Equivalency of conditions involving angular momentum of a rolling ball hitting a wall

Question

(59th Polish Olympiad in Physics)

A ball of mass $m$, radius $r$ and a moment of inertia $I = \frac 25 mr^2$ is rolling on the floor without sliding with the linear velocity $v_0$. It hit the wall perpendicularly. Find out the velocity $v_k$ of the ball's receding from the wall after a long time after the collision.

The friction coefficient between the ball and the floor equals $\mu$, whereas the friction coefficient between the wall and the ball is very big. The collisions are infinitesimally short. All collisions are perfectly elastic and do not undergo deformation. Ignore the rolling resistance and the air resistance.

As the collision is very short, the forces acting between the wall and the ball are very big, so we may neglect the gravity, friction between the floor and the ball and the floor's reaction force. Then the angular momentum wrt to the axis of ball's tangency to the wall is conserved. This means $I' \omega' = \mathrm{const}$, where $I'$ is the moment of intertia of a ball wrt to that axis and $\omega'$ the angular velocity wrt to that axis.

But why is it equivalent to the condition that $I\omega + mv_yr = \mathrm{const}$ where $v_y$ is the vertical component of the velocity of the ball?

/edit: the official solution:

The coordinate system is used of axis $x$ perpendicular to the wall and directed left, the axis $y$ is perpendicular to the floor and is directed upwards. Positive angular velocities mean a counterclockwise motion.

As the ball is rolling without sliding, it approaches the wall with linear velocity $v_0$ and angular $\omega_0$.

The collision of with the wall is very short, so the contact force and the reaction force are very big. This means that during the collision we may neglect the gravity, the reaction of the floor and the friction of the ball with the floor. In this situation the torques wrt to the axis of the ball's tangency to the wall equal 0. So the total angular momentum is conserved wrt to that axis $$I\omega + mv_y r = \mathrm{const} (1)$$

Because the wall's friction coefficient is very big, during the collision, the ball will stop sliding wrt to the wall. This means that right after the collision the vertical component of the ball's velocity $v_{y2}$ and is angular velocity $\omega_2$ fulfill the formula $v_{2y} = \omega_2 r$. Taking in account that before the collision $\omega = v_0 /r, v_y = 0$, from the conservation on angular momentum (1) we get $$\omega_2 = \frac {I}{I + mr^2} \frac {v_0} r = \frac 2 7 \frac {v_0} r$$ $$v_{2y} = \omega_2 r = \frac 2 7 {v_0}$$

The wall and the ball are ideally elastic, the total work done by the reaction forces perpendicular to the wall equals zero, so the kinetic energy in the direction of $x$ is conserved, so $v_{2x} = - v_0$

After the collision the ball's motion is a projectile with initial velocity $(v_{2x}, v_{2y}$. The floor and the ball are ideally elastic, so the ball will jump infinitely long, reaching the same maximum height (it has no importance for finding out the final horizontal velocity).

During the collision with the floor we will have friction until we get $v_{x_{konc}} = \omega_{x_{konc}} r$. On the other hand, at each collision with the floor the angular momentum is conserved wrt to the axis of the ball's tangency to the floor $$I \omega + m v_x r = \mathrm {const}$$

Hence $$v_{x_{konc}} = \frac{I \omega_2 + mrv_{2x}}{I+mr^2}r$$

/edit2: It should give $\omega_2 = 2/7 \omega_0$, indeed.

We have $$\frac {dp_x}{dt} = N(t)$$ The friction gives upwards acceleration $$\frac {dp_y}{dt} = fN(t)$$ And diminishes the angular velocity $$I \frac {d\omega}{dt} = -fN(t)r$$

Hence $$I \frac {d\omega}{dt} + r\frac {dp_y}{dt} = 0 ~~~~(*)$$ So after integrating and finding the constants $$mrv_{2y} - mrv_0 = mrv_{2y} = mr^2 \omega_2 + I \omega_2 - I \omega_0 = 0$$ So $$\omega_2 = \frac 2 7 \omega_0$$

In fact, the formula (*) is the formula I have problems with. But why is it in fact the formula for angular momentum conservation for the axis of tangency.

Answer 1

Assuming it is an elastic collision, you will have, right after it, a ball with the same rolling motion (anticlockwise) but translating in the opposite direction at $v_0$. Now there is friction because the ball is "sliding", and the new equilibrium movement will be when both are moving without sliding again at $v_k$. Let me know if you do not know how to solve this last problem. In short, the effect of the collision is only to change the direction of $v_0$. The friction between the wall and the ball should have no effect at all (as the collision is infinitesimally short)

UPDATE: I assume that the ball reverses $v_0$ and friction starts until the ball stops sliding. Thus the final angular speed will be $v_f/r$. We have

$v_f=v_0-at_s=v_0-\mu gt_s$ (1)

where $t_s$ is the time it takes to stop sliding. You can obtain $t_s$ from the torque:

$\tau.t_s=\Delta L=\frac{2}{5} m r^2(\frac{v_f}{r}+\frac{v_0}{r})=\mu mgrt_s$ (2)

from here you obtain $t_s$ and replace in (1) to obtain:

$v_f=\frac{3}{7}v_0$

UPDATE 2: if we accept the explanation that the ball will roll up until it stops sliding, reaching $w_2$, then we need to change, in eq. (2), the initial angular speed in the previous solution from $\frac{v_0}{r}$ to $\frac{2v_0}{7r}$.

In such a case we get:

$v_f=\frac{31}{49}v_0$

Answer 2

Though it is a bit late for @marmistrz I think this question deserves a better explanation.

To explain why the official solution is correct, it is best to clarify first how the angular momentum is calculated for a rigid body with respect to a point which is non-stationary and different from the centre of mass.

In general, the angular momentum with respect to a point O fixed in inertial space can be expressed as:

$L^O = z_P \times m(v_P + \omega \times r_C) + r_C \times m v_P + J^P \cdot \omega \tag{1} $

where:

• P is a point fixed to the body
• $z_P$ Is the position vector of P with respect to the point O in the inertial reference system
• $r_C$ is the position of the centre of mass as seen from the point P
• $v_P$ the velocity of the point P as seen from inertial space
• $\omega$ the angular velocity of the frame fixed to the body with respect to the inertial frame
• $J^P$ the moment of inertia about P

Note 1: I am using the same notation as in “Dynamics of systems of rigid bodies” p35 by Wittenburg. Check it out to understand where the equation above comes from.

Note 2: this differs from the usual equation used to express the angular momentum $L = J^C \cdot \omega + r_C \times m v_C$ as this is valid if and only if the momentum is calculated with respect to a point which is fixed both in inertial space and in the body frame or is coinciding with the centre of mass of the body. Indeed it can be verified that (1) results in this last equation if P coincides with the centre of mass C.

To have the angular momentum conserved we choose the points O and P both coinciding with the impact point. Angular momentum is indeed conserved here because the impact is supposed instantaneous: in this hypothesis only impulsive forces should be accounted for, that is the horizontal reaction of the wall and the friction force on the wall (as the friction coefficient is infinitely big), and both are applied on P:

$M^{ext}_P = 0$

$L^P = const$

As P is coinciding with O, the position vector $z_P = 0$. In these conditions, the angular momentum formula above simplifies to:

$L^P = r_C \times m v_P + J^P \cdot \omega$

The moment of inertia about P is the moment of inertia about the centre of masses plus the mass times the radius squared (parallel axis theorem):

$J^P = J^C + mr^2 = \frac{2}{5} mr^2 + mr^2 = \frac{7}{5} m r^2$

Before the impact, the point P is moving downwards with a velocity of $v_0/r$, i.e. in inertial frame $r \times v_P$ is directed inside the screen. The angular velocity of the body ($\omega = v_0/r$) however is in the opposite direction, hence by taking the positive direction as going out of the screen, the angular momentum before the impact can be written as:

$L_{before} = \frac{7}{5} m r^2 \frac{v_0}{r} - m r^2 \frac{v_0}{r}$

After the impact, the point P is fixed to the wall by the no slip condition, hence the angular momentum is just:

$L_{after} = \frac{7}{5} m r^2 \frac{v_1}{r}$

Equating the two angular momenta:

$\frac{7}{5} m r^2 \frac{v_0}{r} - m r^2 \frac{v_0}{r} = \frac{7}{5} m r^2\frac{v_1}{r}$

$\frac{2}{5} v_0 = \frac{7}{5} v_1$

$v_1 = \frac{2}{7} v_0$

Which is equal to the result given in the solution.

I believe the rest of the solution is clear enough and there is no more explaining needed.

Though this way of reasoning may be more convoluted than the official answer, I find it hard to convince myself of the solution otherwise.

Hope this can help someone!

Answer 3

The equations (no friction) are:

Translation

\begin{align*} &g1:=m_1\,(v_1-v_0)=-dp\\ &g2:=m_2\,v_2=dp \end{align*} Rotation

\begin{align*} &g3:=I_B\,\omega=dp\,r\quad,I_B=\frac 25\,m_1\,r^2+m_1\,r^2\\ \end{align*}

and the conservation of the kinetic energy

\begin{align*} &g_4:=m_1\,v_1^2+m_2\,v_2^2=m_1\,v_0^2 \end{align*}

where $~m_2~$ is the wall mass .

if you add equation (g1) and (g2) you obtain the conservation of the linear momentum.

solving equation (g1) for $~dp~$ and substitute the solution in (g3) , you obtain the conservation of the angular momentum.

First solve the 4 equations for the unknows $~v_1~,v_2~,\omega~,dp~$ then take the $~\lim(\ldots)_{m_2\mapsto\infty}~$ you obtain

\begin{align*} &v_1=-\frac 16\,v_0\quad,v_2=0\quad,\omega= \frac{5}{6}\frac{v_0}{r} \end{align*}

Theory