Rod swinging and hitting a ball

But that is a useful equation, because it relates the unknown ##v## to the known ##\omega##. You have another equation relating ##\omega## to ##v##, so you can solve them simultaneously for ##v## and ##\omega##.In summary, the problem involves a uniform rod and a uniform ball in an instantaneous collision. Assuming the collision is completely inelastic, the velocity of the rod's center of mass before and after the collision can be related using linear momentum conservation, while the angular momentum expression can be used to relate the angular velocity of the rod before and after the collision. An additional equation is needed to relate the linear and angular velocities of the ball before and after the collision.
  • #1
erfz

Homework Statement


There is a uniform rod of mass m = 1 kg and length L = 0.2 m fixed to the wall by an axis passing through its end.
A uniform ball of mass M = 0.1 kg and radius R = 2.85 cm is on the ground, below the axis of the rod, such that the rod's unfixed end is at the height of the ball's center-of-mass if the rod were to be at an angle of 0 degrees with the vertical.
The rod is set at angle 45 degrees to the left of the vertical and let go -- find the angular velocity of the rod before and after the collision, as well as the ball's translational and angular speeds after the collision. Assume the ball rolls without slipping.

I have written my approach below. I don't have the answer at hand, but I'm sure something is wrong here since the translational speed of the ball I get is 0.

Homework Equations


##\vec{L} = \vec{r} \times m\vec{v}_{cm} + I_{cm}\vec{\omega}##
##L = I\omega##

The Attempt at a Solution


I viewed this as an instantaneous collision and set the axis of rotation at the axis through the rod's end. So, the first step is finding the initial omega ##\omega_i## right before the collision using energy conservation about the rod's COM: $$m_{rod}g\frac{L}{2}(1-cos ~\theta) = 1/2 I_{rod}\omega_i^2$$ The moment of inertia of the rod about its end is ##1/3mL^2##. Plugging in the numbers, I get that ##\omega_i = 6.6 ~\text{rad/s}##

The angular momentum expression is: $$I_{rod}\omega_i = I_{rod}\omega_f + MLv_{ball,cm} + I_{ball,cm}\omega_{ball}$$ This second and third terms on the right come from the first equation I listed in "Relevant equations." I am defining counter-clockwise to be positive, so the translational contribution to the angular momentum of the ball (the second term on the right) should also be positive. However, the third term should be negative overall since the ball will be rotating clockwise. Since the ball isn't slipping, I can replace ##\omega_{ball} = -v_{ball,cm}/R##. So then: $$I_{rod}\omega_i = I_{rod}\omega_f + MLv_{ball,cm} - I_{ball,cm}v_{ball,cm}/R$$ I still need another equation to relate ##\omega_f## and ##v_{cm}##. I believe I can use linear momentum conservation for this: $$m_{rod}v_{rod,cm} = m_{rod}v'_{rod,cm} + Mv_{ball,cm}$$ The velocity of the rod's center-of-mass, I believe, should be such that ##v_{rod,cm} = \omega r##, where r is the distance from the axis to the center-of-mass of the rod (i.e. ##L/2##). So then this becomes $$m_{rod}\omega_i L/2 = m_{rod}\omega_f L/2 + Mv_{ball,cm}$$ Plugging in the values here, I get that relation that $$\omega_f = \omega_i - v_{ball,cm}$$ Now, if I substitute this into the angular momentum expression and solve, I get that ##v_{ball,cm}## is essentially 0, which intuitively cannot be correct. What is wrong with my approach?
 
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  • #2
erfz said:
Assume the ball rolls without slipping.
That is almost impossible. You would have to allow the collision process to take a significant time and for the coefficient of friction with the ground to be very high. But glossing over that...
erfz said:
I believe I can use linear momentum conservation for this:
No. There is a frictional impulse from the ground (or the ball would not be immediately rolling) and a horizontal impulse from the hinge.
You need to know whether the impact is completely elastic or completely inelastic, or, in between, the coefficient of restitution.
 
  • #3
haruspex said:
That is almost impossible. You would have to allow the collision process to take a significant time and for the coefficient of friction with the ground to be very high. But glossing over that...
What is the reasoning behind that? Such a scenario seems very possible in real life without requiring a great time or large coefficient of friction.
haruspex said:
No. There is a frictional impulse from the ground (or the ball would not be immediately rolling) and a horizontal impulse from the hinge.
You need to know whether the impact is completely elastic or completely inelastic, or, in between, the coefficient of restitution.
I see. But even if the impact was elastic, how would that allow me to account for the frictional and hinge impulses?
 
  • #4
erfz said:
What is the reasoning behind that?
In an impact you get large forces acting for very short times. Often it is very hard to know what the actual force profile looks like over that time. It is much easier to calculate the total momentum and work with that.
In the present problem, the impact is horizontal, so does not affect the normal force between ball and ground, so does not affect the maximum frictional force. The peak force in the impact would almost always overwhelm the frictional force.
This might not be apparent because if the frictional coefficient is high then it might quite quickly develop rolling contact.
In other situations, the impact might have a component normal to the ground, thereby generating a corresponding normal impulse reaction and corresponding impulsive friction.
erfz said:
if the impact was elastic, how would that allow me to account for the frictional and hinge impulses?
You just need another equation. If completely elastic, work is conserved; if completely inelastic then the linear motions will have the same velocity immediately after impact (thereby maximising work lost).
 
  • #5
@haruspex So for example, if we assume that collision is (instantaneously?) completely inelastic, then is it true that ##m \omega_i L/2 = (m + M)v = (m + M) \omega_f L/2##? Wouldn't this mean that I don't even need the angular momentum expression?

Even if this is true, I suspect that friction does work to make the ball roll and that ##v## is not precisely ##v_{cm,ball}.## In particular, ##v_{cm,ball} < v##. Realistically, how different would they be and what determines this difference?
 
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  • #6
erfz said:
@haruspex So for example, if we assume that collision is (instantaneously?) completely inelastic, then is it true that ##m \omega_i L/2 = (m + M)v = (m + M) \omega_f L/2##?
No, reread my post. You get an equation about velocities, not momenta.
 
  • #7
haruspex said:
No, reread my post. You get an equation about velocities, not momenta.
So then ##\omega_f L/2 = v_{ball, cm}##?
And if the collision were perfectly elastic, then ##1/2I_{rod}\omega_i^2 = 1/2I_{rod}\omega_f^2 + 1/2Mv_{ball,cm}^2 + 1/2I_{ball}(\frac{v_{ball,cm}}{R})^2##?
 
  • #8
erfz said:
So then ##\omega_f L/2 = v_{ball, cm}##?
Not quite. How fast would the tip of the rod be moving in that case?
 
  • #9
haruspex said:
Not quite. How fast would the tip of the rod be moving in that case?
##v_{tip} = \omega_f L = v_{ball,cm}##? Also, see my edit if the collision is elastic.
 
  • #10
erfz said:
And if the collision were perfectly elastic, then ##1/2I_{rod}\omega_i^2 = 1/2I_{rod}\omega_f^2 + 1/2Mv_{ball,cm}^2 + 1/2I_{ball}(\frac{v_{ball,cm}}{R})^2##?
yes, depending on the axes for the two moments of inertia.
 
  • #11
erfz said:
##v_{tip} = \omega_f L = v_{ball,cm}##?
Yes.
 
  • #12
haruspex said:
yes, depending on the axes for the two moments of inertia.
How do you mean? In my case I would choose the axis passing through the tip of the rod.

Also, as for the coefficient of restitution, I see that it is defined as ##e = \frac{v_b - v_a}{u_a - u_b}## in one-dimension, where ##v## denotes final velocities and ##u## denotes initial velocities of two objects. If I call the velocities of the rod ##v_b## and ##u_b##, does this refer to the velocities of its center-of-mass or of its tip? Also, is it possible to measure this coefficient in such a problem without having the described collision occur? How could I even measure it, practically speaking?
 
  • #13
erfz said:
How do you mean? In my case I would choose the axis passing through the tip of the rod.
I assume you mean through the hinge, but the value is the same.
Since that is the axis of the rod's rotation, not its mass centre, it is not clear from your equation what Iball means. Again, you could choose mass centre of the ball or its instantaneous centre of rotation. I'm pretty sure which you mean, but it should be made clear in the equation.
erfz said:
does this refer to the velocities of its center-of-mass or of its tip?
Clearly not the mass centre since that again would lead to the rod penetrating the ball when e=0. It would have to be the tip.
However, it is not entirely clear that the formula can be directly applied in this case. You could check by plugging in e=1 and seeing if you get the same solution as you got for perfect elasticity. If not, let me know and I will see if I can come up with a working formula.
 
  • #14
haruspex said:
I assume you mean through the hinge, but the value is the same.
Since that is the axis of the rod's rotation, not its mass centre, it is not clear from your equation what Iball means. Again, you could choose mass centre of the ball or its instantaneous centre of rotation. I'm pretty sure which you mean, but it should be made clear in the equation.

Clearly not the mass centre since that again would lead to the rod penetrating the ball when e=0. It would have to be the tip.
However, it is not entirely clear that the formula can be directly applied in this case. You could check by plugging in e=1 and seeing if you get the same solution as you got for perfect elasticity. If not, let me know and I will see if I can come up with a working formula.
I am not able to sit down and check this right now but I do want to bring up another question I thought of.
Since the hinge exerts a horizontal force, making momentum not conserved, doesn't it also do work? Wouldn't energy not be conserved? So then how can the collision be perfectly elastic... Is this from assuming that the work done by the hinge is negligible?
 
  • #15
erfz said:
Since the hinge exerts a horizontal force, making momentum not conserved, doesn't it also do work?
The hinge does not move. Work = force times distance moved in the direction of the force.
 
  • #16
haruspex said:
The hinge does not move. Work = force times distance moved in the direction of the force.
Ah of course
 
  • #17
@haruspex
Using energy conservation (assuming elastic), I get that ##\omega_f = 4.49 ~rad/s## and that ##v_{ball, cm} = 1.49 ~m/s##.
Using $$e = \frac{v_{tip, f} - v_{ball, cm}}{0 - v_{tip, i}} = 1$$ and of course ##v_{tip} = \omega L##, I get that ##\omega_f = 3.69 ~rad/s## and that ##v_{ball, cm} = 2.06 ~m/s##. So it does not seem like ##e## can be used directly here, unless I've made a mistake.
 
  • #18
erfz said:
@haruspex
Using energy conservation (assuming elastic), I get that ##\omega_f = 4.49 ~rad/s## and that ##v_{ball, cm} = 1.49 ~m/s##.
Using $$e = \frac{v_{tip, f} - v_{ball, cm}}{0 - v_{tip, i}} = 1$$ and of course ##v_{tip} = \omega L##, I get that ##\omega_f = 3.69 ~rad/s## and that ##v_{ball, cm} = 2.06 ~m/s##. So it does not seem like ##e## can be used directly here, unless I've made a mistake.
Working purely algebraically (no numbers) I get that the usual coefficient of restitution formula works, using the velocity of the tip of the rod.
Please try it the same way (no numbers) and post your working. You should be able to leave the rod's MoI as just "I", not needing to substitute mL2/3.
 
  • #19
haruspex said:
Working purely algebraically (no numbers) I get that the usual coefficient of restitution formula works, using the velocity of the tip of the rod.
Please try it the same way (no numbers) and post your working. You should be able to leave the rod's MoI as just "I", not needing to substitute mL2/3.
I have decided to upload the image of my written algebraic work. Hopefully it is neat enough for you to read.
On the bottom right, I have used the angular momentum expression, multiplied both sides by ##\omega_i + \omega_f## to utilize the relationship with the coefficient of restitution, and combined. I get a form strikingly similar to the energy expression, but not quite. The denominator of the ball's rot. KE contains ##LR## rather than ##R^2##. Interestingly, the numbers I found before work here.

Where have I gone wrong?
 

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  • #20
erfz said:
I have decided to upload the image of my written algebraic work. Hopefully it is neat enough for you to read.
On the bottom right, I have used the angular momentum expression, multiplied both sides by ##\omega_i + \omega_f## to utilize the relationship with the coefficient of restitution, and combined. I get a form strikingly similar to the energy expression, but not quite. The denominator of the ball's rot. KE contains ##LR## rather than ##R^2##. Interestingly, the numbers I found before work here.

Where have I gone wrong?
I had forgotten that the ball is supposed to start rolling instantly. This might be why the standard equation does not work. (My original reason for thinking it might not work was connected with the hinge on the rod.)
I'll take the rolling into account and get back to you, but it might be 12 hours or so before I get the chance.
 
  • #21
erfz said:
I have decided to upload the image of my written algebraic work. Hopefully it is neat enough for you to read.
On the bottom right, I have used the angular momentum expression, multiplied both sides by ##\omega_i + \omega_f## to utilize the relationship with the coefficient of restitution, and combined. I get a form strikingly similar to the energy expression, but not quite. The denominator of the ball's rot. KE contains ##LR## rather than ##R^2##. Interestingly, the numbers I found before work here.

Where have I gone wrong?
The problem is that for the ball to transition immediately to rolling there must be a horizontal frictional impulse at the ground during the collision. This invalidates your angular momentum equation.
To fix this, suppose the impulse at the collision point is J. In terms of that, you can now write the angular momentum equation for the rod about its hinge and that for the ball about its pivot (the point of contact with the ground). J can then be eliminated between the two equations.
Indeed, we can generalise the problem to two objects with MoIs I1, I2, about their hinge points, some initial angular velocities about those hinges, and respective final angular velocities.
Let the line of action of the collision impulse be r1, r2 respectively from the hinges. (I.e the perpendicular distances to the impulse vector.)
It turns out that the usual linear velocity relationship still arises. If, at the point of impact, the first body's velocity goes from u11r1 to u'1, etc. and KE is conserved then u'2-u'1=u1-u2.
 
  • #22
haruspex said:
The problem is that for the ball to transition immediately to rolling there must be a horizontal frictional impulse at the ground during the collision. This invalidates your angular momentum equation.
To fix this, suppose the impulse at the collision point is J. In terms of that, you can now write the angular momentum equation for the rod about its hinge and that for the ball about its pivot (the point of contact with the ground). J can then be eliminated between the two equations.
Indeed, we can generalise the problem to two objects with MoIs I1, I2, about their hinge points, some initial angular velocities about those hinges, and respective final angular velocities.
Let the line of action of the collision impulse be r1, r2 respectively from the hinges. (I.e the perpendicular distances to the impulse vector.)
It turns out that the usual linear velocity relationship still arises. If, at the point of impact, the first body's velocity goes from u11r1 to u'1, etc. and KE is conserved then u'2-u'1=u1-u2.
Hm that's interesting.
I just realized a few hours ago that I can write ##I_{rm}\omega_i = I_{rm}\omega_f + Mv'L## where ##v'## is the translational velocity of the ball right after the collision and before it starts rotating. This is before the angular impulse has an effect.
Of course, the ball has to start rolling after this. If we assume that the energy lost to sliding is negligible (because the distance it slides is ##\approx 0##), I can simply write ##\frac{1}{2}Mv'^2 = \frac{1}{2}Mv^2 + \frac{1}{2}I_b\omega^2##. ##v## is the new translational velocity after the ball has started rolling.
If I treat the problem like this, using the coefficient of restitution ##e = 1## allows me to obtain the energy equation with the angular momentum expression I've written.
Effectively, I guess this is the same as what you've suggested.
 
  • #23
erfz said:
Hm that's interesting.
I just realized a few hours ago that I can write ##I_{rm}\omega_i = I_{rm}\omega_f + Mv'L## where ##v'## is the translational velocity of the ball right after the collision and before it starts rotating. This is before the angular impulse has an effect.
Of course, the ball has to start rolling after this. If we assume that the energy lost to sliding is negligible (because the distance it slides is ##\approx 0##), I can simply write ##\frac{1}{2}Mv'^2 = \frac{1}{2}Mv^2 + \frac{1}{2}I_b\omega^2##. ##v## is the new translational velocity after the ball has started rolling.
If I treat the problem like this, using the coefficient of restitution ##e = 1## allows me to obtain the energy equation with the angular momentum expression I've written.
Effectively, I guess this is the same as what you've suggested.
I don't think that we can treat the impact and the transition to rolling as separate events. In effect, that would be saying that the ball initially sits on a frictionless surface and only reaches the (extremely) frictional surface later. That could alter the proportion of energy that ends up in the two objects. In the simultaneous treatment, the frictional impulse from the ground increases the collision impulse, which in turn pushes back the rod more.
 
  • #24
haruspex said:
I don't think that we can treat the impact and the transition to rolling as separate events. In effect, that would be saying that the ball initially sits on a frictionless surface and only reaches the (extremely) frictional surface later. That could alter the proportion of energy that ends up in the two objects. In the simultaneous treatment, the frictional impulse from the ground increases the collision impulse, which in turn pushes back the rod more.
In that case, let me see if I've got the equations correct.
haruspex said:
The problem is that for the ball to transition immediately to rolling there must be a horizontal frictional impulse at the ground during the collision. This invalidates your angular momentum equation.
To fix this, suppose the impulse at the collision point is J. In terms of that, you can now write the angular momentum equation for the rod about its hinge and that for the ball about its pivot (the point of contact with the ground). J can then be eliminated between the two equations.
I'm not entirely sure what you mean here.
How I imagine it is:
For the rod: ##I_{rm}\omega_i - NLt = I_{rm}\omega_f## (about hinge)
For the ball: ##-fRt = - I_b v/R## (about ball's COM)
where ##f## is the force of friction; ##N## is the normal force between the rod and ball; ##t## is the collision time.
And then we know ##Ma = N - f## and use that to solve the equations, I think? But that still leaves time to be solved.

I'm not entirely sure how the J's cancel and what J would be here.
What equations would you set up?
 
  • #25
@haruspex
Sorry for bumping; haven't heard from you in a while.
What equations would you use here?
 
  • #26
erfz said:
In that case, let me see if I've got the equations correct.

I'm not entirely sure what you mean here.
How I imagine it is:
For the rod: ##I_{rm}\omega_i - NLt = I_{rm}\omega_f## (about hinge)
For the ball: ##-fRt = - I_b v/R## (about ball's COM)
where ##f## is the force of friction; ##N## is the normal force between the rod and ball; ##t## is the collision time.
And then we know ##Ma = N - f## and use that to solve the equations, I think? But that still leaves time to be solved.

I'm not entirely sure how the J's cancel and what J would be here.
What equations would you set up?
As I posted, it will be easier to take moments for the ball about its point of contact with the ground. That eliminates friction, while producing a second equation with Nt in it.
The time should not matter because N and t always appear as the combination Nt. This is what I was calling J, the impulse between the objects.
 
  • #27
haruspex said:
As I posted, it will be easier to take moments for the ball about its point of contact with the ground. That eliminates friction, while producing a second equation with Nt in it.
The time should not matter because N and t always appear as the combination Nt. This is what I was calling J, the impulse between the objects.
So..
For the rod: ##I_{rm}\omega_i - NLt = I_{rm}\omega_f##
For the ball: ##-NRt = -MRv - I_bv/R##?
(I am not 100% sure on the signs of the angular impulse. How does it depend on my conventions?)

Combining these, I get that ##I_{rm} \omega_i = I_{rm} \omega_f + MLv + I_bvL/R^2##. Is there an intuitive explanation here, assuming this is correct?
 
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  • #28
erfz said:
So..
For the rod: ##I_{rm}\omega_i - NLt = I_{rm}\omega_f##
For the ball: ##-NRt = -MRv - I_bv/R##?
(I am not 100% sure on the signs of the angular impulse. How does it depend on my conventions?)

Combining these, I get that ##I_{rm} \omega_i = I_{rm} \omega_f + MLv + I_bvL/R^2##. Is there an intuitive explanation here, assuming this is correct?
It looks right. If, when combined with work conservation, it leads to the usual "relative velocity after = -relative velocity before" then that's as close as I can see to an intuitive explanation.
 
  • #29
haruspex said:
It looks right. If, when combined with work conservation, it leads to the usual "relative velocity after = -relative velocity before" then that's as close as I can see to an intuitive explanation.
Yes, this expression does allow me to reconstruct work conservation using ##e = 1##.
 
  • #30
@haruspex
I'm thinking now, would it be possible to set your axis on the table that the ball sits on and treat the rod and ball as a single system?
This would eliminate frictional torque, I think.
Do you see anything wrong with that?

Like so:
$$m \omega_i L/2 (R + L/2) - I_r \omega_i = m \omega_f L/2 (R + L/2) - I_r \omega_f + MvR + I_b v/R,$$ where I have used the linear velocities of the rod's COM, ##I_r## is the rod's MOI about its COM, ##I_b## is about the ball's COM.
I don't seem to get the same numbers with this expression; something is wrong here mathematically or conceptually.
 
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  • #31
erfz said:
@haruspex
I'm thinking now, would it be possible to set your axis on the table that the ball sits on and treat the rod and ball as a single system?
This would eliminate frictional torque, I think.
Do you see anything wrong with that?
But then you would have torque from the unknown reaction at the rod's hinge.
You can take anywhere as your reference points for angular momentum, but if that means dragging in the reaction force from the hinge or the friction from the table then you will need to bring in a linear momentum equation as well in order to eliminate the unknown. The great benefit of using the rod's axis for its equation and the point of contact of the ball with the table for its equation is that we never get those impulses in the equations.
 
  • #32
haruspex said:
But then you would have torque from the unknown reaction at the rod's hinge.
You can take anywhere as your reference points for angular momentum, but if that means dragging in the reaction force from the hinge or the friction from the table then you will need to bring in a linear momentum equation as well in order to eliminate the unknown. The great benefit of using the rod's axis for its equation and the point of contact of the ball with the table for its equation is that we never get those impulses in the equations.
Ah, shoot. I forgot about that completely. Thank you very much!
 

Related to Rod swinging and hitting a ball

1. What is the science behind rod swinging and hitting a ball?

Rod swinging and hitting a ball involves the principles of force, motion, and energy. When the rod is swung, it creates a force that is transferred to the ball upon impact. This force causes the ball to move in a certain direction, and the amount of force applied determines the speed and distance of the ball's movement.

2. How does the angle of the rod affect the trajectory of the ball?

The angle of the rod can greatly impact the trajectory of the ball. When the rod is swung at a lower angle, the ball will travel in a lower, more horizontal trajectory. On the other hand, a higher angle of the rod will result in a higher, more vertical trajectory of the ball.

3. What is the role of momentum in rod swinging and hitting a ball?

Momentum is the product of an object's mass and velocity, and it plays a crucial role in rod swinging and hitting a ball. The more momentum the rod has when it hits the ball, the more force will be transferred to the ball, resulting in a stronger hit.

4. How does the type of rod used affect the outcome of hitting a ball?

The type of rod used can greatly impact the outcome of hitting a ball. A heavier rod will have more mass and therefore more momentum, resulting in a stronger hit. A lighter rod may be easier to swing, but it will have less momentum and may result in a weaker hit.

5. What factors can influence the accuracy of rod swinging and hitting a ball?

There are several factors that can influence the accuracy of rod swinging and hitting a ball. These include the angle and speed of the swing, the type and weight of the rod, the type and size of the ball, and external factors such as wind or surface conditions. Practice and proper technique can also greatly improve accuracy.

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