# Tension in a string attached to a bob moving in horizontal circles

1. Aug 13, 2011

### joe465

1. The problem statement, all variables and given/known data

Calculate the tension in a 2m string attached to a 2kg bob that is moving in

2. Relevant equations

I thought it was F=MA but the working out shows all sorts

3. The attempt at a solution

I know the answer is 78.4N but dont know how to get there.

2. Aug 13, 2011

### PeterO

have you worked out this is a conical pendulum?

If your text book has nothing anout analysing them, you could check this wikipedia address if you want to check what it means.

http://en.wikipedia.org/wiki/Conical_pendulum

3. Aug 13, 2011

### joe465

the book gives:

θ = tan-1 0.5/2
= 14º

Tcosθ = mg

T = mg / cosθ

T = 2*9.80 / 0.25

T = 78.4N.

I dont know why its carrying out those particular calculations

Surely this should not be using tan-1 either since we only have the hypotenuse and opposite values.
shouldnt it be Sine?

Last edited: Aug 13, 2011
4. Aug 13, 2011

### joe465

Also just noticed that cos 14 isnt 0.25, its 0.97

5. Aug 13, 2011

### PeterO

I agree, the angle should be worked out with a sine function, not tan.

6. Aug 13, 2011

### Staff: Mentor

It would seem then that your book has some errors in the solution provided. You should be working to a diagram similar to this:

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7. Aug 13, 2011

### joe465

Would i be correct by saying the answer should be:

20.2N and not the one stated?

How come it changes to cosine once i have calculated the angle?

The amount of errors i have found on this course is disgusting, every other page there is an error. I would never reccomend anyone to ICS after what i have seen on this course.

Thanks a lot for the help so far:)

8. Aug 13, 2011

### Darth Frodo

How come it changes to cosine once i have calculated the angle?

The reason it changes is because you're looking for a different side of the triangle.

9. Aug 13, 2011

### joe465

Im looking for another side of the triangle, am i? i didnt think i was

10. Aug 13, 2011

### Darth Frodo

Yes, because that's the only one equal to mg.

11. Aug 13, 2011

### Staff: Mentor

If you want to bypass the trig functions altogether you can use Pythagoras' theorem and similar triangles. Use Pythagoras to find the length of the vertical leg of the big triangle, then construct suitable ratios for the triangle sides.

12. Aug 13, 2011

### joe465

So is 20.2N correct just so i know i'm doing this right?

Cheers

13. Aug 13, 2011

### Staff: Mentor

Yes, 20.2N is good.

14. Aug 13, 2011

### joe465

15. Aug 14, 2011

### joe465

Next question i dont where to start.

Calculate the angular speed for the bob in the previous question?

the working out gives:

cosθ = g/1w2

ω2 = g/1cosθ

ω2 = 9.8/2cos14

ω2 = 9.8/2*0.97

ω2 = 5.05

Now i thought to get angular speed it is w=angular displacement/time

But this doesnt appear to be doing any of that

16. Aug 14, 2011

### Staff: Mentor

They are equating the ratio of the gravitational acceleration of the mass to the centripetal acceleration due to its rotation, with the ratio of the length of the long leg of the large triangle to the radius leg:

In the diagram, $y = L cos(\theta)$. The centripetal acceleration of the mass as it rotates with radius r is given by $a_c = \omega^2 r$.

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