Physics midterm problem: inelastic rotational collision

Physics 205A Midterm 2, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.76

[20 points.] A 0.500 kg point mass is dropped onto a disk rotating at 4.00 rad/s that has a mass of 0.700 kg and a radius of 0.150 m. The point mass eventually rotates with the disk at a distance of 0.100 m from the axis. What is the final kinetic energy of the disk and point mass system? Show your work and explain your reasoning.

Solution and grading rubric:

• p = 20/20:
  Correct. Applies conservation of angular momentum to find the final angular velocity of the disk and point mass system; and then calculates the change in initial and final rotational kinetic energies in this inelastic rotational collision.
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Some attempt at applying conservation of angular momentum before evaluating the final kinetic energy.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Applies kinetic energy conservation to find final rotational kinetic energy, even though this is an inelastic rotational collision (as seen in lab), and if it were elastic, the final rotational kinetic energy would be exactly equal to the initial kinetic energy.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Sections 70854, 70855
p: 1 student
r: 3 students
t: 3 students
v: 31 students
x: 3 students
y: 0 students
z: 2 students

A sample of the sole "p" response, applying angular momentum conservation to find the final angular velocity of the point mass and disk system, and then calculating the final (rotational) kinetic energy of this system (from student 0215):
A sample "t" response (from student 0420), with an attmempt to calculate the final kinetic energy of the system, but with a realization that the final angular velocity is not yet determined:

Physics midterm problem: penny on turntable

Physics 205A Midterm 1, Fall Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Comprehensive Problem 5.80(b)

[20 points.] A Physics 205A student places a penny (mass 2.50 g) on an old phonograph turntable, at a distance of 10.0 cm from the center. The coefficient of static friction between the penny and the turntable is 0.19. The turntable starts with an initial angular velocity of 33.3 rpm, and speeds up to 45.0 rpm. Will the penny stay on the turntable? Neglect air resistance. Show your work and explain your reasoning.

Solution and grading rubric:

• p = 20/20:
  Correct. Argues that the penny will slide off the turntable before reaching its final speed of 45 rpm, by either solving for the required radial (centripetal) net force and showing that it is greater than the maximum static friction force; or by solving for the rotational speed that the penny would still remain on the turntable (when its static friction force is maxed out), and showing that it is less than 45 rpm = 4.7 rad/s. Free-body diagram, identification of forces, application of Newton's laws (first law, and second law for uniform circular motion), and interpretation of results is clearly shown.
  
• r = 16/20:
  Nearly correct, but includes minor math errors.
  
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least has free-body diagram with magnitudes of forces calculated, and/or rational attempt at applying Newton's second law for uniform circular motion to these force magnitudes.
  
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Typically calculates w_i, w_f, f_s, and/or m*r*w^2, but does not implement Newton's second law for uniform circular motion.
  
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Sections 70854, 70855
p: 8 students
r: 1 students
t: 9 students
v: 20 students
x: 6 students
y: 1 student
z: 0 students

A sample of a "p" response, demonstrating that the angular speed that the penny would begin to slide off the turntable is less that the final angular speed (from student 1977):
Another "p" response (from student ), demonstrating that the maximum static friction force is less than the centripetal net force required to keep the penny on the turntable at its final angular speed:

Physics clicker question: empty can derby

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Example 8.12

"All Disks Roll the Same," 1Q1035.mov
University of Minnesota, School of Physics & Astronomy
http://groups.physics.umn.edu/demo/mechanics/1Q1035.html

Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle:

A V-8 can and a Diet Pepsi can (both empty) are placed at the top of an inclined plane, and are released such that they begin to roll down at the same time. (This experiment is set-up, but not yet demonstrated yet for the students until after answers have been compiled.)

[0.6 participation points.] Which empty can do you think will reach downhill first?
(A) V-8 can.
(B) Soda can.
(C) (They will reach the bottom of the slope at approximately the same time.)
(D) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 11 students
(B) : 21 students
(C) : 5 students
(D) : 0 students

Correct answer: (C)

The energy conservation equation for a rolling ring (which approximates an empty beverage can) is:

0 = (1/2)*m*v_f^2 + (1/2)*I*w_f^2 + M*g*h,

where I = m*R^2, such that:

v_f = sqrt(g*h).

Thus both cans will have the same speed as they reach the bottom of the inclined plane, which can plausibly be reasoned to mean that they reach the bottom at the same time.

Physics clicker question: falling stick derby

Physics 5A, Fall Semester 2007
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Comprehensive Problems 8.97, 8.98

"Falling Meter Sticks--Scaling," 1Q2060.mov
University of Minnesota, School of Physics & Astronomy
http://groups.physics.umn.edu/demo/mechanics/1Q2060.html

Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle:

A 1-m stick and a 2-m stick are held vertically upright, and are released such that they begin to fall over at the same time. (This experiment is set-up, but not yet demonstrated yet for the students until after answers have been compiled.)

[0.6 participation points.] Which stick do you think will fall and hit the ground first?
(A) 1 m stick.
(B) 2 m stick.
(C) (They will hit the ground at approximately the same time.)
(D) (I'm lost, and don't know how to answer this.)

Sections 0906, 0907
(A) : 10 students
(B) : 7 students
(C) : 17 students
(D) : 1 student

Correct answer: (A)

The energy conservation equation for a falling stick is:

0 = (1/2)*I*w_f^2 + M*g*(L/2),

where I = (1/3)*M*L^2, such that:

w_f = sqrt(3*g/L).

Thus the shorter stick will have the faster speed as it hits the ground, which can plausibly be reasoned to be the stick that will reach the ground first, before the longer stick does.