A point A is located on the rim of a wheel of radius `R=0.50m` which rolls without slipping along a horizontal surface with velocity `v=1.00m//s`. Find:
(a) the modulus and the direction of the acceleration vector of the point A,
(b) the total distance s traveresed by the point A between the two successive moments at which it touches the surface.

Explanation

The plane motion of a solid can be imagined as the combination of translation of the C.M. and rotation about `C.M`
So, we may write `vec(v_A)=vec(v_C)+vec(v_(AC))`
`=vec(v_C)+vecomegaxxvec(r_(AC))` (1) and
`vec(w_A)=vec(w_C)+vec(w_(AC))`
`=vec(w_C)+omega^2(-vec(r_(AC)))+(vecbetaxxvec(r_(AC)))` (2)
`vec(r_(AC))` is the position of vector of A with respect to C.
In the problem `v_C=v=` constant, and the rolling is without slipping i.e., `v_C=v=omegaR`,
So, `w_C=0` and `beta=0`. Using these conditions in Eq. (2)
`vec(w_A)=omega^2(-vecr_(AC))=omega^2R(-hatu_(AC))=v^2/R(-hatu_(AC))`
Here, `hatu_(AC)` is the unit vector directed along `vecr_(AC)`.
Hence `w_A=v^2/R` and `vecw_A` is directed along `(-hatu_(AC))` or directed toward the centre of the wheel.
(b) Let the centre of the wheel move toward right (positive x-axis) then for pure tolling on the rigid horizontal surface, wheel will have to rotate in clockwise sense. If `omega` be the angular velocity of the wheel then `omega=(v_C)/(R)=(v)/(R)`.
Let the point A touches the horizontal surface at `t=0`, further let us locate the point A at `t=t`,
When it makes `theta=omegat` at the centre of the wheel.
From Eqn. (1) `vecv_A=vec_C+vec(omega)xxvec(r_(AC))`
`=vveci+omega(-veck)xx[Rcostheta(-vecj)+Rsin theta(-veci)]`
or, `vec(v_A)=vveci+omegaR[cos omegat (-veci)+sin omega tvecj]`
`=(v-cos omegat)veci+v sin omega tvecj` (as `v=omegaR`)
So, `v_A=sqrt((v-vcos omega)^2+(v sin omegat)^2)`.
`=vsqrt(2(1-cosomegat))=2vsin (omegat//2)`
Hence distance covered by the point A during `T=2pi//omega`
`s=int v_Adt=underset(0) overset(2pi//omega)int=2v sin (omegat//2)dt=(8v)/(omega)=8R`.