Physics 1: Find an equation relating θ1 and θ2 - in which a particle falls of a cylinder?
A particle moves from rest at point A on the surface of a smooth circular cylinder of radius R. At B the particle leaves the cylinder. Find the equation relating θ1 and θ2.
Diagram description:
Semi-circle of radius R.
Point A - is θ1 degrees past 90° or (from the y-axis)
Note: There is a gap between θ1 and θ2
Point B - is between the θ2 degrees up from 180° or (from the x-axis)
***I hope that my description of the diagram is sufficient.
***Thank for all of your help! I really appreciate it!!***
Update:**EDIT:
Point B - is θ2 degrees up from 180° or (from the x-axis)
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Verified answer
Potential energy at A = mgRcosθ1
Potential energy at B = mgRsinθ2
Gain of kinetic energy = loss of potential energy
=> (1/2) mv^2 = mgR (cosθ1 - sinθ2)
=> mv^2/R = 2mg (cosθ1 - sinθ2) ... ( 1 )
At point B, the particle will lose contact if the centripetal force = radial component of weight
=> mv^2/R = mgsinθ2 ... ( 2 )
From ( 1 ) and ( 2 ),
mgsinθ2 = 2mg (cosθ1 - sinθ2)
=> 2cosθ1 - 3sinθ2 = 0.