Caluclating tension in string that is moving in horizontal circles.

[Dave_1984]

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 learned 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.

 

[grzz]

Which forces do you think are acting on the bob?

 

[Dave_1984]

That would just be gravity?

 

[grzz]

The pull of gravity is not the only force.

Imagine you are the bob. Which pulls or pushes would you feel?

 

[Dave_1984]

I guess a force would be created with the momentum?

 

[grzz]

There are only two forces:

the pull of gravity mg and the tension in the string.

 

[Dave_1984]

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?

 

[grzz]

What kind of force does a particle need so that it can move in a circle?

 

[Dave_1984]

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.

 

[Dave_1984]

centrifugal

 

[grzz]

'...angle = tan-1 0.6/3...'

note that tan(angle) is NOT 0.6/3

0.6/3 = sin(angle)

 

[Dave_1984]

Ok I'm very confused now, that's the way a simular problem just different mass and string lengh is worked out in my learning material - is that wrong then?

 

[grzz]

It all depends on which angle one takes.
A diagram of the forces is very important in a problem like this.

 

[Dave_1984]

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.

 

[grzz]

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

 

[Dave_1984]

Thats great.

So I now have

\theta = sin-1 06/3

\theta = 11.53695903

t = mg/cos\theta

t = 3x9.81/0.9797958971

t = 30.03686797N

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