Q:

Velocity of a point on the surface of a sphere rolling with slipping

Here a sphere is rolling on a frictional surface but slipping takes place. Hence, the velocity of the bottom-most point is not equal to $0$. Hence, $v_{cm} \ne R\omega$. The angular velocity of the sphere is $\omega$.
Let's say I want to find the tangential velocity ($v_t$) and thereafter the total velocity of another point on the surface of the sphere. Is this relation correct? $$v_t = R\omega + v_{cm} \cos(\theta)$$ where $\theta$ is the angle between $v_t$ and $v_{cm}$.
If this is correct, does it mean that even though there is slipping, we can use $v = R\omega$ for finding the linear velocity due to the angular velocity of any other point except the bottom-most one.

A:

1.  The condition for rolling without slipping is $V_{cm}$=.
2. Irrespective of whether a body is slipping or not, velocity due to rotation of any point can be related as $V$=, where R is the radius of curvature and  is the angular velocity. So yes, the relation holds good and the magnitude of tangential velocity can be found using the formula mentioned in your question.
3. The total velocity of a point can be obtained by taking vector sum of $V_{cm}$ and $V$. The magnitude can be found out as
$$V_{total}=\sqrt{V_{cm}^{2}+V^{2}+2V_{cm}.V.cos(\theta )} $$

