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Homework Help: A series of questions about Rotational Dynamics that will be continuously updated.

  1. Mar 18, 2006 #1
    I guess I'll introduce myself as well:

    My name is, well, my username, but you can call me Drew (most people do). I'm 18 (a senior in highschool) and in an AP Physics II class focusing on mechanics. I posted something in the advanced physics forum that should probably have been moved here, so I think I'm going to post my questions about rotational dynamics here from now on. If you want to look at my other thread to understand my "problems", please do so: https://www.physicsforums.com/showthread.php?t=114597"

    Essentially, I have a horrible "text book" (Schaums Outline of Physics for Engineering and Science) that I can't trust, and I have a test on Tuesday. Also, I'm on a break right now and don't have access to the teacher to ask questions. Most of my questions will be focusing on understanding "why" rather than looking for numerical answers. The reason for this is that although I seem to be able to do problems confined to one part of Rotational Dynamics, I cannot seem to integrate all of these "tools" and use them to answer more complicated problems. This is why I think "Why" would help.

    So if you're sympathetic with my cause, I would appreciate your help.:smile:

    My first question is located in the link; here's another:

    There are a mix of capitalized letters and lower case letters that confuse the hell out of me.

    i.e. (M) vs. (m) or (R) vs. (r)

    Sometimes (M) stands for the mass of the entire system and sometimes it just stands for the mass of the object affected by torque. Is there any standard for this or does it vary from different types of teaching? With (r) & (R), (R) seems to stand for the total radius of the object, while (r) I'm told is the radius of the "torque disc". Take a roll of tape, for example; (r) is the measurement from the center to the edge of the inside circle and (R) is the measurement from the center to the outside of the tape. But isn't the whole system affected by torque? What exactly is a "torque disc"?

    Another question:

    How can I find the moment of inertia of an object without using calculus? Is this possible, or am I just supposed to memorize formulas for common shapes?

    Thanks in advance for my petty questions. I used to be excellent at this stuff, but I fell off a cliff last summer and have lost a lot of my processing ability (no joke). It takes me hours to grasp things that I could have in my sleep before. Ugh.
    Last edited by a moderator: Apr 22, 2017
  2. jcsd
  3. Mar 18, 2006 #2

    Chi Meson

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    There is no set convention for the use of M vs. m, or R vs. r. It will entirely depend on the context of each problem. Often, M and m is used instead of using subscripts, just to differentiate the masses of two objects .

    Usually, "r" is a "radial displacement" meaning a distance from the center to a point on the circumference of a circle or sphere. "R" will often be used to be the actual radius of an object, either a disk or a planet or some other thing.

    In problems of torque, when a force is applied non-perpendicularly to the radial line (at point of force application) the sin theta factor comes in. This means that the "lever arm" is shorter than the actual radial distance to point of contact. I do not use the term "torque disk," but I imagine it is referring to the disk that has a radius of the shortened "lever arm." So "R" would be the actual length of the "stick" and torque would be RF sin theta. the lever arm would be length "r" and be equal to R sin theta, so that torque is rF. I never lliked this hair -splitting done in some books. I say, it's rF sin theta, and that's it.

    You really do need calculus for various shapes. All of the specific formulas are based on mr^2. Usually people just look up the necessary formulas from a table. It never hurts to memorize the basic ones in your book though.
  4. Mar 18, 2006 #3
    I posted this in the other thread, but I'll post it again here:

    Say you have an atwood machine (pulley) with mass (M) and radius (R). You have a block of mass (m) hanging from one side of the pulley. Find the linear acceleration (a) of the block.

    Ok. So first guess is to use F=ma. Bingo. Second, plug in the forces. We have the weight of the block (mg) and Tension. So..

    mg - T = ma

    Now, what I substituted for T doesn't make sense to me. It was [tex]\frac{1}{2}[/tex](M)(a). I guess we got this from Torque = I[tex]\alpha[/tex] and plugged (T)(R) for Torque, then did a bunch of algebra and winded up with (.5)(M)(a) being Torque. Then we solved for (a), a=[tex]\frac{mg}{m+(.5)M}[/tex], and all was gravy.

    But I don't understand why the mass of the pully is revelant to the Tension and the mass of the block isn't. It used to be that the mass of the block was all that mattered, and now it's thrown out of the picture? The Tension being subject only to the mass of the pully and the acceleration?
  5. Mar 18, 2006 #4
    I don't understand your argument.

    You have solved for a, if you substitue this into your origional equation and then re-arrange you get:

    [tex]T=mg- \frac{m^2 g}{m+\frac{1}{2}M}=mg(1-\frac{1}{1+\frac{M}{2m}})[/tex]

    This shows that the tension is based on the mass of the block.

    Last edited: Mar 18, 2006
  6. Mar 18, 2006 #5

    To get the Tension of the system, I used:

    [tex]\tau_{net}[/tex] = I [tex]\alpha[/tex]

    then substituted Tension x Radius in for [tex]\tau_{net}[/tex], and [tex]\frac{1}{2}[/tex](mass of the pulley)([tex]radius^2[/tex]) for I and [tex]\frac{a}{R}[/tex] for [tex]\alpha[/tex].

    (T)(R)= [tex]\frac{1}{2}[/tex](mass of the pulley)([tex]radius^2[/tex])([tex]\frac{a}{R}[/tex])

    All of the R's cancel out, and we wind up with

    ([tex]\frac{1}{2}[/tex])(mass of the pulley)(a) = Tension


    There's no mass of the block in there.
    Last edited: Mar 18, 2006
  7. Mar 18, 2006 #6
    Its hidden in a, which you prove in the next step.
  8. Mar 18, 2006 #7
    Excellent. I didn't even think about that.
  9. Mar 18, 2006 #8
    Ok. The parallel-axis theorem.

    The moment of inertia about a parallel axis displaced by a distance "d" from the center of mass axis can be represented by:

    [tex]I_{parallel-axis}[/tex] = [tex]I_{cm}[/tex] + M [tex]d^2[/tex]

    I must be given one of the moments of inertia and the distance in order to solve this, right?
    Last edited: Mar 18, 2006
  10. Mar 18, 2006 #9
    You have a sphere of radius .012m and mass .036kg rolling down a slope with an initial speed of .48 m/s. How fast will it be moving after it has dropped 12 cm in elevation?

    Conservation of energy (taking into account the total KE of a rolling sphere) (w=angular velocity):

    [tex]\frac{1/2}[/tex]I[tex]w_1^2[/tex] + [tex]\frac{1}{2}m[/tex][tex]v_1^2[/tex] + mgh = [tex]\frac{1/2}[/tex]I[tex]w_2^2[/tex] + [tex]\frac{1}{2}[/tex]m[tex]v_1^2[/tex]

    multiply everything by 2 and cancel out the masses:

    [tex]\frac{2}{5}[/tex][tex]r^2[/tex][tex]w_1^2[/tex] + [tex]v_1^2[/tex] + 2gh = [tex]\frac{2}{5}[/tex][tex]r^2[/tex][tex]w_2^2[/tex] + [tex]v_2^2[/tex]

    plug in the numbers for the initial situation and convert "w" on the right to [tex]\frac{v}{r}[/tex] use this to cancel the square on "r" out of [tex]\frac{2}{5}[/tex][tex]r^2[/tex][tex]w_2^2[/tex]

    and I wind up with

    2.72256 = [tex]\frac{2}{5}[/tex](.012)[tex]v_2^2[/tex] + [tex]v_2^2[/tex]


    2.72256 = (2)(.012)([tex]\frac{2}{5}[/tex])[tex]v_2^2[/tex]

    therefore, [tex]v_2[/tex] = [tex]\sqrt{283.6}[/tex] = 16.84 m/s

    That's way too fast. Where did I go wrong? Do I need to factor in the force of friction?
  11. Mar 18, 2006 #10


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    How can you factor in friction when you haven't been given any data regarding it? Your logic seems right to me, I haven't checked the arithmatic and details though.
  12. Mar 18, 2006 #11
    using [tex]\tau[/tex] = I [tex]\alpha[/tex]. I mean, we're not told if it's frictionless or not, but if we assume there is, we can use this equation to find it and plug it into F=ma and solve for a.

    ? The book gives the answer 1.38 m/s

    I've got to go eat dinner. I'll be back.
  13. Mar 18, 2006 #12


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    I'm afraid I've spotted an error in your working

    Upto here is right. However, subsituting [itex]\frac{v}{r}[/itex] in gives (i'm using u for initial velocity);

    [tex]\frac{2}{5}u^2 + u^2 + 2gh = \frac{2}{5}v^2 + v^2[/tex]

    Can you go from here?

    NOTE: The answer I get is different to the answer the book gives, so I'm gona check my work.
    Last edited: Mar 18, 2006
  14. Mar 18, 2006 #13

    Ah! My algebra is horrible. Also, I just solved for the initial angular velocity and plugged it in (40 rad/s) on the left. But still, thank you.
  15. Mar 18, 2006 #14
    I've got it!

    I still solved for a number on the left side (2.72256) and used your correction on the right:

    2.72256 = [tex]\frac{7}{5}[/tex][tex]v_2^2[/tex]

    multiplied both sides by [tex]\frac{5}{7}[/tex]

    1.94469 = [tex]v_2^2[/tex]

    [tex]v_2[/tex] = 1.39

  16. Mar 19, 2006 #15
    A disc is mounted with its axis vertical. It has radius R and mass M. It is initially at rest. A bullet of mass m and velocity v is fired horizontally and tangential to the disc. It lodges in the perimeter of the disk. What angular velocity will the disc aquire?

    So, Angular Momentum is conserved, and the initial angular momentum is just the angular momentum of the bullet, [tex]\vec{L_1}[/tex] = mvR.

    The resultant Angular Momentum [tex]\vec{L_2}[/tex] = m(rw)r + Iw

    [tex]\vec{L_1}[/tex] = [tex]\vec{L_2}[/tex]

    mvr = mw[tex]r^2[/tex] + [tex]\frac{1}{2}[/tex](m+M)[tex]r^2[/tex]w

    Is this the correct set-up?
  17. Mar 19, 2006 #16
    For [itex]L_{2}[/itex] the second term must be only angular momentum of the disc.So you should use [itex]I\omega = \frac{1}{2}Mr^2\omega[/itex]
    and not M+m since you have already included AM of bullet in the first term.
    Last edited: Mar 19, 2006
  18. Mar 19, 2006 #17
    Thank you.
  19. Mar 19, 2006 #18
    The evolution of a star depends on its size. If a star is sufficiently large, the gravity forces holding it together may be large enough to collapse it into a very dense object composed mostly of neutrons. The density of such a neutron star is about [tex]10^14[/tex] times that of the earth. Suppose that a star initially had a radius about that of our sun, 7 x [tex]10^8[/tex] , and that it rotated once every 26 days, as our sun does. What would be the period of rotation (the time for 1 rev) if the star collapsed to a radius of 15km?


    [tex]\vec{L_1}[/tex] = [tex]\vec{L_2}[/tex]

    [tex]\vec{L_1}[/tex] = [tex]I_1[/tex][tex]w_1[/tex] [tex]\vec{L_1}[/tex] = [tex]I_1[/tex][tex]w_1[/tex]

    I = [tex]\frac{2}{5}[/tex]m[tex]r^2[/tex]

    [tex]\frac{2}{5}[/tex]m[tex]r_1^2[/tex][tex]w_1[/tex] = [tex]\frac{2}{5}[/tex]m[tex]r_2^2[/tex][tex]w_2[/tex]

    multiply by [tex]\frac{5}{2}[/tex], cancel out "m";

    [tex]r_1^2[/tex][tex]w_1[/tex] = [tex]r_2^2[/tex][tex]w_2[/tex]

    w = 2[tex]\pi[/tex]f;

    ((7 x [tex]10^8[/tex])^2)([tex]2\pi[/tex])(26) divided by ([tex]15^2[/tex])(2)([tex]\pi[/tex])= ([tex]15^2[/tex])(2)([tex]\pi[/tex])([tex]f_2[/tex]) divided by ([tex]15^2[/tex])(2)([tex]\pi[/tex])

    cancel out 2[tex]\pi[/tex] on both sides;

    [tex]f_2[/tex] = ([tex]\(7)[tex]10^8[/tex]^2[/tex](26) divided by ([tex]15^2[/tex])

    [tex]f_2[/tex] = 56622222220000000 rad/day


    After I fix this, where do I go from here?
    Last edited: Mar 19, 2006
  20. Mar 19, 2006 #19


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    I haven't checked your working, so I can't say whether your values are correct but for your next step you need to think about the relationship between frequency and time period.
  21. Mar 19, 2006 #20
    sorry about the latex. it's really not coming out correctly
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