# Conversion of linear velocity to angular velocity and vice versa

This is very similar to this question, with the main difference being that it doesn't assume that $$\vec v=\mathbf0$$.

Assume that you have a rolling and sliding sphere with radius and mass 1 with a uniform density of 1, on a frictionless surface with linear velocity $$\vec v$$ and angular velocity $$\vec\omega$$.

At time $$t=0$$, the surface has a coefficient of friction of $$\mu$$ and coefficient of rolling friction $$C_{rr}$$.

1. What are the linear and angular velocities of the sphere at time $$t$$?
2. (Less important) Assume there is another pulling force $$\vec g$$. What is the new answer to 1?
• if you have a sliding ball on a surface with friction "No energy is lost to heat or anything like that" is impossible.It is only possible if you have no sliding so $\omega= \frac{v}{r}$ Commented Feb 24 at 18:12
• If it is sliding and there is a coefficient of friction then indeed energy is lost to heat and friction. Commented Feb 24 at 22:39
• Making the mass and radius equal to one and setting the density equal to one is a conflicting set of constraints that are not needed as it does not simplify the problem at all. Commented Feb 24 at 22:40
• @JohnAlexiou I want the general case. That was on purpose. Commented Feb 25 at 6:42
• @nonhuman If you want the general case, then you want mass$=m$, radius$=r$, and so on, you are instead getting the highly unusual special case where everything equals 1. However, since density equals mass divided by area, or in your case $1/\pi$; you can't have mass and radius equal to one while also having density equal to one. It is contradictory as opposed to "general". Commented Feb 25 at 12:05

## Dynamics

The object of mass $$m$$, and mass movement of inertia (sphere) of $${\rm I} = \tfrac{2}{5} m R^2$$ has the following momentum values

\begin{aligned} \vec{p} & = m \vec{v} & & \text{linear momentum} \\ \vec{L} & = {\rm I}\, \vec{\omega} & & \text{angular momentum} \\ \end{aligned} \tag{1}

where $$\vec{v}$$ is the velocity vector of the center of mass, and $$\vec{\omega}$$ is the rotational velocity vector of the body.

Note for a general shape $$\rm I$$ is a 3×3 matrix, and $${\rm I}^{-1}$$ the matrix inverse.

Since Newton's 2nd law connects external forces/torques applied to a body, to the changes in its momentum with

\begin{aligned} \vec{F} & = \tfrac{\rm d}{{\rm d}t} \vec{p} & & \text{net force} \\ \vec{\tau} & = \tfrac{\rm d}{{\rm d}t} \vec{L} & & \text{net torque} \\ \end{aligned} \tag{2}

we track the body's momentum state through time, and we extract the velocity by reversing (1) as

\begin{aligned} \vec{v} & = \tfrac{1}{m} \vec{p} & & \text{velocity} \\ \vec{\omega} & = {\rm I}^{-1} \vec{L} & & \text{rotation}\\ \end{aligned} \tag{3}

## Kinematics

Suppose the ball is sliding on a horizontal surface with normal direction vector $$\hat{n}$$ pointing upwards against gravity. In that case, the ball's velocity at the contact point is restricted to be in-plane only.

$$\hat{n} \cdot ( \vec{v} + R \hat{n} \times \vec{\omega}) = 0$$

simplified to

$$\hat{n} \cdot \vec{v} = 0 \tag{4}$$

which is intuitive as the velocity of the center of mass must be in-plane also.

## Friction

Sliding friction acts in a direction opposite of the relative slipping of the contact, and in this case

$$\vec{F} = -\mu m g \frac{\vec{v}+R\hat{n}\times\vec{\omega}}{\|\vec{v}+R\hat{n}\times\vec{\omega}\|}$$

and corresponding equipollent torque

$$\vec{\tau} = \vec{F} \times \hat{n} R$$

At this point, it might be worth establishing some direction vectors, and looking at these relationships in terms of the vector components along these directions.

At any time frame, the direction of travel is designated $$\hat{e}$$ and the perpendicular direction is $$\hat{b}$$. By definition $$\vec{v} = v \hat{e}$$ at all times. Decompose the angular velocity vector into three components, denoting the roll, spin, and yaw as

$$\vec{\omega} = \omega_{\rm spin} \hat{e} + \omega_{\rm roll} \hat{b} + \omega_{\rm yaw} \hat{n}$$

and calculate the velocity of the contact point

$$\vec{v} + R \hat{n} \times \vec{\omega} = ( v + R \omega_{\rm roll}) \hat{e} - (R \omega_{\rm spin}) \hat{b} \tag{5}$$

This makes the sliding frictional force equal to

$$\vec{F} = -\mu m g \frac{ ( v + R \omega_{\rm roll}) \hat{e} - (R \omega_{\rm spin}) \hat{b}}{\sqrt{ ( v + R \omega_{\rm roll})^2 + (R \omega_{\rm spin})^2}} \tag{6}$$

This illustrates the difficulty here because the direction of $$\vec{F}$$ depends on $$\omega_{\rm spin}$$. We see this by calculating the torque due to friction

$$\vec{\tau} = -\left( \frac{\mu m g R^2 \omega_{\rm spin}}{\sqrt{ ( v + R \omega_{\rm roll})^2 + (R \omega_{\rm spin})^2}} \right) \hat{e} - \left(\frac{\mu m g R ( v + \omega_{\rm roll}R)}{\sqrt{ ( v + R \omega_{\rm roll})^2 + (R \omega_{\rm spin})^2}} \right) \hat{b} \tag{7}$$

and since $$\vec{\tau} = {\rm I} \tfrac{\rm d}{{\rm d}t} \vec{\omega}$$ we can find the change in rotation as

\begin{aligned} \tfrac{\rm d}{{\rm d}t} \omega_{\rm roll} & = - \frac{5 \mu m g (v+R \omega_{\rm roll})}{2 R \sqrt{ ( v + R \omega_{\rm roll})^2 + (R \omega_{\rm spin})^2}} \\ \tfrac{\rm d}{{\rm d}t} \omega_{\rm spin} & = - \frac{5 \mu m g \omega_{\rm spin}}{2 \sqrt{ ( v + R \omega_{\rm roll})^2 + (R \omega_{\rm spin})^2}} \\ \tfrac{\rm d}{{\rm d}t} \omega_{\rm yaw} & = 0 \end{aligned}

In summary, due to the nature of friction, it is not possible to integrate the above system into a later time frame and predict analytically the motion of the ball.