Question

2.
For a particle in uniform circular motion, the acceleration à at a point P (R, 0) on the circle of radius R
is (Here is measured from the x-axis)
[AIEEE - 2010, 4/144]
(1) - coso i +ë sin oj
(2) -
sino + p cos oj
(3) -
cos e irsin e
of macose me and my are moving in circles of radii r1 and r2, respectively. Their speeds are​

Answer 1

Given info : For a particle in uniform circular motion the acceleration a at a point P(R,θ) on the circle of radius R is (here 'v' is the speed of the particle and θ is measured from the x-axis)

Solution : position of particle in vector form is given by, s = Rcosθ i + Rsinθ j

Here, θ = ωt

So, s = Rcosωt i + Rsinωt j

Now differentiating with respect to time,

ds/dt = -ωRsinωt i + ωR cosωt j

Again differentiating with respect to time we get,

d²s/dt² = -ω²R cosωt i - ω²R sinωt j

we know, v = ωR ⇒ω = v/R

So, d²s/dt² = a = -v²/R cosωt i - v²/R sinωt j

Now putting ωt = θ

So, a = -v²/R cosθ i - v²/R sinθ j

Therefore the acceleration a at a point P(R,θ) on the circle of radius R is -v²/R cosθ i - v²/R sinθ j