- #1
Wavefunction
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- 4
Homework Statement
A yo-yo of mass M is composed of two disks of radius R separated by a shaft of radius r. 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
vertical.
Homework Equations
J_{ij}=\int_{V}[\rho(r)\delta_{ij}\sum_{k}x_{k}^2-x_ix_j]dV
I_{ij}=J_{ij}-M[\delta_{ij}\sum_{k}a_{k}^2-a_ia_j]
L_i=\sum_{j}I_{ij}\omega_{j}
The Attempt at a Solution
Setup: First I'll define the x_3 axis to be the axis of rotation of the yo-yo. Then \vec{\omega} will have the components \begin{pmatrix}0\\0\\\omega_{3}\end{pmatrix}. Now I need to find \mathbb{I} in order to find \vec{L}. Finally, because I can neglect the mass of the small shaft I will be calculating the inertia tensor of a solid cylinder.
Calculation: Using J_{ij}=\int_{V}[\rho(r)\delta_{ij}\sum_{k}x_{k}^2-x_ix_j]dV I'll find the inertia tensor at the bottom of the cylinder. So I'll need the volume of the region in question:
0\leq z\leq a, 0\leq r\leq R, 0\leq \theta\leq 2\pi
Case 1) \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
where x_1 = rcos(\theta), x_2 = rsin(\theta), x_3 =z
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
= \rho[\frac{a\pi R^4}{4}+\frac{\pi R^2a^3}{3}] \rightarrow<br /> 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}
i=2; the only that changes in this calculation is that sin(\theta) becomes cos(\theta). Since \int_{0}^{2\pi}sin^2(\theta)d\theta = \int_{0}^{2\pi}cos^2(\theta)d\theta \Rightarrow J_{11}=J_{22}
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
= \rho[\frac{a\pi R^4}{2}] \rightarrow M\frac{\frac{a\pi R^4}{2}}{a\pi R^2} = \frac{MR^2}{2}
Case 2)\delta_{ij}=0 \Rightarrow J_{ij}=\int_{0}^{a}\int_{0}^{2\pi}\int_{0}^{R}[-x_ix_j]rdrd\theta dz
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
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
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 <br />
So \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}
Now I want to find the inertia tensor about the center of mass \mathbb{I}. Using I_{ij}=J_{ij}-M\delta_{ij}\sum_{k}a_{k}^2-a_ia_j where \vec{a} = \begin{pmatrix}0\\0\\-\frac{a}{2}\end{pmatrix}
Case 1) \delta_{ij}=1 \rightarrow I_{ij}=J_{ij}-M[a_1^2+a_2^2+a_3^2-a_i^2]
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}
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}
i=3; I_{33} = J_{33}-M[a_1^2+a_2^2] \rightarrow I_{33}=J_{33}
Case 2) \delta_{ij}=0 \rightarrow I_{ij}=J_{ij}+M[a_ia_j] since j≠i \Rightarrow I_{ij} = J_{ij} Since the only non-zero component of \vec{a} is a_3.
Therefore: \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}
Then: \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}
Clearly the only non-zero component of \vec{L} is L_3 = \frac{MR^2}{2}\omega_3 So what is \omega_3? its the angular velocity \dot{\theta} about the x_3 axis so:
\vec{L} = \begin{pmatrix}0\\0\\\frac{MR^2}{2}\dot{\theta}\end{pmatrix}. Also I know that \vec{\omega} is parallel to \vec{L} so \frac{\partial \vec{L}}{\partial t}_{inertial} =\frac{\partial \vec{L}}{\partial t}_{body}=\vec{\tau} And \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}
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.
