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Homework Help: Finding the string tension and acceleration of a yo-yo.

  1. Apr 21, 2014 #1
    1. The problem statement, all variables and given/known data

    A yo-yo of mass [itex]M[/itex] is composed of two disks of radius [itex]R[/itex] separated by a shaft of radius [itex]r[/itex]. A
    massless string is would on the shaft, and the loose end is held in the hand. Upon release the
    yo-yo descends until the string is unwound. The string the begins to rewind, and the yo-yo
    climbs. Find the string tension and acceleration of the yo-yo in descent and ascent. Neglect the
    mass of the shaft and assume the shaft radius is sufficiently small so that the string is essentially

    2. Relevant equations

    [itex]J_{ij}=\int_{V}[\rho(r)\delta_{ij}\sum_{k}x_{k}^2-x_ix_j]dV [/itex]

    [itex]I_{ij}=J_{ij}-M[\delta_{ij}\sum_{k}a_{k}^2-a_ia_j] [/itex]


    3. The attempt at a solution

    Setup: First I'll define the [itex] x_3 [/itex] axis to be the axis of rotation of the yo-yo. Then [itex]\vec{\omega}[/itex] will have the components [itex]\begin{pmatrix}0\\0\\\omega_{3}\end{pmatrix}[/itex]. Now I need to find [itex]\mathbb{I}[/itex] in order to find [itex]\vec{L}[/itex]. Finally, because I can neglect the mass of the small shaft I will be calculating the inertia tensor of a solid cylinder.

    Calculation: Using [itex]J_{ij}=\int_{V}[\rho(r)\delta_{ij}\sum_{k}x_{k}^2-x_ix_j]dV [/itex] I'll find the inertia tensor at the bottom of the cylinder. So I'll need the volume of the region in question:

    [itex] 0\leq z\leq a, 0\leq r\leq R, 0\leq \theta\leq 2\pi [/itex]

    Case 1) [itex] \delta_{ij}=1 \Rightarrow J_{ii}=\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[x_1^2+x_2^2+x_3^2-x_i^2]rdrd\theta dz[/itex]

    where [itex] x_1 = rcos(\theta), x_2 = rsin(\theta), x_3 =z [/itex]

    [itex] i=1; \rightarrow J_{11}=\rho\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[(rsin(\theta))^2+(z)^2]rdrd\theta dz = \rho\int_{0}^{a}\int_{0}^{2\pi}[\frac{R^4sin^2(\theta)}{4}+\frac{R^2z^2}{2}]d\theta dz = \rho\int_{0}^{a}[\frac{\pi R^4}{4}+2\pi \frac{R^2z^2}{2}]dz [/itex]
    [itex]= \rho[\frac{a\pi R^4}{4}+\frac{\pi R^2a^3}{3}] \rightarrow
    M\frac{\frac{a\pi R^4}{4}+\frac{\pi R^2a^3}{3}}{a\pi R^2} = \frac{MR^2}{4}+\frac{Ma^2}{3}[/itex]

    [itex] i=2; [/itex] the only that changes in this calculation is that [itex] sin(\theta) [/itex] becomes [itex] cos(\theta) [/itex]. Since [itex] \int_{0}^{2\pi}sin^2(\theta)d\theta = \int_{0}^{2\pi}cos^2(\theta)d\theta \Rightarrow J_{11}=J_{22} [/itex]

    [itex] i=3; \rightarrow J_{11}=\rho\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[(r)^2]rdrd\theta dz = \rho\int_{0}^{a}\int_{0}^{2\pi}[\frac{R^4}{4}]d\theta dz = \rho\int_{0}^{a}[2\pi\frac{R^4}{4}]dz [/itex]
    [itex] = \rho[\frac{a\pi R^4}{2}] \rightarrow M\frac{\frac{a\pi R^4}{2}}{a\pi R^2} = \frac{MR^2}{2} [/itex]

    Case 2)[itex] \delta_{ij}=0 \Rightarrow J_{ij}=\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[-x_ix_j]rdrd\theta dz[/itex]

    [itex] i=1,j=2 \rightarrow \int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[-r^2cos(\theta)sin(\theta)]rdrd\theta dz = \frac{1}{2}\int_{0}^{2\pi}sin(2\theta)d\theta[\int_{0}^{a}\int_{0}^{R}[-r^3]dr dz] = 0 [/itex]

    [itex] i=1,j=3 \rightarrow \int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[-rcos(\theta)z]rdrd\theta dz = \int_{0}^{2\pi}cos(\theta)d\theta[\int_{0}^{a}\int_{0}^{R}[-zr^2]dr dz] = 0 [/itex]

    [itex]i=2,j=3 \rightarrow \int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[-rsin(\theta)z]rdrd\theta dz = \int_{0}^{2\pi}sin(\theta)d\theta[\int_{0}^{a}\int_{0}^{R}[-zr^2]dr dz] = 0

    So [itex] \mathbb{J}=\begin{pmatrix}\frac{MR^2}{4}+\frac{Ma^2}{3}&0&0\\0&\frac{MR^2}{4}+\frac{Ma^2}{3}&0\\0&0&\frac{MR^2}{2}\end{pmatrix}[/itex]

    Now I want to find the inertia tensor about the center of mass [itex] \mathbb{I} [/itex]. Using [itex] I_{ij}=J_{ij}-M\delta_{ij}\sum_{k}a_{k}^2-a_ia_j [/itex] where [itex] \vec{a} = \begin{pmatrix}0\\0\\-\frac{a}{2}\end{pmatrix} [/itex]

    Case 1) [itex] \delta_{ij}=1 \rightarrow I_{ij}=J_{ij}-M[a_1^2+a_2^2+a_3^2-a_i^2] [/itex]

    [itex] i=1; I_{11}=J_{11}-M[a_2^2+a_3^2] \rightarrow I_{11} = \frac{MR^2}{4}+\frac{Ma^2}{3}-M\frac{a^2}{4} = \frac{MR^2}{4}+\frac{Ma^2}{12} = M\frac{3R^2+a^2}{12} [/itex]

    [itex] i=2; I_{22} =J_{22}-M[a_1^2+a_3^2] \rightarrow I_{22} = \frac{MR^2}{4}+\frac{Ma^2}{3}-M\frac{a^2}{4} = \frac{MR^2}{4}+\frac{Ma^2}{12} = M\frac{3R^2+a^2}{12} [/itex]

    [itex] i=3; I_{33} = J_{33}-M[a_1^2+a_2^2] \rightarrow I_{33}=J_{33} [/itex]

    Case 2) [itex] \delta_{ij}=0 \rightarrow I_{ij}=J_{ij}+M[a_ia_j] [/itex] since [itex] j≠i \Rightarrow I_{ij} = J_{ij}[/itex] Since the only non-zero component of [itex]\vec{a}[/itex] is [itex] a_3 [/itex].

    Therefore: [itex] \mathbb{I} = \begin{pmatrix}M\frac{3R^2+a^2}{12}&0&0\\0&M\frac{3R^2+a^2}{12}&0\\0&0&\frac{MR^2}{2}\end{pmatrix}[/itex]

    Then: [itex] \vec{L}=\mathbb{I}\vec{\omega} \rightarrow \begin{pmatrix}L_1\\L_2\\L_3\end{pmatrix} = \begin{pmatrix}M\frac{3R^2+a^2}{12}&0&0\\0&M\frac{3R^2+a^2}{12}&0\\0&0&\frac{MR^2}{2}\end{pmatrix} \begin{pmatrix}0\\0\\\omega_3\end{pmatrix}[/itex]

    Clearly the only non-zero component of [itex] \vec{L} [/itex] is [itex] L_3 = \frac{MR^2}{2}\omega_3 [/itex] So what is [itex] \omega_3 [/itex]? its the angular velocity [itex] \dot{\theta} [/itex] about the [itex] x_3 [/itex] axis so:

    [itex] \vec{L} = \begin{pmatrix}0\\0\\\frac{MR^2}{2}\dot{\theta}\end{pmatrix} [/itex]. Also I know that [itex] \vec{\omega} [/itex] is parallel to [itex] \vec{L} [/itex] so [itex] \frac{\partial \vec{L}}{\partial t}_{inertial} =\frac{\partial \vec{L}}{\partial t}_{body}=\vec{\tau} [/itex] And [itex]\dot{\vec{L}} = \begin{pmatrix}0\\0\\\frac{MR^2}{2}\ddot{\theta}\end{pmatrix} = \begin{pmatrix}\tau_1\\\tau_2\\\tau_3\end{pmatrix} [/itex]

    At this point I think I should look at the sum of the forces and torques on the yo-yo but I'm not quite sure, and I'd like to make sure the above work is correct before moving forward. Thanks for your help in advance.
  2. jcsd
  3. Apr 21, 2014 #2
    I think you may be making this way more complicated than it needs to be. First off, you haven't defined this mysterious 'a' length. If I am understanding this setup correctly, all motion is restricted to a plane, so you don't really have to worry about any tensors. The moment of inertia should just be that of two disks, i.e. I=2(1/2MR^2)=MR^2. All you really need to solve this problem is torque and force balance.
  4. Apr 21, 2014 #3
    I know it seems i made it more complicated, but I need practice in setting up the more general case i.e dealing with tensors. Oh and the [itex]\vec{a}[/itex] is the vector pointing from the center of mass to the origin of the body-frame coordinates. Oh and a quick question: why is the moment of inertia of two disks instead of one solid cylinder if the thin shaft joining both parts of the yo-yo is massless?
  5. Apr 21, 2014 #4
    The moment of inertia is that of two disks because, if you neglect the mass of the axle, that's what a yo-yo is made out of: two disks with empty space between them. Also, if that's what 'a' is, then your integrals don't make any sense. The center of mass should be your origin, right between the two disks, so |a|=0. If you really want practice with tensors, do the problem first using simple equations, and then if you must, go back and redo it with tensors.
  6. Apr 21, 2014 #5
    Okay I see what you're saying about the moment of inertia, but [itex]a[/itex] is the height of the cylinder, but [itex] \vec{a} [/itex] is the vector that points from the center of mass to the origin which is originally at the bottom of the cylinder( I know it was bad notation) I should have called the upper limit in my integral [itex] h [/itex] but the tensor i got is correct.

  7. Apr 21, 2014 #6
    What cylinder? You have two disks. Their moments of inertia depend only the the mass and the radius.
  8. Apr 21, 2014 #7
    I was thinking that the two parts of the yo-yo had thickness but I went back and re-read the problem and it did say disks.
  9. Apr 21, 2014 #8
    But I could just take the limit as [itex]h[/itex] goes to zero which would give me the inertia tensor for a disk, correct?
  10. Apr 21, 2014 #9
    As I said before, the moment of inertia of a disk depends only on its mass and its radius.
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