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

(`a`) Since the speed of wheel is constant, acceleration will be due to rotation (in the form of centripetal acceleration, because the direction of velocity is changing)
`a_(c)=(v^(2))/(R ) =((1)^(2))/(0.5)=2m//s^(2)`, always directed towards the center of wheel.
(`b`) When the point has rotated an angle `theta`, its velocity can be calculated in the following manner:

(`i`) Assuming rolling without slipping about the center of mass `O`




`v_(A)=sqrt(v^(2)+v^(2)+2vxxvcostheta(180-theta))`
`=sqrt(2v^(2)-2v^(2)costheta)`
`=sqrt2vsqrt(1-(1-2sin^(2)((theta)/(2))))=2vsin((theta)/(2))`
`theta=omega t`
`v_(A)=2vsin((omega t)/(2))=2Romegasin((omegat)/(2))`
Alternatively the velocity of point `A` can be calculated as follows.
(`ii`) Rolling without slipping about the center of mass is equivalent to pure rotation about the contact point of wheel and surface

`v_(A)=(AP)omega, alpha=90-(theta)/(2)`
By the sine rule
`(AP)/(sintheta)=(OA)/(sinalpha)=(OA)/(sin(90-(theta//2))`
`(AP)/(2sin((theta)/(2))cos((theta)/(2)))=(R)/(cos((theta)/(2)))`
`AP=2Rsin((theta)/(2))`
`v_(A)2Romegasin((theta)/(2))=2Romegasin((omegat)/(2))`
Distance traveled `A` in small time `dt`
`ds=v_(A)dt=2Romegasin((omegat)/(2))dt`
Total distance travelved in one rotation, i.e. `T=2pi//omega`
`s=int_(0)^(2pi//omega)2Romegasin((omegat)/(2))dt`
`=2Romega|-cos((omegat)/(2))|_(0)^(2pi//omega)/((omega)/(2))`
`=4R{[-cos((omegaxx2pi)/(2omega))]-[(-costheta)]}`
`=4R(1+1)=8R`

Distance `A_(1)A_(2)A_(3)A_(4)A_(5)=8R`