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Caluclating tension in string that is moving in horizontal circles.

  • Thread starter Dave_1984
  • Start date
  • #1
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Hi all,

First of all I'd like to say thank you to you all for providing a great forum which has been a great help to me working through my ics home learning course.

As i have learnt on here and other places on the internet the information I have in my course materials is not always correct and can be very confusing causing hours of unnecessary extra work.

The question I am very stuck on is: Calculate the tension in a 3m string attached to a 3kg bob that is moving in horizontal circles of 0.6m radius.

All the information I have on working this out shows that tan should be used to work out the angle, I have questioned this with my tutor and he doesn’t seem to have an answer for me. (What chance have I got?)

I have always been very interested in physics and have enjoyed learning throughout the course. I just can’t seem to find a simple clear formula for working this question out?

Any help greatly appreciated
Many thanks,Dave.
 

Answers and Replies

  • #2
993
13
Which forces do you think are acting on the bob?
 
  • #3
13
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That would just be gravity?
 
  • #4
993
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The pull of gravity is not the only force.

Imagine you are the bob. Which pulls or pushes would you feel?
 
  • #5
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I guess a force would be created with the momentum?
 
  • #6
993
13
There are only two forces:

the pull of gravity mg and the tension in the string.
 
  • #7
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Ok so I think I have got it:

mg=f
3x9.81=29.43N (force acting down on the bob)

So now I need to work out the angle and get the force acting on the string?
 
  • #8
993
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What kind of force does a particle need so that it can move in a circle?
 
  • #9
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I think I finally have the answer.

angle = tan-1 0.6/3
angle = 11.30993247

t=3x9.81/cos11.30993247
t=29.43/0.9805806757
t=30.01282886

So the tension on the string will be 30N?

The main thing that has confused me is that i read on here that tan shouldent be used also there is a mistake in my learning material.
 
  • #10
13
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centrifugal
 
  • #11
993
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'...angle = tan-1 0.6/3...'

note that tan(angle) is NOT 0.6/3

0.6/3 = sin(angle)
 
  • #12
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Ok i'm very confused now, thats the way a simular problem just different mass and string lengh is worked out in my learning material - is that wrong then?
 
  • #13
993
13
It all depends on which angle one takes.
A diagram of the forces is very important in a problem like this.
 
  • #14
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I have drawn a diagram that definitely helps.

I have attached an example of the same question with different figures (notice the mistake cos14 is not 0.25?) I have also used this example to work out my problem and have written it to the side.

Would you mind taking a look at this?

Many Thanks
Dave.
 

Attachments

  • #15
993
13
I always start with a diagram of the forces acting so that I try to understand how the equations are obtained.
 

Attachments

  • #16
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Thats great.

So I now have

[itex]\theta[/itex] = sin-1 06/3

[itex]\theta[/itex] = 11.53695903

t = mg/cos[itex]\theta[/itex]

t = 3x9.81/0.9797958971

t = 30.03686797N

That definitely clears up everything, Thankyou very much for your help.
 

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