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# M3 panic! (help with a few questions) watch

1. My exam's in a few hours, and I just figured out I'm extremely crappy at horizontal circular motion questions.

Here are a couple I can't do:

A particle, of mass m, is suspended from a fixed point O by a light elastic string, of natural length l and modulus X. The particle moves with constant angular speed W in a horizontal circular path with the string making a constant angle @ with the downward direction of the vertical. Show that e, the extension of the string, is given by
Xe = W²lm(l+e)
Deduce that the motion described cannot take place unless W²<X/(lm)
Show further that, for a given value of W, the depth of the horizontal circle below O is independant of X.

A conical pendulum consists of a light inextensible string which has one end attached to a fixed point A. A particle P of mass m is attached to the other end of the string. The particle P moves with constant speed comleting 2 orbits of its circular path every second and the tension in the string is 2mg.
Find to the nearest cm
a) the radius of the circular path of P
b) the length of the string

Help!!!
2. if it's from one of the past papers why dont you try neos site if it is back to operate.
3. A particle, of mass m, is suspended from a fixed point O by a light elastic string, of natural length l and modulus X. The particle moves with constant angular speed W in a horizontal circular path with the string making a constant angle @ with the downward direction of the vertical. Show that e, the extension of the string, is given by
Xe = W²lm(l+e)
Resolving up: [email protected] = mg
Resolving horiontally: F=ma
[email protected] = mrW^2
Hookes law gives: T=Xe/l

Substitution gives: (Xe/l)[email protected] = mrW^2.
Taking the total length of the string to be l+e, hence the radius is (l+e)[email protected] (resolving dimensions horizontally)

(Xe/l)[email protected] = m(l+e)[email protected]^2
Xe/l = m(l+e)w^2
Xe = lm(l+e)w^2

Deduce that the motion described cannot take place unless W²<X/(lm)
For a tension to be present there must be an extension. Hence e>0.
Taking Xe=lmw^2(l+e)
Xe = l^2mw^2 + lmw^2e
Xe - lmw^2e = l^2mw^2
e[X-lmw^2] = l^2mw^2
e = [l^2.mw^2]/[X-lmw^2]
E>0
Hence: [l^2.mw^2]/[X-lmw^2]>0
As the numerator is always >0:
X - lmw^2 > 0
X > lmw^2
or, X/lm > w^2 and w^2 < x/lm

Show further that, for a given value of W, the depth of the horizontal circle below O is independant of X.
Xe=lm(l+e)w^2
w^2 = Xe/lm(l+e), T=mg/[email protected], Xe/l = mg/[email protected], X = mgl/[email protected]
w^2 = [e/lm(l+e)][mgl/[email protected]]
w^2 = [g/(l+e)[email protected]]
[email protected]^2 = g/(l+e)
Using trigonometry, let the depth by y. [email protected] = y/(l+e)
yw^2/(l+e) = g/(l+e)
yw^2 = g
y = g/w^2.
4. A conical pendulum consists of a light inextensible string which has one end attached to a fixed point A. A particle P of mass m is attached to the other end of the string. The particle P moves with constant speed comleting 2 orbits of its circular path every second and the tension in the string is 2mg.
Find to the nearest cm
a) the radius of the circular path of P
2orbits per sec means 4pi radians per sec [remember w is always in radians per second]
w=4pi.
Resolving vertically, let the angle theta be the angle between the string and the vertical: [email protected] = mg
[email protected] = 1/2.
Resolving horizontally: f=ma
[email protected] = mr(4pi)^2
[email protected] = r(4pi)^2
[email protected]/(4pi)^2
r=0.107489...
r=11cm.

b) the length of the string
[email protected] = r
y = r/[email protected]
y = 2g/(4pi)^2
y =0.12412m
y=12cm.

Feel free to ask if you have any other questions.

Edit: Btw, you don't need to panic. We proabably won't have anything like the first question in the exam. The most they'll ask is for you to use the fact T>0 to prove an inequality, which is good because it serves as a quick answer check.
5. Thanks Gaz! You're a life saver. I just tried doing a few more horizontal circular motion questions, and I got the right answers.

Two questions, though... I don't really understand what "depth" is. Why are you using [email protected]? And in the second question, where'd the [email protected] come from?
6. (Original post by shift3)
Thanks Gaz! You're a life saver. I just tried doing a few more horizontal circular motion questions, and I got the right answers.

One question though... I don't really understand what "depth" is. Why are you using [email protected] (both questions)?
No worries, it's all practice for me too.
Depth is the distance of the centre of the horizontal circle below the point the string is fastened to.
I've let y=the length of the string.
Hence [email protected] would be the depth and [email protected] would be the radius.
7. Oh. I thought y was some other variable.

Thanks again... I'm repping you ASAP.
8. (Original post by shift3)
Oh. I thought y was some other variable.

Thanks again... I'm repping you ASAP.
Heh, no worries. I find that if i'm getting into trouble with a question i'll often make up a variable to simplify things as i can always cancel it out later.
I hope the exam goes well. The butterflies are setting in now.

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