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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.

## 4 Answers

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.

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) ]

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

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