# Movement of a ball rolling and slipping on the ground

Suppose there is a ball on the ground. At initial state, the ball has no velocity but it's rolling, so the ball is sliding on the ground. The friction force is applied tangentially to the contact point between the ball and the ground.

How to calculate the acceleration and the angular acceleration? Empirically, The ball will start rolling forward and spin slower.

I know that when the pivot point is fixed, the torque is $$\vec{\tau} = \vec{r} \times \vec{F}$$ and the angular acceleration is $$\vec{\alpha} = \frac{\vec{\tau}} {I}$$ . But what about a non-fixed object?

• "Suppose there is a ball on the ground. At initial state, the ball has no velocity but it's rolling" If it is rolling, then it has velocity. Do you mean spinning in place? Aug 16, 2023 at 11:25

But what about a non-fixed object?

The system under consideration is the ball and it has one horizontal force acting on it, the frictional force $$\vec F$$.

If you want to use Newton's second law you must be careful about the frame of reference.
If you are using a non-inertial frame of reference you must add appropriate pseudoforce(s) acting through the centre of mass so that Newton's second law can be used.

In this example the centre of mass has a horizontal translational acceleration $$a$$ to the right so the pseudoforce is $$ma$$ to the left with its line of action passing through the centre of mass, and $$m$$ is the mass of the ball.

If $$\vec r$$ is the position vector from the centre of mass of the ball to the point of contact between the ball and the ground, the torque about the centre of mass of the ball is $$\vec r \times \vec F$$.
Note that the pseudoforce does not contribute to the torque about the centre of mass.

So what is happening to the ball?

The frictional torque is reducing the clockwise rotational speed of the ball and the combination of the frictional force and the pseudoforce (which is in the opposite direction to the frictional force) is increasing the translational speed of the ball's centre of mass to the right.
Both effects are reducing the relative speed of the ball and the ground at the point of contact with an ultimate goal of there being no slipping at the point of contact.

A similar analysis can be done about any point but note that when using $$\tau = I \alpha$$ the appropriate value of the moment of inertia must be used.

A ball cannot "roll" unless it has some velocity. The direction of friction (+x direction) here implies that the body has a tendency to slip or perform relative motion in the other direction.

Moreover, the direction in which the body appears to rotate seems wrong, because there is no force that can apply torque on the body causing it to rotate that way. Assuming that the frictional force is the only force capable of producing torque in the body, it would appear to rotate in the anticlockwise sense.

A general method to figure out the angular acceleration is to use the force equation equivalent - which is 𝜏=𝐼𝛼. If it is specified that the object is NOT slipping, or sliding on the surface, you can equate the acceleration at the lowest point to zero (or simply perform 𝑟⃗𝛼= acceleration of the center of mass.)

You may be over thinking this problem a little bit.

The only force $$F$$ will accelerate the ball.

$$F=m\ a=m\frac{dv}{dt}$$

If you put yourself at the center of the ball, then from your standpoint the ball is fixed. You will feel a pseudo force from your acceleration, but that force will produce no torque. The only force producing a torque is $$F$$ so that

$$r\ F=-I\ \alpha=-I\frac{d\omega}{dt}$$

The fact that it's moving does not effect the torque equation. The negative sign assures that $$\omega$$ is decreasing. This looks like the classic problem where you have a rotating ball and you gently set it on a floor and calculate its final velocity at which it rolls away (when sliding stops and pure rolling sets in.

With these equations you can eliminate $$F$$ and $$dt$$ to get

$$m\ dv=- \frac{I}{r}d\omega$$

Integrate the left side from $$0$$ to $$vf$$ and the right side from $$\omega0$$ to $$\omega f$$. Combine with the pure rolling equation

$$vf=r\ \omega f$$

and solve for $$vf$$ and $$\omega f$$ to get the final velocity

$$vf=\frac{I\ r\ \omega 0}{I+m\ r^2}$$

Some examples

Solid ball, $$I=\frac{2 m r^2}{5}$$, $$vf=\frac{2\ r\ \omega0}{7}$$

Ping Pong ball, $$I=\frac{2 m r^2}{3}$$, $$vf=\frac{2\ r\ \omega0}{5}$$

Disk, $$I=\frac{m r^2}{2}$$, $$vf=\frac{r\ \omega0}{3}$$

Hula Hoop, $$I=m r^2$$, $$vf=\frac{r\ \omega0}{2}$$

Note that the final velocity does not depend on $$F$$, but given $$F$$ you could determine the time it takes to reach pure rolling.

Note also that if you take the angular momentum about a point on the floor, which is $$m\ v\ r+I\ \omega$$, you will find that it is constant. You could have used that to find the final velocity also.

There are three cases to consider. Forward spin, back spin, and pure rolling.

First, describe the kinematics for the general case, and then apply the specifics to the equations of motion for each case.

## General Case

The object moves to the right with the geometric center having translational velocity $$v$$ and rotational velocity $$\omega$$.

### Conventions

The convention for positive values is below for $$v$$ and $$\omega$$ is shown below. Similarly for accelerations $$a$$ and $$\alpha$$, and the net forces $$F_{\rm net}$$ and torques $$\tau_{\rm net}$$

### Kinematics

As a result, the slip velocity of the ball on contact with the floor (point A) is derived from the 3D equation $$\vec{v}_A = \vec{v} - \vec{r} \times \vec{\omega}$$ with $$\vec{r}$$ the location of the contact point relative to the center of mass. Since $$|\vec{r}| = R$$, then the velocity kinematics are

$$v_{\rm slip} = v + \omega\; R$$

NOTE: the cross product $$-\pmatrix{0 & -R & 0} \times \pmatrix{0 & 0 & \omega} = \pmatrix{\omega\,R & 0 & 0}$$ is where the positive sign in the kinematics comes from.

the first derivative of the above gives out the acceleration kinematics

$$\dot{v}_{\rm slip} = a + \alpha\; R$$

### Free Body Diagram

The force applied to the ball is shown below.

Friction force $$F_{\rm friction}$$ is applied to the ball, and an equal and opposite force is applied to the ground. The weight of the ball $$m g$$ is assumed to be completely counter-acted by the normal force $$N$$ and any applied force to the center (like air resistance) can be included in $$F_{\rm apply}$$

As a result the net force $$F_{\rm net}$$ and $$\tau_{\rm net}$$ are calculated from the 3D equation $$\vec{\tau}_{\rm net} = \vec{r} \times \vec{F}_{\rm friction}$$ is

$$F_{\rm net} = F_{\rm friction} + F_{\rm apply}$$ $$\tau_{\rm net} = R\;F_{\rm friction}$$

NOTE: The only force that produces a torque is friction.

### Equations of Motion

The net forces and torques relate to the acceleration quantities as follows

$$F_{\rm friction} + F_{\rm apply} = m a$$ $$R\;F_{\rm friction} = I_{\rm ball} \alpha$$

## Specific Cases

1. Forward Spin occurs when $$v_{\rm slip} > 0$$ and thus friction is defined as opposing motion and limited by traction to

$$F_{\rm friction} = -\mu m g$$

Since the translational and rotational motion are not kinematically linked the resulting motion is

$$a = \frac{F_{\rm apply} - \mu m g}{m}$$ $$\alpha = -\frac{R\;\mu m g}{I_{\rm ball}}$$

As a result, the change in slip velocity is

$$\dot{v}_{\rm slip} = \frac{F_{\rm apply}}{m} - \mu g \left( 1 + \frac{m R^2}{I_{\rm ball}} \right)$$

NOTE: that for spherical ball $$I_{\rm ball} = \frac{2}{5} m R^2$$.

As you can see, when zero external force is applied, the slip velocity is reduced from a positive value (initially) down to zero after time finite time $$t_{\rm slip} = v_{\rm slip}/\dot{v}_{\rm slip}$$

2. Pure Rolling occurs when $$v_{\rm slip}=0$$ which means that the translational motion and rotational are linked together, but the amount of friction needed is not predetermined.

You can use $$\alpha = -\frac{a}{R}$$ from the acceleration kinematics to solve the equations of motion

$$a = \frac{F_{\rm apply}}{m+\frac{ I_{\rm ball}}{R^2}}$$ $$F_{\rm friction} = -\frac{F_{\rm apply}}{1+\frac{m R^2}{I_{\rm ball}}}$$

3. Backspin occurs when $$v_{\rm slip} < 0$$ and it is the same as forward spin, but the direction of friction is positive this time

$$F_{\rm friction} = \mu m g$$

Since the translational and rotational motion are again not kinematically linked the resulting motion is

$$a = \frac{F_{\rm apply} + \mu m g}{m}$$ $$\alpha = \frac{R\;\mu m g}{I_{\rm ball}}$$

As a result, the change in slip velocity is

$$\dot{v}_{\rm slip} = \frac{F_{\rm apply}}{m} + \mu g \left( 1 + \frac{m R^2}{I_{\rm ball}} \right)$$

which makes increase the value from a negative up to zero, resulting in pure rolling again after time finite time.