8)/sin 53.1 = 1.23
What you call Fc is the tension in the string. The vertical component of that force must support the weight of the bead, as you stated.

This is where I come to problems. For the horizontal component, it seems like Fnet would equal the centripital force pulling inwards towards A plus the horizontal component of the force from BC string. Would the equation be Fc(cosθ) = mv2/r? How would I solve this without knowing Fc horizontal or velocity?
Firstly, the 'centripetal force' is Fnet in the horizontal direction. It's not a separate force.

The only forces acting on the bead in the horizontal direction are due to the string. What's the total horizontal force from both sections of string on the bead? Set that equal to ma, where a is the centripetal acceleration.
 
  • #3
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Firstly, the 'centripetal force' is Fnet in the horizontal direction. It's not a separate force.
Okay, that makes sense. So like you said, the horizontal tension from CB and AB will equal the net force, which equals maR.

To find the horizontal tension in CB, I went mg(cosθ) = (9.8)(0.1)(cos53.1) to find horiztonal CB = 0.59.
Now, to find AB I do the same thing of mg(cosθ) = (9.8)(0.1)(cos53.1) to get the same value for AB= 0.59. Should these be the same? It seems too easy.

Going on this, 0.59+0.59 = maR
1.18 = 0.1 (v2/.3)
√3.54 = v
v= 1.88 m/s

Then:
FR= mv2/r
FR= (0.1)(1.882)/(0.3)
FR= 1.178

Is this correct, or did I go completely off base in the horizontal forces part? Thanks so much for your help, sorry if I seem like a numbskull!
 
  • #4
Doc Al
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Okay, that makes sense. So like you said, the horizontal tension from CB and AB will equal the net force, which equals maR.
Good.

To find the horizontal tension in CB, I went mg(cosθ) = (9.8)(0.1)(cos53.1) to find horiztonal CB = 0.59.
Now, to find AB I do the same thing of mg(cosθ) = (9.8)(0.1)(cos53.1) to get the same value for AB= 0.59. Should these be the same? It seems too easy.
The string has a single tension throughout, which you already found above. (What you called Fc.) For CB, you'll need to calculate the horizontal component of that tension force at the given angle; for BA no calculation is needed.

Give it another shot.
 
  • #5
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The string has a single tension throughout, which you already found above.
Nice.

For CB, you'll need to calculate the horizontal component of that tension force at the given angle; for BA no calculation is needed.
Ok, so to find the horizontal force from the tension I went:
tan53.1= 1.23/x
x= 0.92

I'm not quite sure what you mean by no calculation being needed for BA. Is is the same value as the horizontal force from the tension? Would the equation FT=Ffrom a to b/2sinθ be applicable here, or is that only when the strings are of equal length?
 
  • #6
Doc Al
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Ok, so to find the horizontal force from the tension I went:
tan53.1= 1.23/x
x= 0.92
Not sure why you used tanθ. The tension force in CB acts at angle θ. How do you find the horizontal component of a vector, given the angle?
I'm not quite sure what you mean by no calculation being needed for BA.
Since the string segment BA is already horizontal, the horizontal component is just the tension itself. (No need to calculate anything--you already know the tension.)
Would the equation FT=Ffrom a to b/2sinθ be applicable here, or is that only when the strings are of equal length?
Not sure what that equation represents. See my tips above.
 
  • #7
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Ohhhh, thanks Doc Al, I think I'm understanding better now. I've found the tension, but that tension is acting on the bead from two sources. Full value of tension is found in AB, and this value added to the strictly horizontal part of the tension from BC would equal total force in the x direction, right? I guess I thought that if total tension was 1.23, then it would need to be "split" between the two segments of string or something.

So horizontal value for BC = mgsinθ
= (.1)(9.8)(sin53.1)
= 0.78

So then velocity equals:
horizontal AB + horizontal BC = mv2/r
1.23 + 0.78 = (0.1)(v2)/0.3
v2)/= 2.01(0.3)/.0
v = 2.46m/s

Am I finally on to something here?
 
  • #8
Doc Al
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45,093
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Ohhhh, thanks Doc Al, I think I'm understanding better now. I've found the tension, but that tension is acting on the bead from two sources. Full value of tension is found in AB, and this value added to the strictly horizontal part of the tension from BC would equal total force in the x direction, right?
Right.
I guess I thought that if total tension was 1.23, then it would need to be "split" between the two segments of string or something.
Yes, that idea was messing you up.

So horizontal value for BC = mgsinθ
= (.1)(9.8)(sin53.1)
= 0.78
I don't understand your thinking here. What's the tension in the string? What's the horizontal component of that tension?

So then velocity equals:
horizontal AB + horizontal BC = mv2/r
1.23 + 0.78 = (0.1)(v2)/0.3
v2)/= 2.01(0.3)/.0
v = 2.46m/s

Am I finally on to something here?
You're almost there. Redo your calculation of the horizontal component of BC tension.
 
  • #9
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I don't understand your thinking here. What's the tension in the string? What's the horizontal component of that tension?

Mmm yes I think I multiplied by mg because I remembered doing that in previous problems...but now I realize those problems involved normal force which this one doesn't :bugeye: I'm doing this course by correspondence and I don't think I'm grasping some of these concepts as strongly as I should be.

Ok so the horizontal component of BC tension would equal:
cos 53.1 = BChorizontal/1.23
BChorizontal= 0.74

Velocity: 1.23 + 0.74 = (0.1)(v2)/0.3
v2= 1.97(0.3)/0.1
v = 2.43m/s
 
  • #10
Doc Al
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Looks good to me.
 
  • #11
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GREAT! Thanks so much for your help and patience!!:smile:
 

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