1. Oct 20, 2012

### GandhiReborn

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

A particle of mass 15g is attached to the end of a string of length 50cm, rotating at 6rads-1 to form a conical pendulum.

Find a) The tension in the string
Find b) The angle

2. The attempt at a solution

Okay I get that Tcosθ= mg
& TSinθ= mv^2/r
which => tanθ= v^2/rg

and I get why, but I don't understand how to find tension..... I'm really blooming confused and have been banging my head for over two hours now.

I also know I need to find r using the length of the string and the angular velocity, I just don't get how.

Basically, I need r to find both and I can't work out how to get r? Please, someone help me??

Thanks!

2. Oct 20, 2012

### TSny

Hello and welcome.

You are right that you will need to somehow use the given length, L, of the string. See if you can set up another equation that relates L, r, and θ. Also, you'll need the relation between v and the angular speed, ω.

3. Oct 20, 2012

### mishek

I mistyped, but I cannot delete it.....

Last edited: Oct 20, 2012
4. Oct 20, 2012

### GandhiReborn

Well R = 0.5Sinθ, but I still don't know θ at this point...
and v=rω but I only know ω right?

Don't worry mate. I know I'm missing something blindingly obvious here, and I'm sorry for missing it.

Thank you both of you, I really do appreciate your responses. But I'm still not getting that lightbulb moment, I'm pretty shocking at resolving forces. :/ Please continue to help me..?

5. Oct 20, 2012

### mishek

Check Your third equation (tanθ= v^2/rg), and consider the fact that you know the following relationships:R = 0.5Sinθ & v=rω.

6. Oct 20, 2012

### TSny

You now have two equations with sinθ. Can you combine them somehow?

7. Oct 20, 2012

### GandhiReborn

Okay, you sir are a legend but there is no way that the working out I just did is what they were looking for?? (the answer I got was right btw, as it were in my notes which had poor explanation):

I said tanθ = v^2/rg; given v=rω and r=0.5sinθ ∴ tanθ=ω^2/g

subbed in values and rearranged gave me:

g/0.5×ω^2 = cosθ & took arccos of that, giving θ = 57° (2 s.f.)

You are also a legend.

Tsinθ = mv^2/r
r=0.5sinθ
r/0.5=sinθ
∴ Tr/0.5=mv^2/r
because v^2=r^2(ω^2)
T(r^2)=0.5m(r^2)(ω^2)

cancel and resolve

T= 0.27N - again absolutely correct.

But seriously, is that the working out they were looking for or is there a neater way of doing things??

Thank you you absolute geniuses, you have put me out of my misery whilst letting me think a bit. I owe you massively.

8. Oct 20, 2012

### TSny

A neat geometrical way of solving it is to draw two right triangles. One triangle will show the pendulum with L as hypotenuse and r as the side opposite θ. The other triangle will be a vector addition of force triangle. If you add the tension force and mg force as vectors, the resultant net force must point horizontally (toward the center of the circular motion) and the net force must have a magnitude of mω2r. So, you get a right triangle with T as hypotenuse and mω2r as side opposite θ.

Since the two triangles are similar, you can say the ratio of T to r in the first triangle equals the ratio of the corresponding sides in the force triangle. That should allow you to get T easily. Now that you know two sides in the force triangle, you can get θ. Then r from either triangle.

9. Oct 20, 2012

### GandhiReborn

I actually thought this and drew the two triangles, but I guess my mind got crowded and I freaked out so I quickly stopped and came on here. Thank you, going back and tying it to this concept, having already worked out how to get my values with you guys' prior awesome help has allowed me to write it down really clearly to myself.

To anyone this might help in the future:

Part a) *to find T, we must know what F is.

Consider two triangles, one of forces T, F and mg (1) and the other of lengths r and L (2).

Using (1)

Tsinθ = F
F = mω2r
∴ Tsinθ = mω2r

and Using (2)

r = 0.5sinθ

so subbing in gives -->

Tsinθ = mω2(0.5)(sinθ)

cancel sinθ and plug in remaining values to give

T = 0.27N

Part b)

Using (1)

Tcosθ=mg
∴ arc cos(mg/T) = 57°

Many thanks indeed, I really owe you two, one ;)

Last edited: Oct 20, 2012