This problem took me awhile as well. Mainly, its second part took me the longest because I could not get the same answer as was in the book (Savchenko et al, 1981, problem 6.26). If you have any ideas about the second part of the problem, please leave me your feedback in the comments below the post.
Problem statement
A ring of radius $R$ is sliding with a speed $v$ on a floor and hits a corner of a step of height $h$. Determine the height $H$ to which the ring will bounce in the following 2 cases:
- (a) No friction between the ring and the step.
- (b) Friction is present between the ring and the step, but no sliding.
Note: There is no rotation before the collision.
Link to the latex source of the figures
Solution
The solution is based on energy and angular momentum conservation during the collision. For the energy conservation I have convinced myself that if there is no displacement of the center of mass during the collision and no sliding, that should be OK. On the other hand, angular momentum conservation means that there is no torques acting on the ring during the collision (i.e. $\vec{\tau}_{\rm net} = \vec{0}$ with respect to a selected point).
If we consider torques with respect to point $O$ (see Fig. 1). The torques of reaction of the step corner and of the friction force with the corner are 0. There are the force of gravity and the force from the floor. I assume they cancel each other and the sum of their torques is approximately 0 (this might need more detailed discussion).
In both cases the maximum height of the ring after the collision can be expressed using the vertical component of velocity at the last moment of the collision $v_z$ through the energy conservation:
\begin{equation*} H = \frac{v_z^2}{2g} \tag{1} \end{equation*}It is convenient to express the velocity just after the collision ($\vec{v}_{ec}$) as a sum of two components:
$$ \vec{v}_{ec}=\vec{v}_1 + \vec{v}_2 $$where $\vec{v}_2$ is parallel and $\vec{v}_1$ is perpendicular to $OC$ (see Fig. 2).
In both cases $v_z$ (vertical component of the velocity just after the collision) is expressed as (see Fig. 2):
$$ v_z = v_2 \cos\alpha + v_1 \sin\alpha \tag{2} $$where $\cos\alpha=(R-h)/R$. Therefore, in the following part, I will consider $\alpha$ as a known parameter.
Case (a): no friction
Energy conservation during the collision yields:
$$ v^2=v_1^2 + v_2^2 $$where $v$ is the ring speed before the collision.
Conservation of the angular momentum (with respect to the point $O$) gives:
$$ v(R-h) = v_1 R \Rightarrow v_1 = v\cos\alpha $$From the two previous equations $v_2$ can be found as:
$$ v_2 = v \sin\alpha $$Then by inserting the above expressions for $v_1$ and $v_2$ into (2), we get:
$$ v_z= v_2 \cos\alpha + v_1 \sin\alpha = 2 v \cos\alpha\sin\alpha $$And by using equation (1) we find the final result for the part (a) of the problem: $$ H_a = \frac{v^2\sin^2 2\alpha}{2g} $$
Case (b): friction, no sliding
Here, the friction force will rotate the ring and we need to take the rotation into account in our energy and angular momentum conservation equations.
Rotational kinetic energy around the center of mass of the ring is $$ K_r = \frac{mR^2\omega^2}{2} $$
Therefore the energy conservation yields:
$$ v^2 = v_1^2 + v_2^2 + \omega^2R^2 $$Using the non-sliding condition $v_1=\omega R$, we can simplify the above equation as follows:
$$ v^2 = 2v_1^2 + v_2^2 $$Angular momentum conservation will give us the following relation:
$$ v(R-h) = v_1 R + \omega R^2 = 2 v_1 R \Rightarrow v_1 = \frac{v\cos\alpha}{2} $$Then using the above two equations $v_2$ can be found as: $$ v_2 = v\left( 1 - \frac{\cos^2\alpha}{2} \right)^{1/2} $$
Finally, using equations (1) and (2), as in the previous part we get the following expression for the maximum height of the ring:
$$ H_b = \frac{v^2\cos^2\alpha}{8g} \left[ \sin\alpha + 2 \left( 1 - \frac{\cos^2\alpha}{2} \right)^{1/2} \right]^2 $$Below is my attempt to explain why the answer above is different from the one in the problem book.
I think the above answer is the correct answer. Although the expression in the problem book is different. I suspect that the authors of the book made a pen-slip when they were working out $v_2$. If they, by accident, ommited the $1/2$ factor, the expression for $v_2$ would then be:
$$ v_2=v\sin\alpha $$and from here $v_z$ would be:
$$ v_z = v_2 \cos\alpha + v_1 \sin\alpha = v \sin\alpha\cos\alpha + \frac{1}{2} v \cos\alpha\sin\alpha = \frac{3}{4}v\sin2\alpha $$Then the above expression for $v_z$ would result in the following: $$ H_b = \frac{9}{16}v^2 \frac{\sin^2 2\alpha}{2g}=\frac{9}{16}H_a $$
