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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.

Vinay Arya
37 Points
10 years ago

(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.

509 Points
10 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) ]

509 Points
10 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

Aiswarya Ram Gupta
35 Points
10 years ago

thnx a lot