# Coin inclined at fixed angle toward the horizontal dropped onto table and rebounds [closed]

Here is my question:

A circular coin of radius $a$ falls at speed $u$ (without rotating) onto a smooth horizontal table. The perpendicular to its face makes an angle $\theta$ with the vertical.

Determine the state of motion of the coin just after it strikes the table, assuming that the collision is elastic. Show that, when $θ$ is small, the coin strikes the table a second time at an angle of $5 \hspace{.5mm} \theta/11$.

I think that after landing on the table the impulse $\delta p$ gives the centre of mass of the coin an upwards linear velocity $v$.

The impulse also provides an angular impulse $=\delta p* a*cos(\theta)$

which sets the coin rotating about an axis in its plane through its centre of mass, so I imagine the other end of the coin will hit the table, it has downward velocity due to the rotation $\omega*r*cos(\theta)$

and an upward velocity due to the upward motion of the centre of mass. This end of the coin starts at a height of $2\hspace{.5mm} a* sin(\theta)$ so takes a time T to reach the table, which is this distance over its velocity.

Setting angular momentum to the angular impulse $=\delta p* a cos(\theta)$ gives an expression for $\omega$, which can be integrated (note that the cosine term above is a constant!) to find the change in angle over a time T. Doing this and using small angles gives an expression for changes in theta in terms of $u$ and $v$. Using conservation of KE gives an expression $v$ in u, $v\approx3u/5$.

Is this approach ok?

• Could I suggest a diagram and also layout your math in line by whitespace line. I think a block of text is greatly reducing your chances of users reading your post correctly. Best of luck with it. – user140606 Feb 2 '17 at 10:30
• I agree with the above comment. I am also fairly certain there's a number of mistakes in the above line of argument, but that's hard to tell without a complete outline in rigorous language (=math). But, for starters, the quantity you call "angular impulse" cannot have the same units as the impulse, and you obviously cannot set it equal to angular momentum in the form your narrative seems to suggest. – Pirx Feb 2 '17 at 12:47
• Whoops you're right, didn't write the factor of a. It can obviously now be set equal to angular momentum. – Tradrad0997 Feb 2 '17 at 12:59
• You think so? Me, I wouldn't dare making statements like that without having done the math. But, more power to you if your physical intuition is such that the correct result is obvious to you. – Pirx Feb 2 '17 at 13:40
• Angular impulse is the change in angular momentum, just like linear impulse is the change of linear momentum. If the coin isn't intially rotating (we are told it isn't) then it starts with no angular momentum, so it's final angular momentum is just the angular impulse, so the only maths to do is to say that: change in angular momentum = final angular momentum- initial angular momentum= final angular momentum since initial is zero. – Tradrad0997 Feb 2 '17 at 13:48

This kind of problem becomes much easier if you consider the effective mass of the body along the contact direction. Once the effective mass $m^\star$ is calculated the impulse magnitude is simply $$J = (1+\epsilon)\, m^\star\, v_{impact}$$

where $\epsilon=1$ is the coefficient of restitution and $v_{impact}$ the relative velocity before the impact along the contact direction

Coming up with the effective mass for a 2D problem is easy. If the distance between the line of action of the contact force and the center of mass is $c=a \cos \theta$ and the mass moment of inertia about the center of mass $I_C = \frac{m}{4} a^2$ and the mass is $m$ then

$$m^\star = \left( \frac{1}{m} + \frac{c^2}{I_C} \right)^{-1}$$

You can prove that from the equations of motion (or see this answer)

In fact, in your case with some assumptions you can make the 3D problem into a 2D one and use the equations above. You need to assume that the instant axis of rotation does not change orientation and the plane to consider would be perpendicular to said axis.

The impact speed is then $$v_{impact} = v + \dot{\theta} a \cos \theta$$ or $$v_{impact} = v - \dot{\theta} a \cos \theta$$ depending on which side of the rotation the impact happens.

The effect if the impulse $J$ to the velocities is

\begin{align} \Delta v & = \frac{1}{m} J \\ \Delta \dot{\theta} & = \frac{a \cos\theta}{I_C}J \end{align}

Now you systematically go through each impact (like in your posting) and find the ballistic flight time to the next impact and evaluate the angle $\theta$ before impact. The repeat the equations above for the next impact and so on.