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Grade: 11

                        

A block of mass M with semicircular track of radius R ,rests on a horizontal frictionless surface.A uniform cyllinder of radius r and mass m is released at the top (pont A)The cyllinder slips on the semicircular frictionless track.How far has the block moved when the cyllinder reaches the bottom (point B)of the track?[Ans :-m(R-r)/(M+m)] How fast is the block moving when the cyllinder reaches the bottom of the track?

4 years ago

Answers : (1)

Archit Uttarwar
35 Points
							
Here external force is 0 therefore,
Let x be the displacement of the wedge, 
And r is the radius of cylinder and R is radius of semicircular path
0= m(R-r-x) - Mx
x(M+m) =m(R-r)
x=m(R-r)/(M+m) 
2 years ago
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