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

  • #1
giodude
30
1
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!
 

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  • #2
giodude said:
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.
giodude said:
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.
 
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  • #3
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.
 
  • #4
giodude said:
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
 
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  • #5
PeroK said:
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.
 
  • #6
PeroK said:
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?
 
  • #7
giodude said:
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.
 
  • #8
kuruman said:
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.
 
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  • #9
PeroK said:
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.
 
  • #10
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.
 
  • #11
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!
 

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

What is the main concept behind the problem of an elastic collision between a ball and a pivoted rod?

The main concept involves understanding the principles of conservation of linear momentum, angular momentum, and kinetic energy during an elastic collision. The problem typically requires analyzing how these quantities are conserved when a ball collides elastically with a rod that is free to rotate about a pivot point.

How do you apply the conservation of linear momentum in this problem?

In an elastic collision, the total linear momentum of the system (ball plus rod) is conserved. This means that the sum of the linear momenta of the ball and the rod before the collision must equal the sum of their linear momenta after the collision. Since the rod is initially at rest, its initial linear momentum is zero.

What role does angular momentum play in the collision between the ball and the pivoted rod?

Angular momentum is crucial because the rod is pivoted and can rotate. The conservation of angular momentum about the pivot point must be applied. The angular momentum of the ball about the pivot point before the collision must equal the total angular momentum of the ball and the rod about the pivot point after the collision.

How is kinetic energy conserved in this elastic collision?

In an elastic collision, the total kinetic energy of the system is conserved. This means that the sum of the kinetic energies of the ball and the rod before the collision is equal to the sum of their kinetic energies after the collision. This includes both translational kinetic energy of the ball and the rotational kinetic energy of the rod.

What equations are typically used to solve for the final velocities and angular velocities in this problem?

The key equations include the conservation of linear momentum, the conservation of angular momentum, and the conservation of kinetic energy. These equations are:1. \( m_b v_{b,i} = m_b v_{b,f} + m_r v_{r,f} \)2. \( m_b v_{b,i} l = I_r \omega_r + m_b v_{b,f} l \)3. \( \frac{1}{2} m_b v_{b,i}^2 = \frac{1}{2} m_b v_{b,f}^2 + \frac{1}{2} I_r \omega_r^2 \)where \( m_b \) is the mass of the ball, \( v_{b,i} \) and \( v_{b,f} \) are the initial and final velocities of the ball, \( m_r \) is the mass of the rod, \( v_{r,f} \) is the final velocity of the rod's center of mass, \( l \) is the distance from the pivot to the point of impact, \( I_r \

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