Bengaltiger
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Report Thread starter 13 years ago
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A small bead B of mass m can slide on a smooth circular wire of radus a which is fixed in a vertical plane. B is attached to one end of a light elastic string, of natural length (3/2)a and modulus of elasticity mg. The other end of the string attached to a fixed point A which is vertically above the centre C of the circular wire with AC = 3a.

Show that there is a stable equilibrium position of the system when \theta, the angle which the raduis through B makes with the downward vertical, satisfies cos\theta = - (1/6)
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dvs
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Report 13 years ago
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Take the reference line to be the bottom of the circle. Then:
PE = mga(1 - cos(t))

Now let's find the extension x in the sting. Using the cosine law:
AB^2 = a^2 + AC^2 - 2a(AC)cos(pi - t)
= 10a^2 + 6a^2 cos(t)

But AB = 1.5a + x, so:
x = sqrt(10a^2 + 6a^2 cos(t)) - 1.5a
x^2 = 10a^2 + 6a^2 - 3a^2 sqrt(10 + 6cos(t)) + 2.25a

Hence:
EPE = (mg/3a) * (10a^2 + 6a^2 - 3a^2 sqrt(10 + 6cos(t)) + 2.25a^2)
= mga ((73/12) - sqrt(10 + 6cos(t)))

So, the total potential energy of the system is:
V = mga(1 - cos(t)) + mga ((73/12) - sqrt(10 + 6cos(t)))
=> dV/dt = mga sin(t) - 3 [(mga sin(t))/sqrt(10 + 6cos(t))]

If cos(t)=-1/6, then sin(t)=sqrt(35/36). So:
dV/dt = mga sqrt(35/36) - 3 [(mga sqrt(35/36))/3] = 0

Hence it gives a position of equilibrium. I'll let you verify that it's stable.
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Bengaltiger
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#3
Report Thread starter 13 years ago
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(Original post by dvs)
But AB = 1.5a + x, so:
x = sqrt(10a^2 + 6a^2 cos(t)) - 1.5a
x^2 = 10a^2 + 6a^2 - 3a^2 sqrt(10 + 6cos(t)) + 2.25a

If x = \sqrt (10a^2 + 6a^2 cos\theta) - \frac {3a}{2}

Shouldnt x2 = (10a^2 + 6a^2 cos\theta) - 3a^2 \sqrt( 10 + 6cos\theta) + \frac {9a^2}{4}

and not be x2 = (10a^2 + 6a^2) - 3a^2 \sqrt( 10 + 6cos\theta) + \frac {9a^2}{4}

If that is correct, then i cant get my correct awnser :confused:

EDIT: I just realised it does simplyfy and works.
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