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

Let us fix the co-ordinate axis xyz as shown in the figure. As the ball rolls without slipping along the rigid surface so, on the basis of the solution of problem 1.52:
Thus `{:(vecv_0=vecv_c+vecomegaxxvecr_(oc)=0,,,,),(v_c=omegaR and vecomegauarruarr(-veck),as,vecv_cuarruarrveci,,):}}` (1)
and `{:(vecomega_c+vecbetaxxvecr_(oc)=0,,,,),(w_c=betaR and vecbetauarruarr(-veck),as,vecw_cuarruarrveci,,):}}`
At the position corresponding to that of figure., in accordance with the problem,
`w_c=w`, so `v_c=wt`
and `omega=(v_c)/(R)=(wt)/(R)` and `beta=w/R` (using 1)
(a) Let us fix the co-ordinate system with the frame attached with the rigid surface as shown in the figure.

As point O is the instantaneous centre of rotation of the ball at the moment shown in figure.
so, `vecv_0=0`,
Now, `vecv_A=vecv_C+vecomegaxxvecr_(AC)`
`=v_Cveci+omega(-veck)xxR(vecj)=(v_C+omegaR)veci`
So, `vecv_A=2v_Cveci=2wt veci` (using 1)
Similary `vecv_B=vecv_C+vecomegaxxvecr_(BC)=vecv_Cveci+omega(-veck)xxR(veci)`
`=v_Cveci+omegaR(-vecj)=v_Cveci+v_C(-vecj)`
So, `v_B=sqrt2v_c=sqrt2wt` and `vecv_B` is at an angle `45^@` from both `veci` and `vecj` (figure)

(b) `vecw_0=vecw_C+omega^2(-vecr_(oc))+vecbetaxxvecr_(OC)`
`=omega^2(-vecr_(OC))=(v_C^2)/(R)(-hatu_(OC))` (using 1)
where `hatu_(OC)` is the unit vector along `vecr_(OC)`
so, `w_0=(v_0^2)/(R)=(w^2t^2)/(R)` (using 2) and `vecw_0` is directed towards the centre of the ball
Now `vecw_A=vecw_C+omega^2(-vecr_(AC))+vecbetaxxvecr_(AC)`
`=wveci+omega^2R(-vecj)+beta(-veck)xxRvecj`
`=(w+betaR)veci+(v_c^2)/(R)(-vecj)` (using 1) `=2wveci+(w^2t^2)/(R)(-vecj)`
So, `w_A=sqrt(4w^2+(w^4t^4)/(R^2))=2wsqrt(1+((wt^2)/(2R))^2)`
Similarly `vecw_B=vecw_c+omega^2(-vecr_(BC))+vecbetaxxvecr_(BC)`
`=wveci+omega^2R(-veci)+beta(-veck)xxR(veci)`
`=(v-(v_C^2)/(R))veci+betaR(-vecj)` (using 1)
`=(w-(w^2t^2)/(R))veci+w(-vecj)` (using 2)
So, `w_B=sqrt((w-(w^2t^2)/(R))^2+w^2)`