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DISCUSSION THREAD
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DISCUSSION and HINTS
Recall the definition of angular momentum of a particle P (of mass m) about a fixed point O: HO = m rP/O x vP.
For this problem, use this equation to find the angular momentum for each particle and add these together. As you work the problem, consider the number of cancellations that occur among these terms and consider why these cancellations occur. This will help you get insights on the meanings of angular momentum.
The equation from the book is HO = rP/O x mvP. Should we use the one above instead?
m is a constant so the answer will be the same no matter where you put it in the equation.
Would the units for this problem be slugs-rad/s?
The units would be slugs*ft^2/s
Remember that components along the axis that travels towards the center O will go to zero. Breaking up velocities into components will make this more clear in the cross products.
Should we define a constant axis system for the whole system, or will it change particle to particle (with the radial unit vector always pointing away from the origin) as we draw each FBD?
All of your position vectors should be referenced to the same point O. Do not change from point to point.
I would recommend using cartesian coordinates and not polar for this problem. In that case you would not have to worry about radial unit vectors with polar.
Is it correct that gravity does not affect this system and therefore does not need to be accounted for in the calculations?
Calculating angular momentum is strictly a kinematics thing: rx(m*v). Forces are not included in the calculation of H_O.
Although forces do not affect the values of H_O, they do influence the rate of change of H_O through the moments created about O: M_O = dH_O/dt.
Don't forget to remember that angular momentum is still dependent on mass. So don't forget to make sure that you are using the right mass for each particle when doing your calculations.
I found it helpful for my first step to do a little bit of trig to help solve for angles tied to B and C.
You can also look at it and see that it is a 3-4-5 triangle which you can use to avoid long decimals in angles
Since we are finding angular momentum, would the units be in rads or ft/s^2?
Just look at what terms make up angular momentum: rx(m*v) = (ft)*(slugs)*(ft/s) = ft^2*slugs/s.
Yeah this entire problem is one big sum of the angular velocity equation. It gets reduced when you cancel out all of the cross products in the same direction because that is zero. Then just make sure your cross products are right and the final answer will be in the k direction because it is a rotational value assuming you set your plan as i and j.
While I recommend using the the cross product formula to get used to it, you don't actually need to use it for most particles. You can just use m * w where w is the (r_perpendicular * v) angular velocity formula.
Is it necessary to create free body diagrams for each of the particles? I think it might be a waste of time because it probably wouldn't give much if any new information about the problem. Also, when summing the angular momentums, would it be correct to say that the angular momentum from D will cancel out with the angular momentum from E? I believe this is true because they are the same distance, mass and speed, in exactly opposite rotational directions.
This problem is purely kinematics along with the definition of angular momentum about a point. If we were asked about how the angular momentum changes in time, then we would need to know and use the forces to first find the moments about point O. Since that was not asked, then FBDs would not be needed.
There will be considerable cancellation of terms in the answer for this problem. In each case, think about why this cancellation occurs.
This problem is a fairly simple application of the angular momentum formula. A large number of the cross products cancel out due to a direction being crossed with itself. Also note that the mass of G is 2m and not m. Basic trig will also be needed when considering the velocity vectors of B & C.
For anyone needing to get this problem done quickly, this problem is best accomplished by taking a large system including all points moving around point O. Because point O is stationary, the moment about it is zero as its reaction forces act on a fixed point.
This detail allows the total angular momentum to be found by adding the angular momentum of each point with the known cross-product equation. Many of the cross-products involve identical unit vectors, allowing many to be canceled out, and the remaining cross-products will be easy to add algebraically. The velocities at points C and B move at an angle, but their components can be found with basic trig as the needed lengths are given.
This problem is essentially just a sum of many singular cross products. Just keep in mind that there is 2m at G and the fact that the angles used at B and C to make components are different, where there sum adds to 90 degrees.
The trickiest part about this problem is simply doing the cross products correctly especially for particles B and C. The velocity vector for both particles should be equal but opposite signs. Besides this you shouldn’t have many complications and should see that your values for angular momentum end up canceling out with each other.
The best way I found to approach this problem is to find the angular momentum for each individual component about O. Then adding them up together to find the total will give you your answer.