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(B) For a disc rolling purely with constant angular velocity $\omega$,the velocity of the center of mass is $v_{cm} = R\omega$. For point $P$ at the s

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$45^o$

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(B) For a disc rolling purely with constant angular velocity $\omega$,the velocity of the center of mass is $v_{cm} = R\omega$. For point $P$ at the same horizontal level as the center $C$,the velocity vector $\vec{v}_P$ is the vector sum of the translational velocity $\vec{v}_{cm}$ and the tangential velocity $\vec{v}_{rot} = \vec{\omega} 	imes \vec{r}$. Since $\vec{v}_{cm}$ is horizontal and $\vec{v}_{rot}$ is vertical at point $P$,the magnitude is $v_P = \sqrt{v_{cm}^2 + (R\omega)^2} = \sqrt{2} v_{cm}$. The velocity vector makes an angle of $45^o$ with the horizontal. The acceleration of point $P$ is purely centripetal,directed towards the center $C$,which is horizontal. Thus,the angle between the velocity vector (at $45^o$ to the horizontal) and the acceleration vector (horizontal) is $45^o$.

$A$ disc of radius $R$ is rolling purely on a flat horizontal surface with a constant angular velocity $\omega$. The angle between the velocity and acceleration vectors of point $P$ (which is at the same horizontal level as the center $C$) is

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