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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.```
8 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.

```
8 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) ]
```
8 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
```
8 years ago
```							thnx a lot
```
8 years ago
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