- #1
etotheipi
- Homework Statement
- A polygon is rolling down a hill, find the angular velocity after the ##N^{\text{th}}## impact with the hill (assume polygon has slight concavity)
- Relevant Equations
- N/A
An ##n##-sided regular polygon is rolling down a frictional ramp at angle ##\theta## to the horizontal. I define ##\beta := \frac{2\pi}{n}## as the angle at the top of each of the ##n## isosceles triangles that make up the polygon. Let ##\omega_{k, 1}## be the angular velocity just after the ##k^{\text{th}}## impact with the ramp, and ##\omega_{k, 2}## be the angular velocity just before the ##(k+1)^{\text{th}}## impact with the ramp. Finally, let the moment of inertia about the centre of the polygon be ##I_{cm} := pmr^2## and the moment of inertia about one of the vertices be ##I_{v} := qmr^2##, where ##r## is the distance from the centre to a vertex. The side length of the polygon is ##l = \sqrt{2r^2(1-\cos{\beta})}##.
At the ##(k+1)^\text{th}## impact, we conserve angular momentum about the point on the ground with which the next vertex is just about to come into contact,$$I_{cm} \omega_{k,2} + mr^2 \omega_{k,2} \cos{\beta} = I_v \omega_{k+1, 1}$$ $$(p+\cos{\beta})\omega_{k,2} = q\omega_{1, k+1}$$Between impacts, the polygon gains kinetic energy ##mgl\sin{\theta}## due to the decrease in height of its centre of mass, so we can also say that$$\frac{1}{2}I_v \omega_{k, 1}^2 + mgl\sin{\theta} = \frac{1}{2}I_v \omega_{k,2}^2$$ $$\omega_{k,2}^2 - \omega_{k, 1}^2 = \frac{2g\sin{\theta}}{qr}\sqrt{2(1-\cos{\beta})}$$I put this into the first equation to obtain$$\omega_{k+1,1} = \frac{p +\cos{\beta}}{q} \sqrt{\omega_{k,1}^2 + \frac{2g\sin{\theta}}{qr}\sqrt{2(1-\cos{\beta})}}$$I wondered if anyone can help me to solve this recurrence relation? Thanks
At the ##(k+1)^\text{th}## impact, we conserve angular momentum about the point on the ground with which the next vertex is just about to come into contact,$$I_{cm} \omega_{k,2} + mr^2 \omega_{k,2} \cos{\beta} = I_v \omega_{k+1, 1}$$ $$(p+\cos{\beta})\omega_{k,2} = q\omega_{1, k+1}$$Between impacts, the polygon gains kinetic energy ##mgl\sin{\theta}## due to the decrease in height of its centre of mass, so we can also say that$$\frac{1}{2}I_v \omega_{k, 1}^2 + mgl\sin{\theta} = \frac{1}{2}I_v \omega_{k,2}^2$$ $$\omega_{k,2}^2 - \omega_{k, 1}^2 = \frac{2g\sin{\theta}}{qr}\sqrt{2(1-\cos{\beta})}$$I put this into the first equation to obtain$$\omega_{k+1,1} = \frac{p +\cos{\beta}}{q} \sqrt{\omega_{k,1}^2 + \frac{2g\sin{\theta}}{qr}\sqrt{2(1-\cos{\beta})}}$$I wondered if anyone can help me to solve this recurrence relation? Thanks
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