| Problem statement Solution video |
DISCUSSION THREAD
Ask your questions here. Or, answer questions of others here. Either way, you can learn.
DISCUSSION and HINTS
The animation below shows the impact of the particle with the rigid bar. As stated in the problem statement, the particle sticks to the bar during the short impact time.
Considering the system made up of the particle and the bar, we see that there are no fixed points that are easily recognized and determining the location of the center of the mass requires some calculation. Because of this, it is advisable to consider the particle and the bar in separate FBDs in your analysis.
Recall the following four-step plan outline in the lecture book and discussed in lecture:
Step 1: FBDs
Draw individual free body diagrams (FBDs) of the particle and bar. Be sure to draw the impact force on both FBDs.
Step 2: Kinetics (impulse/momentum)
Consider the linear impulse/momentum equation for the particle and the angular and linear impulse/momentum equations for the bar. Note that each of these equations will include the impulse of the impact force.
Eliminate the impulse of the impact force from the above three equations. This will leave you with two equations in terms of the post-impact velocity of the particle, the post-impact velocity of the bar's center of mass, and the post-impact angular velocity of the bar.
Step 3: Kinematics
Since the particle sticks to the bar during impact, you can relate the post-impact velocities above through the rigid body kinematics equation:
vB= vG + ωbar x rB/G
where B is the top point on the bar where the particle impacts and sticks.
Step 4: Solve
From your equations in Steps 2 and 3, solve for the velocity of G and the angular velocity of the bar immediately after impact.
Is there a way to solve this problem by treating both the particle and the bar as part of the system for our LIM and AIM equations? (I.e. not using the impulse of the impact force by considering it to be an internal force)
There could be, but I personally think its easier to consider the linear momentum of the system before and after the collision and the moment created by the collision to the new combined system (because the particle sticks).
In the angular impulse momentum equation for the bar only, would the moment of inertia simply be the moment of inertia for a homogenous bar or do you have to use the parallel axis theorem to take into account the particle sticking after impact?
The angular momentum for the bar is: I_G*omega,
where I_G = (1/12)*M*L^2 (just the bar)
Because the particle sticks to the bar during the impact time being analyzed, can we assume the total momentum is conserved at point B?
As recommended above in the HINTS, it is recommended that you treat the bar and the particle individually. With this choice, the impact force will not be internal, and it will create a moment about point G on the bar. Therefore, angular momentum of the bar about G is not conserved.
You will also have LIM equation for the particle to use.
Would the moment at G just be the normal force of the particle? How should we find that?
Yes, if you use individual FBDs for the bar and the particle.
The AIM equation for the bar and the LIM equation for the particle will both include this normal contact force. Use those two equations to eliminate the contact force for those equations.
Would we need to also consider the mass of the particle inside our equation for moment of inertia?
It is recommended above that you treat the bar and the particle individually. With that, the AIM equation for the bar is just for the bar alone.
Thank you.
just to clarify B is at the top of the bar, where the particle is colliding with the bar correct?
Yeah, I believe we should be comparing the top of the bar (where the particle collides and what the hint refers to as B) to the center of mass of the bar for our kinematics.
Yes B is at the top of the bar right where the particle hits
From the animation, it looks like the center of mass and particle only have i components of velocity. Is this true?
after the impact I mean
Yes, I believe so from the cross product of kinematics I am confirming that will only vectors in the i direction
Is the linear momentum of the particle conserved?
The linear momentum isn't conserved but we still use it, you just have to include the impulse force from it hitting the bar in the equation. While we don't know what that force is, we can cancel it out using the other equations we obtain for the bar.
Do we need an equation for the linear impulse momentum of the bar also?
I think so, or at least that's what I did. Otherwise, I don't think we have enough equations to solve it.
In office hours I was told that I needed to integrate for one side of the equation when summing the masses around G. Why is that?
If you are referring to the linear impulse momentum equations it's because the forces aren't conserved. So to account for the force acting on point b you can use the integral of the sum of the moments about G to account for the force. This should eventually cancel out.
If the bar initially is moving only translationally, does this mean the initial angular velocity of the bar is 0? This would then make the initial angular momentum of the bar 0.
correct
Would the velocity of Vm continue to be in the i direction after impact or would it be perpendicular to the bar as it rotates
To my belief I think that it would be perpendicular the instant of/right after impact. Otherwise I think that adds a few more complexities.
It is both since the angle of the bar does not change during the short impact time. That is, the particle maintains only an x-component of velocity as well as the corresponding point on the bar.
Would it be effective to make a system of equations from our Linear Momentum and Angular momentum to solve. Also, should our answers be vectors?
Yes, that should be the plan. Since the problem asks for velocities and since velocities are vectors, your answers should be in vector form.
Is it the correct approach to have two linear impulse momentum equations and one angular impulse momentum equations? At least for the LIM, I am almost treating the two bodies as separate after the collision, just that they share the same velocity and position at the end of the bar.
I reread the directions, I appears that I am following the correct approach.
Would using the coefficient of restitution equation be helpful in this problem if we know that e is one because of the perfect elastic collision?
In this problem, e is not one; in fact, it is zero. This says that the n component of velocity for the particle and the n component of velocity of the contacting point on the bar are the same.
Could angular momentum have been taken at a different point like the impact point instead of the center of mass?
The point of contact is neither the center of mass nor a fixed point. Therefore, the short form of the AIM is not possible. Stick with G here on this problem.