A ball of radius `R=10.0cm` rolls without slipping down an inclined plane so that its centre moves with constant acceleration
`w=2.50cm//s^2`, `t=2.00s` after the beginning of motion its position corresponds to that shown in figure. Find:
(a) the velocities of the points A, B, and O,
(b) the accelerations of these points.

Explanation

Linear motion: velocity of center of mass after time `t`
`v_(c)=v_(c.m.)=0+a_(c.m.)t=a_(0)t`
Angular motion: `omega=omega_(0)+alpha t=0+alpha t=(a_(0))/(R ) t`
(In pure rolling, `a_(c.m.)=Ralphaimpliesa_(0)=Ralpha`)
Rolling without slipping about center of mass `C`

(`a`)
`A:` `= v_(0)+Romega=v_(0)+v_(0)=2 a_(0)t`
`O:` `=v_(0)-Romega=v_(0)-v_(0)=0`
`B:` `= v_(B)=sqrt2 v_(0)=sqrt a_(0)t`
`theta=45^(@)`
(`b`) Tangential acceleration:

`A:` `a_(A)=a_(0)+R alpha=2 a_(0)`
`O:` `a_(0)=0`
`O:` `a_(B)=sqrt2 a_(0)`
`theta=45^(@)`
Centripetal acceleration

Total acceleration

`A`: `a_(A)=sqrt((2a_(0))^(2)+((v_(0)^(2))/(R))^(2))=sqrt(4a_(0)^(2)+((a_(0)^(2)t^(2))/(R))^(2))`
`O`: `a_(0)=(v_(0)^(2))/(R)=(a_(0)^(2)t^(2))/(R)`
`B`: `a_(B)=sqrt(a_(0)^(2)+(a_(0)-(v_(0)^(2))/(R))^(2))=sqrt(a_(0)^(2)+(a_(0)-(a_(0)^(2)t^(2))/(R))^(2))`
Alternative approach to find velocities and acceleration of points `A`, `B` and `O`:
Rolling without slipping is equivalent to pure rotation at the contact point
`v_(C)= R omega=a_(0)t`
`v_(A)=2Romega=2v_(0)=2a_(0)t`
`v_(B)=sqrt2 R omega =sqrt2v_(0)=sqrt2a_(0)t`
`v_(O)=0`
`a_(c.m.)=a_(c)=a_(0)=R alpha`
`a_(A)=2R alpha=2a_(0)`
`a_(B)=sqrt2 R alpha=sqrt2a_(0)`
`a_(O)=0`