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AravindG
 4 years ago
a sphere of mass m is attached to a spring and placed on an inclined plane as shown in fig .If a sphere is left free what is maximum extension of the spring if friction allows only rolling of sphere about a horizontal diameter
AravindG
 4 years ago
a sphere of mass m is attached to a spring and placed on an inclined plane as shown in fig .If a sphere is left free what is maximum extension of the spring if friction allows only rolling of sphere about a horizontal diameter

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if so, F = g/sin theta

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1What exactly is the question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327765928827:dw mgsin@ is the net force*x=1/2kx^2dw:1327765781010:dw

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0salini right side not clear

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327709932298:dw Let Initial height be h. Two methods 1) The longer one with forces. At any instant, ma = mgsin@  kx  f And Torque, fr = (2/5 m r ^2) * a/r => f = 2/5ma 7/5 m vdv/dx = mgsin@  kx Integrating from initial v = 0 to finally when again v = 0 for max extension and x (displacement and extension) from 0 to x max , => x = (2mgsin@/k) ^1/2 Or you can simply obtain this expression frrom energy conservation. As change in Gravitational P.E. = Gain in Spring P.E mgxsin@ = 1/2kx^2 Only to get same answer. Just remember that mac extension is not the point at which forces are balanced because at that instant the body possess velocity hence goes further down only to perform SHM  you could have solved it using that concept too to find the amplitude.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And I assumed it to be a solid sphere. hence Moment of inertia was 2/5 mr^2
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