Mechanics problem

A horizontal circular disc of mass MM is free to rotate about a vertical axis through a point on its rim. If a dog(or pig if you wish) of mass mm walks once around the rim, by what angle does the rim turn through? Given that Mm=83\frac{M}{m}=\frac{8}{3}.

Note by Azimuddin Sheikh
7 months, 3 weeks ago

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A uniform circular disk of mass MM and radius RR is confined to the xyxy plane. It rotates with no frictional losses about the zz axis through a point on its edge. Gravity is into the page, and is thus irrelevant to this situation.

A bead of mass mm slides along a friction-less track on the circumference of the disk. The angle θ\theta is the angle between the positive xx axis and a line from the origin to the center of the disk. The angle ϕ\phi is the angle between the positive xx axis and a line from the the center of the disk to the bead.

Coordinates and velocities of the bead:

x=Rcosθ+Rcosϕy=Rsinθ+Rsinϕx˙=Rsinθθ˙Rsinϕϕ˙y˙=Rcosθθ˙+Rcosϕϕ˙x = R \, cos \theta + R \, cos \phi \\ y = R \, sin \theta + R \, sin \phi \\ \dot{x} = -R \, sin \theta \, \dot{\theta} - R \, sin \phi \, \dot{\phi} \\ \dot{y} = R \, cos \theta \, \dot{\theta} + R \, cos \phi \, \dot{\phi}

Kinetic Energy of the system:

E=12m(x˙2+y˙2)+12(12MR2+MR2)θ˙2=(12mR2+34MR2)θ˙2+12mR2ϕ˙2+mR2sinθsinϕθ˙ϕ˙+mR2cosθcosϕθ˙ϕ˙E = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) + \frac{1}{2} \Big(\frac{1}{2} M R^2 + M R^2 \Big ) \dot{\theta}^2 \\ = \Big (\frac{1}{2} m R^2 + \frac{3}{4} M R^2 \Big) \dot{\theta}^2 + \frac{1}{2} m R^2 \dot{\phi}^2 + m R^2 \, sin\theta \, sin\phi \, \dot{\theta} \, \dot{\phi} + m R^2 \, cos\theta \, cos\phi \, \dot{\theta} \, \dot{\phi}

Since all of the mass is at the same height with respect to gravity, the system Lagrangian is equal to the kinetic energy.

L=EL = E

Equations of motion:

ddtLθ˙=LθddtLϕ˙=Lϕ\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\phi}}} = \frac{\partial{L}}{\partial{\phi}}

Evaluating the equations of motion yields the dynamics:

(m+32M)θ¨+m(sinθsinϕ+cosθcosϕ)ϕ¨=m(cosθsinϕsinθcosϕ)ϕ˙2m(sinθsinϕ+cosθcosϕ)θ¨+mϕ¨=m(sinθcosϕcosθsinϕ)θ˙2\Big(m + \frac{3}{2} M \Big) \, \ddot{\theta} + m(sin\theta \, sin \phi + cos\theta \, cos\phi) \, \ddot{\phi} = m (cos\theta \, sin\phi - sin\theta \, cos\phi) \dot{\phi}^2 \\ m(sin\theta \, sin \phi + cos\theta \, cos\phi) \, \ddot{\theta} + m \ddot{\phi} = m (sin\theta \, cos\phi - cos\theta \, sin\phi) \dot{\theta}^2

Initial conditions:

θ0=0θ˙0=0ϕ0=0ϕ˙00\theta_0 = 0 \\ \dot{\theta}_0 = 0 \\ \phi_0 = 0 \\ \dot{\phi}_0 \neq 0

We are interested in the angle θ\theta which has elapsed by the time ϕθ=2π\phi - \theta = 2 \pi, meaning that the bead has made a full revolution around the disk circumference. For M=83mM = \frac{8}{3} m, the simulation is not very well behaved for some reason (it doesn't converge very well as the time step is varied). I will post a problem later asking for the results for the M=5mM = 5 m case, since this case is very well behaved.

Steven Chase - 7 months, 2 weeks ago

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Yeah u r right sir , thx a lot.

Azimuddin Sheikh - 7 months, 2 weeks ago

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@Steven Chase sir u got it ?

Azimuddin Sheikh - 7 months, 2 weeks ago

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Yes, I'm going to do one more derivation to make sure I have it right. Then I'll post it. A bit of warning in advance; it doesn't boil down to a nice simple expression.

Steven Chase - 7 months, 2 weeks ago

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I didn't know the answer , likely some other info must be given to cause it to become easy .

Azimuddin Sheikh - 7 months, 2 weeks ago

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Any ideas to solve this problem?? @Mark Hennings Sir @Steven Chase sir pls share your ideas @Aaghaz Mahajan pls help ?

Azimuddin Sheikh - 7 months, 3 weeks ago

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Let's say we have a smooth track on the edge of the disk that a bead slides on. Do you regard that as being an equivalent problem?

Steven Chase - 7 months, 3 weeks ago

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Yeah exactly same @Steven Chase sir.

Azimuddin Sheikh - 7 months, 3 weeks ago

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