ES201 – Conservation and Accounting Principles – Fall 2003-2004
Exam 2 - Instructor reactions, suggestions, and lessons to be learned.
Problem 1:
Part (c): A common conceptual mistake for this problem is that the net force is NOT equal to zero.
However, the net impulse (time-integral of the force) is equal to zero.
Part (e): The m 2V2 d term is an outflow term, hence, there should be a negative sign in front of it. Its
clockwise direction gives it another negative sign. Double negative makes it a positive in the final
result.
Problem 2:
I was pleased to see that almost everyone realized that conservation of linear momentum was needed to
solve this problem. Having said that, here are some general thoughts on students’ performance on the
problem in general:
o Many people made part (a) much too difficult. All that that was really required was a quick
conservation of mass and the realization that m = ρ AV = ρ V . The lesson: Whenever you
encounter an open system, it is a good idea to look at conservation of mass and its auxiliary
relations first.
o Several students applied linear momentum incorrectly. This was due to two major factors, namely,
incorrect or incomplete free body diagrams (system identification) and confusion as to the vector
nature of linear momentum. The lesson: Always draw a detailed sketch of your system with all the
relevant transport terms. Also, write conservation of linear momentum as a vector equation first
and then focus on its component(s). (An alternative is to clearly state that you are applying the soand-so component of the conservation of linear momentum equation.)
Problem 3:
You guys rock on impulses! You did a great job on this problem.
Problem 4:
This was a challenging problem. If I could revise the grading for this problem (and the entire course), it
would be
• 1/3 identify the system and carefully draw an accurate free body diagram,
• 1/3 write the proper conservation principle and give reasoning for simplifying equations, and
• 1/3 for rest of work (math, calculations, answer).
Those students who identified the system, drew a complete FBD, and gave the conservation principles;
thus, did very well on the problem. Note: F=ma is NOT a conservation principle. It is the result of
applying conservation of linear momentum to a closed system with a single force or just the net force
acting on the system.
Typical errors on this problem included the following:
•
“lbm = lbf” and other assorted units mistakes. This is a serious error! Please learn how to handle
units. These represent some of the largest numerical errors you can make.
•
Placing the friction force acting on the bus in the wrong direction. Because the pad is moving to the
left and the bus is moving is because of the friction force of the pad on the bus, it must point in the
same direction as the motion of the pad, i.e. to the left.
•
Incorrect tipping point. If the bus is being pulled to the left by a force at the bottom of the wheels,
the tipping point must be about the wheel on the right of the picture. When the bus begins to tip the
only normal force acting on the bus is under the right wheel and equals the weight of the bus. This
can and should be shown by conservation of linear momentum in the vertical direction.
•
Failing to give any direction indicators when you apply conservation of linear and angular
momentum. Without explicit direction indicators, your solutions are seriously flawed and getting
the right answer and communicating your answer, right or wrong, is very difficult. Is this a serious
error? Not if you feel the up can be down, right can be left, and counter clockwise can be
clockwise. YES, it is a serious error.
•
Placing VG and dVG /dt on the free body diagram. This can lead to serious errors. The free body
diagram (or linear momentum interaction diagram) should only show transports of linear
momentum. m(dVG /dt) can more appropriately be shown on a kinetic diagram. If your instructor
has not discussed this you may ask them; however, it is not required and some of us purposely do
not draw a kinetic diagram.
•
Mixing systems and forces. Many of you drew a system for the bus as a closed system and then
proceeded to draw both the force of the ram on the pad, Fram, and the friction force of the pad on
the bus Ffrcition, on the bus free-body diagram. Only the body forces and surface forces that
physically act on a system should be drawn on its free-body diagram.
•
Inertia forces. Many of you placed a force at the center of mass of the bus which was written as ma.
There is no such force acting on the system as we have formulated the conservation of linear
momentum.
•
Angular momentum can be calculated about any non-accelerating point; however for this problem
there are two prime candidates: Case I—Angular momentum about the center of mass of the bus G
or Case II—angular momentum about the point where the right tire touches the ground. For Case I,
the time-derivative of the AM goes to zero because rG × mVG and its derivative is always zero, i.e.
rG , VG , and aG are parallel vectors so the cross product is identically zero. For Case II, the time
derivative of the AM has a finite value and can be found by first finding the angular momentum of
the bus with respect to the point where the right tire touches the ground and then differentiating it.
•
Place forces where they act! Experience has
shown that in applying conservation of linear
and angular momentum to system all forces
should be placed on the free-body where they
act. This means either the tail or the head of the
arrow should be placed where the force acts.
By following this practice, you have a better
feel for what the various forces are doing to
your system.
mbus g
mPad g
Ffriction
FPad
Ffriction
FNL
FN
FN
FNR