A fixed hollow frictionless cone is positioned with its tip pointing down. A par

Question

A fixed hollow frictionless cone is positioned with its tip pointing down. A particle is released from rest on the inside surface. After it has slid part way down to the tip, it bounces elastically off a platform. The platform is positioned at a 45 degree angle along the surface of the cone, so the particle ends up being deflected horizontally along the surface (in other words, into the page in Fig. 5.30). If the resulting motion of the particle is a horizontal circle around the cone, what is the ratio of initial height of the particle to the height of the platform?

Please show all steps and calculations.

If you upload an image please make sure everything is clear and your writing is legible.

A fixed hollow frictionless cone is positioned with its tip pointing down. A particle is released from rest on the inside surface. After it has slid part way down to the tip, it bounces elastically off a platform. The platform is positioned at a 45 degree angle along the surface of the cone, so the particle ends up being deflected horizontally along the surface (in other words, into the page in Fig. 5.30). If the resulting motion of the particle is a horizontal circle around the cone, what is the ratio of initial height of the particle to the height of the platform? If you upload an image please make sure everything is clear and your writing is legible.

Explanation / Answer

m g (H - h) = 1/2 m v^2

Where H and h are measured vertically from the tip of the cone

H - h = v^2 / g

Let theta = 1/2 the angle of the vertex of the cone

Fc = m g tan theta     where Fc is the centripetal force component of the weight m g

m v^2 / R = m g tan theta

v^2 / g = R tan theta

H - h = R tan theta

Also R / h = tan theta from the geometry or R = h tan theta

H - h = h tan^2 theta

H / h = 1 + tan^2 theta

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