Tension in simple pendulum.

  • Thread starter PhoniexGuy
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  • #1
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Homework Statement



http://img32.imageshack.us/img32/1551/filevno.jpg [Broken]

I have to find the tension for the rope when angle is 20 degrees, the object is moving back and forth. I know the mass is 2kg, gravity is 10N/kg

Homework Equations



F_t * cos θ = mg.

So F_t = mg/cos θ

The Attempt at a Solution



The problem seemed trivial, but I wanted to make sure I got it right: Using F_t = mg/cos θ

F_t = 2*10/cos θ = 20/cos20 = 21.3 N

Is this right or wrong?
 
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Answers and Replies

  • #2
haruspex
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No, you have it inverted. The block will be accelerating, partly downwards, so T is not overcoming the whole of g.
For now I'll assume the tension is wanted at max amplitude. The trick is to realise that the string length is constant, so when the velocity is momentarily 0 there's no radial acceleration. So resolve forces in the radial direction instead.
If it's not at max amplitude then there is a centripetal acceleration, increasing the tension, but there's not enough info provided to calculate that.
 
  • #3
ehild
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The object moves in the horizontal direction, back and forth. So there can be horizontal acceleration: The resultant force is horizontal. See attachment: I drew the forces, weight (G), T (tension) and the horizontal resultant (F) The shaded angles are equal.G =T cos(20) is right.

ehild
 

Attachments

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  • #4
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Homework Statement



http://img32.imageshack.us/img32/1551/filevno.jpg [Broken]

I have to find the tension for the rope when angle is 20 degrees, the object is moving back and forth. I know the mass is 2kg, gravity is 10N/kg

Homework Equations



F_t * cos θ = mg.

So F_t = mg/cos θ

The Attempt at a Solution



The problem seemed trivial, but I wanted to make sure I got it right: Using F_t = mg/cos θ

F_t = 2*10/cos θ = 20/cos20 = 21.3 N

Is this right or wrong?
Hi!
Your first equation: Tension*cosθ = mg is wrong.
Instead you can write the equation in the centripetal direction (ie. towards the center).It will be in equilibrium in that direction.
If the block in the pic is in the highest position its' velocity is 0.(if its any generic position velocity should be given)
Now you will get
tension= mgcosθ
Hence you can calculate the tension.
 
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  • #5
ehild
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Hi!
Your first equation: Tension*cosθ = mg is wrong.
Instead you can write the equation in the centripetal direction (ie. towards the center).It will be in equilibrium in that direction.
If the block in the pic is in the highest position its' velocity is 0.(if its any generic position velocity should be given)
Now you will get
tension= mgcosθ
Hence you can calculate the tension.
The object moves horizontally. There is no "highest position". As the object moves along a straight line, there is no centripetal force. The resultant of the tension and gravity is horizontal.
Mg = Ftcosθ.
The OP got the correct solution.

ehild
 
  • #6
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Hmm, perhaps the diagram isn't clear, the object is moving back and forth on the pendulum, but in sort of an arc like so:

http://img11.imageshack.us/img11/9210/85456919.png [Broken]

Does this change anything?
 
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  • #7
ehild
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Hmm, perhaps the diagram isn't clear, the object is moving back and forth on the pendulum, but in sort of an arc like so:

http://img11.imageshack.us/img11/9210/85456919.png [Broken]

Does this change anything?
In this case, Ft-mgcos(θ)=mv2/L, the centripetal force. If the object is in rest at θ=20° Ft=mgcos(θ), as haruspex and Vineeth said.


ehild
 
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