MIT OCW 8.01 PS11.3: Elastic Collision Between Ball and Pivoted Rod

[giodude]

Homework Statement

A rigid rod of length $d$ and mass $m$ is lying on a horizontal frictionless table and pivoted at the point $P$ on one end (shown in figure below). A point-like object of the same mass $m$ is moving to the right (see figure below). with speed $v_{i}$. It collides elastically with the rod at the midpoint of the rod and rebounds backwards with speed $v_{f}$. After the collision, the rod rotates clockwise about its pivot point $P$ with angular speed $\omega_{f}$. The moment of inertia of a rod about the center of mass is $I_{cm} = \frac{1}{12}md^{2}$.

Find the angular speed $\omega_{f}$. Express your answer in terms of $d$, $m$, and $v_{i}$ as needed.

Relevant Equations

Elastic collision: $\Delta K = 0$
Parallel axis theorem: $I_{P} = I_{cm} + md^{2}$
Angular kinetic energy: $K = \frac{1}{2}I_{P}^{sys} \omega_{f}^{2}$

Given that we're working with an elastic collision we want to populate the following system:
$k_{i} = k_{f}$
$p_{i} = p_{f}$

Solve for kinetic energy just before and after the collision:
$k_{i} = \frac{1}{2}mv_{i}^{2}$
$k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}I_{P}^{sys} \omega_{f}^{2}$

Solve for $I_{P}$ using parallel axis theorem:
$I_{P} = I_{cm} + md^{2} = \frac{1}{12}md^{2} + m(\frac{1}{2}d)^{2} = \frac{1}{12}md^{2} + \frac{1}{4}md^{2}$
$\frac{1}{12}md^{2} + \frac{1}{4}md^{2} = \frac{1}{3}md^{2}$

Plug moment of inertia about $P$, $I_{P}$ in to our equation for final kinetic energy:
$k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}\frac{1}{3}md^{2} \omega_{f}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}md^{2} \omega_{f}^{2}$

Simplify $k_{i} = k_{f}$:
$\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}md^{2} \omega_{f}^{2}$
$v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}$

Find $p_{i}, p_{f}$:
$p_{i} = mv_{i}$
$p_{f} = mv_{f} + mv_{cm}$

The rod has a pivot P, so upon collision it's center of mass will have angular speed such that $v_{cm} = \frac{d}{2}\omega_{f}$:
$p_{i} = mv_{i}$
$p_{f} = mv_{f} + m\frac{d}{2}\omega_{f}$

Solve the system:
$mv_{i} = mv_{f} + m\frac{d}{2}\omega_{f}$
$v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}$

$v_{f} = v_{i} - \frac{d}{2}w_{f}$
$v_{i}^{2} = (v_{i} - \frac{d}{2}\omega_{f})^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}$
$v_{i}^{2} = v_{i}^{2} - dv_{i}\omega_{f} + \frac{1}{4}d^{2} \omega_{f}^{2} + \frac{1}{3}d^{2} \omega_{f}^{2}$
$dv_{i}\omega_{f} = \frac{7}{12}d^{2}\omega_{f}^{2}$
$\omega_{f} = \frac{12v_{i}}{7d}$

Hello, I know elastic collisions have $\Delta K = 0$ which is why it was used. However, I'm curious when we're allowed to use the coefficient of restitution, $e$, to set the ratio $\frac{v_{f}}{v_{i}} = 1$. I learned in textbook reading that this is usually a feature of elastic collisions however it doesn't seem to be the case here. So I'm curious (a) if this solution is correct and (b) when do I know what scenarios allow for the use of $e = \frac{v_{f}}{v_{i}} = 1$. Thank you in advance!

 

[PeroK]

Hello, I know elastic collisions have $\Delta K = 0$ which is why it was used. However, I'm curious when we're allowed to use the coefficient of restitution, $e$, to set the ratio $\frac{v_{f}}{v_{i}} = 1$. I learned in textbook reading that this is usually a feature of elastic collisions however it doesn't seem to be the case here. So I'm curious (a) if this solution is correct

The answer looks correct. You could also have used conservation of angular momentum about the pivot.

and (b) when do I know what scenarios allow for the use of $e = \frac{v_{f}}{v_{i}} = 1$. Thank you in advance!

That's a special case of particle collisions. The analysis in this case shows that it doesn't hold more generally.

 

[PeroK]

PS You should have used conservation of angular momentum. Linear momentum is not conserved because of the force at the pivot. If we use your calculation, using the answer for $w_f$, we get:
$$v_f = v_i - \frac d 2 w_f = \frac 1 7 v_i$$But, in that case angular momentum about the pivot is not conserved. So, you got the right answer by an invalid method.

 

[nasu]

I'm curious when we're allowed to use the coefficient of restitution, $e$, to set the ratio $\frac{v_{f}}{v_{i}} = 1$. I learned in textbook reading that this is usually a feature of elastic collisions however it doesn't seem to be the case here. So I'm curious (a) if this solution is correct and (b) when do I know what scenarios allow for the use of $e = \frac{v_{f}}{v_{i}} = 1$. Thank you in advance!

The coefficient of restitution is the ratio between the relative velocities of the two colliding objects before and after collision (ratio of magnitudes). You wrote it as the ratio of velocities of the same object which is valid only when the second one is at rest all the time. Which is not the case here, anyway

 

[kuruman]

So, you got the right answer by an invalid method.

The answer that OP got in post#1 using (erroneously) linear momentum conservation is $\omega_{f} = \dfrac{12v_{i}}{7d}.$ This is not right.

I derived the correct answer using angular momentum about the pivot but, as per our rules, this is OP's job with our assistance if needed. I also point out that the colliding object rebounds backwards after the collision which means that both its linear and angular momentum after the collision change sign. This is a point that OP missed when setting up the linear momentum conservation equation and should be taken into account when setting up the angular momentum conservation equation.

 

[giodude]

PS You should have used conservation of angular momentum. Linear momentum is not conserved because of the force at the pivot. If we use your calculation, using the answer for $w_f$, we get:
$$v_f = v_i - \frac d 2 w_f = \frac 1 7 v_i$$But, in that case angular momentum about the pivot is not conserved. So, you got the right answer by an invalid method.

Fixed it using conservation of angular momentum:

System:
$L_{P,i}^{sys} = L_{P,f}^{sys}$
$k_{i} = k_{f}$

Solve for initial and final angular momentum about $P$:
$L_{P,i}^{sys} = -\frac{d}{2}mv_{i}$
$L_{P,f}^{sys} = I_{cm} + md^{2} + \frac{d}{2}mv_{f} = \frac{1}{3}md^{2}\omega_{f} + \frac{1}{2}dmv_{f}$

Set $L_{P,i}^{sys} = L_{P,f}^{sys}$ and isolate $v_{f}$ with respect to $v_{i}, \omega_{f}$:
$L_{P,i}^{sys} = L_{P,f}^{sys}$
$-\frac{d}{2}mv_{i} = \frac{1}{3}md^{2}\omega_{f} + \frac{1}{2}dmv_{f}$
$v_{i} = -\frac{2}{3}md\omega_{f} - v_{f}$
$v_{f} = -v_{i} - \frac{2}{3}md\omega_{f}$

Plug $v_{f}$ into $k_{i} = k_{f}$:
$k_{i} = k_{f}$
$\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}d^{2}\omega_{f}^{2}$
$v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$v_{i}^{2} = (-v_{i} - \frac{2}{3}d\omega_{f})^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$v_{i}^{2} = v_{i}^{2} + \frac{4}{3}d\omega_{f}v_{i} + \frac{4}{9}d^{2}\omega_{f}^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$-\frac{4}{3}d\omega_{f}v_{i} = \frac{4}{9}d^{2}\omega_{f}^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$-\frac{4}{3}v_{i} = \frac{4}{9}d\omega_{f} + \frac{1}{3}d\omega_{f}$
$-\frac{4}{3}v_{i} = \frac{7}{9}d\omega_{f}$
$\omega_{f} = -\frac{12}{7}v_{i}$

Is this the correct procedure?

 

[PeroK]

Fixed it using conservation of angular momentum:

System:
$L_{P,i}^{sys} = L_{P,f}^{sys}$
$k_{i} = k_{f}$

Solve for initial and final angular momentum about $P$:
$L_{P,i}^{sys} = -\frac{d}{2}mv_{i}$
$L_{P,f}^{sys} = I_{cm} + md^{2} + \frac{d}{2}mv_{f} = \frac{1}{3}md^{2}\omega_{f} + \frac{1}{2}dmv_{f}$

Set $L_{P,i}^{sys} = L_{P,f}^{sys}$ and isolate $v_{f}$ with respect to $v_{i}, \omega_{f}$:
$L_{P,i}^{sys} = L_{P,f}^{sys}$
$-\frac{d}{2}mv_{i} = \frac{1}{3}md^{2}\omega_{f} + \frac{1}{2}dmv_{f}$
$v_{i} = -\frac{2}{3}md\omega_{f} - v_{f}$
$v_{f} = -v_{i} - \frac{2}{3}md\omega_{f}$

Plug $v_{f}$ into $k_{i} = k_{f}$:
$k_{i} = k_{f}$
$\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{6}d^{2}\omega_{f}^{2}$
$v_{i}^{2} = v_{f}^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$v_{i}^{2} = (-v_{i} - \frac{2}{3}d\omega_{f})^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$v_{i}^{2} = v_{i}^{2} + \frac{4}{3}d\omega_{f}v_{i} + \frac{4}{9}d^{2}\omega_{f}^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$-\frac{4}{3}d\omega_{f}v_{i} = \frac{4}{9}d^{2}\omega_{f}^{2} + \frac{1}{3}d^{2}\omega_{f}^{2}$
$-\frac{4}{3}v_{i} = \frac{4}{9}d\omega_{f} + \frac{1}{3}d\omega_{f}$
$-\frac{4}{3}v_{i} = \frac{7}{9}d\omega_{f}$
$\omega_{f} = -\frac{12}{7}v_{i}$

Is this the correct procedure?

You've dropped $d$ from your final answer. The question does say angular speed $w_f$. In any case, you can take clockwise to be positive.

 

[PeroK]

The answer that OP got in post#1 using (erroneously) linear momentum conservation is $\omega_{f} = \dfrac{12v_{i}}{7d}.$ This is not right.

I derived the correct answer using angular momentum about the pivot but, as per our rules, this is OP's job with our assistance if needed. I also point out that the colliding object rebounds backwards after the collision which means that both its linear and angular momentum after the collision change sign. This is a point that OP missed when setting up the linear momentum conservation equation and should be taken into account when setting up the angular momentum conservation equation.

If you assume conservation of energy and linear momentum, then the particle does not rebound. I think the OP's calculations were correct, in that sense. That said, the diagram showed $v_f$ in the opposite direction. The OP's use of $v_f$ didn't match the diagram. And the question said explicitly that the particle rebounds, although that probably ought to be calculated. If the mass of the rod were less than the mass of the particle, then the particle may not rebound.

 

[kuruman]

If you assume conservation of energy and linear momentum, then the particle does not rebound. I think the OP's calculations were correct, in that sense. That said, the diagram showed $v_f$ in the opposite direction. The OP's use of $v_f$ didn't match the diagram.

OK, I agree with OP's answer after discovering a sign error in my algebra. Yes, the particle does not rebound. One can see that if one assumes that the rod is not pivoted. Then, because the collision is elastic at the middle of the rod and the masses are equal, the particle will stop and the rod will translate without spinning.

 

[giodude]

Ah, I think my sign error is a result of improperly setting up my cross product calculation with respect to the final velocity. Once fixed it looks like the sign is fixed because what was previously $L_{P, f}^{sys} = I_{cm} + md^{2} + \frac{1}{2}dmv_{f}$ becomes $L_{P, f}^{sys} = I_{cm} + md^{2} - \frac{1}{2}dmv_{f}$ which fixes up the sign error in the final answer.

 

[PeroK]

I'm must admit, I didn't follow the diagram either. I don't like unnecessary subscripts, so I had $u, v$ as the initial and final velocities (with the right being positive) and $w$ as the final angular velocity, with clockwise being positive. Here's how I did it.

The moment of inertia of the rod about the pivot is:$$I_P = \frac 1 3 m d^2$$Conservation of energy gives:
$$\frac 1 2 mu^2 = \frac 1 2mv^2+ \frac 1 2 (\frac 1 3 m d^2)w^2$$$$u^2 - v^2 = \frac 1 3 d^2w^2 \ \ \ \ \ (1)$$Conservation of angular momentum about the pivot gives:
$$mu\frac d 2 = mv\frac d 2 + (\frac 1 3 m d^2)w$$$$u-v = \frac 2 3 dw \ \ \ \ \ (2)$$Now, here's a useful trick, using $u^2 - v^2 = (u-v)(u+v)$ equations (1) and (2) give:
$$\frac 1 3 d^2w^2 = (\frac 2 3 dw)(u+v)$$$$u + v = \frac 1 2dw \ \ \ \ \ (3)$$Adding equations (2) and (3) gives:$$2u = \frac 7 6 dw$$$$w = \frac{12u}{7d} \ \ \ \ \ (4)$$Equations (3) and (4) allow us to calculate $v$:
$$v = \frac 1 2dw - u = \frac 1 2d(\frac{12u}{7d}) - u = -\frac 1 7 u$$$$v = -\frac 1 7 u \ \ \ \ \ (5)$$And the particle does indeed rebound!