- #1
Clari
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A particle m is moving in a vertical circle of radius R inside a track. There is no friction. When m is at its lowest position, its speed is v0.
Suppose v0 is 0.775 vmin [ vmin = root(5gR) ]. The particle will move up the track to some point at P at which it will lose contact with the track and travel along a path roughly like a line connected to the centre of the circle. Find the angular position theta of point P.
I solved the question by conservation of energy. energy at the bottom = 1/2 mv^2...1/2m (0.775)(5gR)
energy at the top, when the particle is momentarily at rest...mgh...
1/2m(0.775^2)(5gR) = mgh
h= 1/2(0.775^2)(5R)
h is less than 2R, so h-R will give the height of a triangle form , where theta is the angle in the triangle...
(h-R) = R sin(theta)...and I get theta = 30 degrees...but the answer is 19.5 degrees...what's wrong with my steps? please tell me...
Suppose v0 is 0.775 vmin [ vmin = root(5gR) ]. The particle will move up the track to some point at P at which it will lose contact with the track and travel along a path roughly like a line connected to the centre of the circle. Find the angular position theta of point P.
I solved the question by conservation of energy. energy at the bottom = 1/2 mv^2...1/2m (0.775)(5gR)
energy at the top, when the particle is momentarily at rest...mgh...
1/2m(0.775^2)(5gR) = mgh
h= 1/2(0.775^2)(5R)
h is less than 2R, so h-R will give the height of a triangle form , where theta is the angle in the triangle...
(h-R) = R sin(theta)...and I get theta = 30 degrees...but the answer is 19.5 degrees...what's wrong with my steps? please tell me...