A particle A moves along a circle of radius `R=50cm` so that its radius vector r relative to the point O(figure) rotates with the constant angular velocity `omega=0.40rad//s`. Find the modulus of the velocity of the particle, and the modulus and direction of its total acceleration.

Explanation

Let us fix the co-ordinate system at the point O as shown in figure, such that the radius vector `vecr` of point A makes an angle `theta` with x axis at the moment shown.
Note that the radius vector of the particle A rotates clockwise and we here take line ox as reference line, so in this case obviously the angular velocity `omega=(-(d theta)/(dt))` taking anticlockwise sense of angular displacement as positive.
Also from the geometry of the triangle OAC
`(R)/(sin theta)=(r)/(sin (pi-2 theta))` or, `r=2 R cos theta`.
Let us write,
`vecr=rcosthetaveci+rsin theta vecj=2R cos^2 theta veci+R sin 2 theta vecj`
Differentiating with respect to time.
`(vec(dr))/(dt)` or `vecv=2R2cos theta(-sin theta)(d theta)/(dt)veci+2Rcos 2 theta(d theta)/(dt)vecj`
or, `vecv=2R((-d theta)/(dt))[sin 2 theta veci-cos 2 thetavecj]`
or, `vecv=2R omega (sin 2 theta veci-cos^2 theta vecj)`
So, `|vecv|` or `v=2omegaR=0*4m//s`.
As `omega` is constant, v is also constant and `w_t=(dv)/(dt)=0`,
So, `w=w_n=(v^2)/(R)=((2omegaR)^2)/(R)=4omega^2R=0*32m//s^2`
Alternate: From the fig. the angular velocity of the point A, with respect to centre of the circle C becomes
`-(d(2 theta))/(dt)=2((-d theta)/(dt))=2omega=const ant`
Thus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius R with constant angular velocity `2omega`.
Hence `v=2omegaR` and
`w=w_n=v^2/R=((2omegaR)^2)/(R)=4omega^2R`