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A particle of mass 'm' moves along a circle of radius 'R'. Find the moduus of the average vector of the force acting on the particle over the distance equal to a quarter of the circle, if the particle moves


(a) uniformly with velocity 'v'


(b) with constant tabgential acceleration 'wt' , the initial velocity being equal to 0.

5 years ago

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Answers : (4)

										

(A)There is only one force acting on the particle which is centripetal force towards the centre of the circle.


                              Fc=mv2/R


This should be equal to the average force acting on the particle.


 

5 years ago
										

force = change in momentam =dp/dt = mdv/dt      ................1


 


let circle is in xy plane with center at origin & initiall particle is directed towards +x axis the


U = vi       (in vector)


finally when it complets one quater circle then its velocity becomes perpendicular to x axis ...


V = vj      (j is unit vector along y direction)


change in momentam = m(vi - vj)    ............2


 


we can use speed = d/time                 (for constant speed)


distance = piR/2(arc of quater circle) ,


 speed = v


so time (t) = piR/4v        ..........3


now plugging values from eq 2,3 in 1


F = (4vm/piR) (vi-vj)


modF = [ 4root2(v2m/piR) ]

5 years ago
										

if accleration is constant then we can use v = u + at


u = o , a = wt so


v = (wt)t 


force = mdv/dt = m (v-u)/t = mwt(t/t) = mwt

5 years ago
										

thnx a lot

5 years ago

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