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# Motion in a circle Watch

1. A particle of mass m moves with constant speed u in a horizontal circle of radius 3a/2 on the inside of a fixed smooth horizontal sphere of radius 2a. Show that 9ag=2sqrt(7)u^2.

I've done:
Rcos(theta) = mg
Rsin(theta) = 2mu^2/3a

dividing both equations gives
tan(theta) = 2u^2/3ag

now the problem is I can't find tan(theta) If it was 3/sqrt(7) this would complete the question but I don't understand how to get it, its 3a/2/something
and this something I can't find..it is obvious what it is since 3a/2/sqrt(7)a/2 = 3/sqrt(7), but how do I find that it is sqrt(7)a/2 for the other side. Sorry might be clearer with a diagram..
2. I haven't checked your working, but it looks like you haven't used the fact that the particle is moving in a circle of radius 3a/2 inside a sphere of radius 2a, they'll give you theta.
3. (Original post by Hopple)
I haven't checked your working, but it looks like you haven't used the fact that the particle is moving in a circle of radius 3a/2 inside a sphere of radius 2a, they'll give you theta.
I don't understand, ive used
Rsin(theta)=2mu^2/3a this uses the fact that it is moving with speed u in a horizontal circle.
And I can't really see how to use the radius of 2a
4. (Original post by Ayakashi)
I don't understand, ive used
Rsin(theta)=2mu^2/3a this uses the fact that it is moving with speed u in a horizontal circle.
And I can't really see how to use the radius of 2a
2a is the distance from the particle to the centre of the sphere, and 3a/2 is the distance from the particle to the centre of its 'orbit'. Draw this out and you should be able to get theta.
5. (Original post by Hopple)
2a is the distance from the particle to the centre of the sphere, and 3a/2 is the distance from the particle to the centre of its 'orbit'. Draw this out and you should be able to get theta.
this is what I have done, still can't see it..
Attached Images

6. (Original post by Ayakashi)
this is what I have done, still can't see it..
2a is from the centre of the sphere to the particle as well as the radius you've drawn, it's the hypotenuse. The 3a/2 is the 'opposite'.
7. (Original post by Hopple)
2a is from the centre of the sphere to the particle as well as the radius you've drawn, it's the hypotenuse. The 3a/2 is the 'opposite'.
I tried this initially,
tan(theta) = 3a/2/X

(2a)^2 = (3a/2)^2 + X^2
X^2 = 4a^2 - 9a^2/4
=> X^2 = 7a^2/4
X = sqrt(7)a/2 ..
O, thanks, im sure I was doing this initially, but I wasn't getting the sqrt(7)a/2 o well, thanks.

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Updated: April 14, 2013
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