# Two Circular Motion Questions. Pendulums

1. Jan 25, 2014

### navm1

1. The problem statement, all variables and given/known data
Got my first physics assignment and Ive been able to work out (I think) all of them except these two.

1. A conical pendulum has length 1.5m and rotates at 4ms^-1. What is its angle to the vertical?

2. A mass moves in a vertical circle attached to a fixed point by a string of length 2m. How fast must it be moving at the lowest point if it completes circles? What would be the speed necessary at the lowest point if the mass were attached by a light rod instead of a string?

2. Relevant equations
F=ma
F=mv^2/r
v=rω
ω=θ/r

3. The attempt at a solution

1. I've drawn a diagram and understand that i have gravity acting downwards and tension up the string. I think i could find θ by using tanθ=v^2/rg but im not sure of how to work the radius out if i only have the length of the string, the velocity, and gravity.

2.
I'm only given the length and gravity on this question. I think i might have to work out the tension but im finding it difficult without knowing a mass of the object.

I understand my attempts didn't go too far. Tension, work, and energy were omitted from my course for some reason so I've got some catching up to do with it. Thanks

2. Jan 25, 2014

### tiny-tim

hi navm1! welcome to pf!

(try using the X2 button just above the Reply box )
either eliminate T by doing F = ma for both the x and y directions,

or ignore T by choosing … which direction?
call the mass "m" … it'll cancel out in the end

3. Jan 26, 2014

### navm1

Thanks
I'm still really struggling with these. Any pointers on where to start? I was told the first one will include a quadratic equation somewhere

4. Jan 27, 2014

### ehild

Look at your diagram (or the yellow triangle on the attached picture): how is the radius r related to the length of the string and the angle θ?

ehild

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5. Jan 27, 2014

### tiny-tim

hi navm1!

(just got up :zzz:)
start with F = ma for both the x and y directions (separately) …

show us what you get

6. Jan 27, 2014

### navm1

me too. i'm sick so its a perfect day to stay in and do physics all day.

so the force of gravity is (m)(-9.8) only in the y direction which means the y component of the tension must be (m)(9.8). the only two forces acting are gravity and tension so the x component of the tension must be providing the centripetal force on its own. so i drew a diagram for tension and ended up with (9.8)tanθ=v2/r and thats as far as i made it so far without knowing angle θ

7. Jan 27, 2014

### haruspex

Right so far. You know v and the length of the string. What other equation can you write relating these with θ and r?

8. Jan 27, 2014

### tiny-tim

you know L = 1.5, so what's the relation between L r and θ ?

9. Jan 27, 2014

### navm1

i dont know if im going down the route you guys have suggested but i ended up changing my equation from

(g)(tanθ) = v2/r

to

sinθ/cosθ = v2/(l*sinθ)(g)

then i think i could square the sinθ and start to build a quadratic equation from there?

if i multiplied both sides by sinθ would i end up with

sin2θ/cosθ = v2/lg ?

edit: or could i replace sin2θ with 1-cos2θ

Last edited: Jan 27, 2014
10. Jan 27, 2014

### navm1

so if x = cosθ
then my equation is x2+(v2/lg)*x-1=0. v2/lg

i came out with x=0.6 so cos-1 would be 53.1°
which would make the vertical angle 36.9°

am i on the right track here? thanks

11. Jan 27, 2014

### tiny-tim

yes!

how are you getting on with question 2?

12. Jan 27, 2014

### navm1

so did I get the correct angle for the vertical angle? thanks a lot for the help

Gonna start question 2 as soon as i know i've got question 1 right then I'll let you know how i get on

13. Jan 27, 2014

### tiny-tim

looks right to me

14. Feb 4, 2014

### navm1

so for the second question i have the equations

v=√vi2 - 2gl(1-cosθ) and T(tension)= (mvi2 / l ) - 2mg+3mg*cosθ

are these the right equations to be using given that i only have the length of the string and nothing else?

Last edited: Feb 4, 2014
15. Feb 4, 2014

### tiny-tim

(= mv2/l + mg*cosθ)

yes

16. Feb 4, 2014

### navm1

thanks. im havin a little trouble working out how i'd find a velocity if i dont have an initial or final velocity. if i want to find out the speed it needs to be at 180 degrees id have theta as well as l and g but no velocity

edit: i had 0=u^2-2gl(1-cosphi) then switched final velocity to make initial velocity on the outside to try and find the velocity needed for it to reach the top where its velocity would be zero but i got a math error

Last edited: Feb 4, 2014
17. Feb 4, 2014

### tiny-tim

what is the lowest possible value of the tension if the mass moves in a complete circle?

18. Feb 4, 2014

### navm1

does it have to be at least 9.8N?

edit: i think if it reaches 0 near the top then the string will slack and it will start free falling affected only by mg

Last edited: Feb 4, 2014
19. Feb 4, 2014

### haruspex

Those are two different answers. Does the tension need to be at least mg or at least zero (+ a tiny bit)?

20. Feb 5, 2014

### tiny-tim

(just got up :zzz:)
that's much closer!

but the string will be slack only if the tension is exactly zero …

(it can't be less than zero, and) if it's greater than zero, then the string must be its full length: which means that the mass stays on the circle!

(this is very similar to the fact that something will leave a surface only if the normal force is exactly zero)