Mechanics - Circle Geometry (1 Viewer)

[ishq]

A topic I simple detest. Why can't we breeze through with only resisted motion? *sigh*

Anyhow - question:

A particle hangs by a light inextensible string of length a from a fixed point O, and a second particle of the same mass hangs from the first by an equal string. The whole system moves with constant angular speed omega about the vertical through O. The upper and lower strings make constant angles alpha and beta respectively with the vertical.

Show that:
tan alpha = p(sin alpha + 0.5sin beta), and that tan beta = p(sin alpha + sin beta), where p = a(omega^2)/g.


I don't have a clue - I've got the solution, but I don't understand that either. Please help! Thanks!

 

[KFunk]

Hmm, I'd help you out if I knew what they were describing . By the sounds of it the second particle is attached to the first particle by a string. I don't see how the second particle could have a circular motion parallel to the whole system if alpha and beta are different angles. In summary, I don't have a clue. Anyone up for making a diagram ?

 

[ishq]

Hmm, I'd help you out if I knew what they were describing . By the sounds of it the second particle is attached to the first particle by a string. I don't see how the second particle could have a circular motion parallel to the whole system if alpha and beta are different angles. In summary, I don't have a clue. Anyone up for making a diagram ?

I made one - which miraculously was correct according to the solutions...

let me scan + attach...


Its too large.......

 

[KFunk]

I made one - which miraculously was correct according to the solutions...

let me scan + attach...

Thanks, you're doing a better job than me if you can get the diagram part.

 

[ishq]

Here it is:

http://photobucket.com/albums/b363/vasu_vasu/?action=view&current=circlegeometry.jpg

 

[Sepulchres]

Hmm, I did one of these with 3 masses attached to each other. From the patel book I htink it was. Let me find it.

 

[underthebridge]

Ok, I'll write alpha as A and beta as B, the length of the string is a. The radius that the first particle is spinning around is r and the larger radiu that the second particle is spinning around is R.

Using right angled trig,
r = asinA
R = a(sinA + sinB)

Now, I'll let the tension in the first string be T and the tension in the second string be U. Note that tensions U and T are both operating on the first particle. Now, by resolving forces on both particles horizontally and vertically:

First Particle:
TsinA - UsinB = mrw^2 .........1
TcosA - UcosB = mg .............2

Second Particle:
UsinB = mRw^2 ..................3
UcosB = mg ...................4

now sub 4 into 2,
TcosA - mg = mg
TcosA = 2mg
T= 2mg/cosA now sub this and 3 into 1 and you get:

2mg.tanA - mRw^2 = mrw^2
2mg.tanA = mw^2 (R + r)
2tanA = (w^2/g)(2asinA + asinB) remembering what we worked out for R and r
tanA = p(sinA + 0.5sinB)

Now, for the other one, from 4 we get:
U= mg/cosB sub this into 3
mg.tanB = mw^2(asinA + asinB)
tanB = (aw^2/g)(sinA + sinB)
tanB = p(sinA + sinB)

As required.

 

[Sepulchres]

Now that I look at it, its quite different to the one I was talking about... Nicely done though.

 

[underthebridge]

Hmmm, with your one, I would have probly done a similar method, although it would probably involve some nastier simultaneous equations work.

 

[ishq]

thanks underthebridge!!

i was having problems with the initial resolving of forces - you cleared it up!

merci!