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

[joe465]

Homework Statement

Calculate the tension in a 2m string attached to a 2kg bob that is moving in
horizontal circles of 0.5m radius.

Homework Equations

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

The Attempt at a Solution

I know the answer is 78.4N but don't know how to get there.

 

[PeterO]

Homework Statement

Calculate the tension in a 2m string attached to a 2kg bob that is moving in
horizontal circles of 0.5m radius.

Homework Equations

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

The Attempt at a Solution

I know the answer is 78.4N but don't know how to get there.

have you worked out this is a conical pendulum?

If your textbook 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

 

[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 don't 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?

 

[joe465]

Also just noticed that cos 14 isn't 0.25, its 0.97

 

[PeterO]

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 don't 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?

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

 

[gneill]

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

 

[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:)

 

[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.

 

[joe465]

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.

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

 

[Darth Frodo]

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

 

[gneill]

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.

 

[joe465]

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

Cheers

 

[gneill]

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

Cheers

Yes, 20.2N is good.

 

[joe465]

thanks a lot, i shall be having some nice words with ICS about this:)

 

[joe465]

Next question i don't 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
ω = 2.25 rad s-1.

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

But this doesn't appear to be doing any of that

Thanks in advanced

 

[gneill]

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