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I have the following multiple choice question that I am attempting to answer:

enter image description here

For B1.3, (a) appears to be the most logical choice. It is my understanding that an object whose shape is asymmetric about the spin axis will result in an angular momentum vector not aligned with the spin axis which makes (c), (d) nonviable answers.

Additionally, for (c) nonzero linear or angular momentum means there is kinetic energy so that is definitely not correct.

Kinetic energy is a scalar, not a vector thus it has no direction which discounts (b) and (e) as answers.

B1.4 (a) or (b)?

Changing the thickness of one sheet would change the angular momentum but the rotation axis would still not be symmetrical so the spin axis and angular momentum would still not be aligned which makes (c), (d) and (e) invalid (as well as (e) making the assertion that kinetic energy has direction).

I'm a little confused as to what is implied by "dynamic bearing reactions". The way the cube is rotated about the spin axis means there is a couple and changing the thickness of one sheet would unbalance this couple - am I on the right track with this line of thought and have I made any conceptual errors in my reasoning so far?

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  • $\begingroup$ Your answer for the first part is incorrect. Some hints: With regard to the first question, what is the inertia tensor for this cube? How is it any different from that of a sphere? With regard to the second question, does the axis of rotation pass through the center of mass? $\endgroup$ – David Hammen Nov 15 '15 at 16:56
  • $\begingroup$ Given the engineering nature of the second part of the question, this question might be better suited for the engineering stack exchange site, engineering.stackexchange.com . $\endgroup$ – David Hammen Nov 15 '15 at 16:57
  • $\begingroup$ You can eliminate some of the choices in B1.3 because kinetic energy is a scalar (it has no direction) and because the kinetic energy in the student's frame is non-zero. $\endgroup$ – David Hammen Nov 15 '15 at 17:00
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With regard to question B1.3, one can immediately eliminate options b and e because energy is a scalar quantity, something without direction. One can also eliminate option c because the work of art has non-zero kinetic energy from the perspective of the (implied but obvious) observer frame of reference. That leaves options a, d, and f. The second conditions of options a and d are tautological true, for different reasons. The first conditions of options a and d are diametrically opposed, which means one can eliminate option f ("none of the above"). This question comes down to determining whether angular momentum and angular velocity are parallel in this case.

A hollow cube constructed from six identical thin square plates has an inertia tensor (e.g., see Tensor of inertia of a hollow cube) that looks just like that of a sphere, a diagonal matrix with all three diagonal elements equal to one another. Angular momentum is always parallel to angular velocity in this special case. Therefore, option a is incorrect. The only remaining answer is option d, "The angular momentum is aligned with the spin axis and kinetic energy is non-zero."


With regard to question B1.4, I first have to caveat my answer. I was trained to think as a physicist first and foremost, then as a mathematician, then as a computer scientist, and lastly as an engineer. (I regularly have to restrain myself from asking my engineering cohorts, "Why do you make things so convoluted?")

From what I've read, "bearing reactions" result from rotating about an axis where the center of mass is not on the axis of rotation or where the axis of rotation is not a principal axis of rotation (i.e., angular momentum is not parallel to angular velocity). Both conditions are true in the context of question B1.4: The rotation axis does not contain the center of mass, and angular momentum is not parallel to angular velocity.

As this is a homework question, I'll leave deriving that the rotation axis (as posed on the textbook question) does not pass through the center of mass as something for you to prove. I'll also leave deriving that angular momentum is not parallel to angular velocity in this case as something for you to prove.

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  • $\begingroup$ As a physicist, how would you rephrase "bearing reactions" to convey the point to either engineers or physicists? $\endgroup$ – user155876 Nov 16 '15 at 14:11
  • $\begingroup$ Both off-center-of-mass and off principle axis of rotation result in bearing reactions, but they have different characters in the sense that the former results in the force being in the same direction at both bearing (accelerating the CoM) and the latter generates opposite direction reactions (providing the requisite torque). You should be able to have combinations of the two as well. $\endgroup$ – dmckee --- ex-moderator kitten Nov 16 '15 at 18:55
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Rotation of a cube around its diagonal axis

The angular momentum is always aligned with the spin axis by definition and the kinetic energy of a spinning mass is non zero.

enter image description here

Source

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    $\begingroup$ That is not true. Angular momentum is not always aligned with the spin axis. Also the kinetic energy of a spinning point mass (without any translation) is zero. Otherwise it is $KE=\frac{1}{2} m v^2$ $\endgroup$ – John Alexiou Nov 15 '15 at 16:54
  • $\begingroup$ ja72: "Also the kinetic energy of a spinning point mass (without any translation) is zero."?!! By definition, a point mass can not spin about its center. $\endgroup$ – Energizer777 Nov 15 '15 at 17:09
  • $\begingroup$ My answer is correct. The minuses are there just to discredit me. There is a troublemaker on this forum who, for unknown reasons, follows me and gives me minuses. $\endgroup$ – Energizer777 Nov 15 '15 at 17:38
  • $\begingroup$ @ja72: he didn't say point mass. Can you show an example where angular momentum vector isn't aligned with axis of rotation. Just curious. $\endgroup$ – Gert Nov 15 '15 at 18:59
  • $\begingroup$ @Gert, take a barbell (two equal point masses at the ends of a light rod), run an axis through the center of the rod but neither parallel nor perpendicular to the rod. Spin around that axis. Remember to compute angular momentum as a vector by $\vec{L}_\text{point mass} = \vec{r} \times \vec{p}$. The contribution from both point masses point in the same direction and not co-linear with the axis. The basic assertion in this post requires a degree of symmetry not present in general mass distributions. $\endgroup$ – dmckee --- ex-moderator kitten Nov 15 '15 at 19:59

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