 #1
lorenz0
 140
 26
 Homework Statement:

A disk of mass ##m## and radius ##r## is free to rotate around an axis passing through its center O. A rod of mass ##m## and length ##l## is attached radially to the edge of the disk at an angle ##\theta##. Find:
1) The moment of inertia with respect to the center O and the position of the center of mass (also with respect to O).
If the system is left free to move starting from rest from the angular position ##\theta=\pi/2##, find, for a generic angle ##0\leq \theta\leq \frac{\pi}{2}##:
2) The formula for ##\frac{d\theta}{dt}## as a function of ##\theta##;
3) The formula for ##\frac{d^2\theta}{dt^2}## as a function of ##\theta##;
4) The vector acceleration ##\vec{a}_{CM}## of the center of mass and the vector ##\vec{N}##, the reaction force of the axis.
 Relevant Equations:
 ##I=\int r_{\perp}^2 dm,\ \vec{r}_{cm}=\frac{\int \vec{r}dm}{M}##
1) Since the rod is uniform, with mass m and length l, it has a linear mass density of ##\lambda=\frac{m}{l}##, so ##I_{rod_O}=\int_{x=r}^{x=r+l}x^2 \lambda dx=\frac{\lambda}{3}[(r+l)^3r^3]=\frac{\lambda r^3}{3}[(1+\frac{l}{r})^31]=\frac{1}{3}mr^2[3+\frac{3l}{r}+\frac{l^2}{r^2}].##
##I_{O}=I_{disk_O}+I_{rod_O}=\frac{1}{2}mr^2+\frac{1}{3}mr^2[3+\frac{3l}{r}+\frac{l^2}{r^2}]=mr^2[\frac{3}{2}+\frac{l}{r}+\frac{l^2}{3r^2}].##
##\vec{r}_{cm}=\frac{1}{2m}((r+\frac{l}{2})\sin(\theta), (r+\frac{l}{2})\cos(\theta))=\frac{2r+l}{4}(\sin(\theta),\cos(\theta))## so ##r_{cm}=\frac{2r+l}{4}.##
Now, the problem is that I don't understand how to find ##\omega=\frac{d\theta}{dt}## and ##\alpha=\frac{d^2\theta}{dt^2}## as a function of ##\theta##. What I have done is: ##\tau_O=I_O\alpha \Leftrightarrow  r_{cm}2mg\sin(\theta)=mr^2[\frac{3}{2}+\frac{l}{r}+\frac{l^2}{3r^2}] \frac{d^2\theta}{dt^2}## so ##\frac{d^2\theta}{dt^2}=\frac{2r+l}{4r^2[\frac{3}{2}+\frac{l}{r}+\frac{l^2}{3r^2}]}g\sin(\theta)## but I don't see how to go from this to ##\frac{d\theta}{dt}## (even knowing the initial conditions ##\theta(0)=\frac{\pi}{2}, \omega(0)=0##), since there is a ##\sin(\theta)## involved in the integration, where ##\theta=\theta(t)##.
So, I would be grateful if someone could help me understand how to find these two quantities, thanks.

By knowing ##\omega## and ##\alpha## then I can find ##\vec{a}_{cm}=\omega^2 r_{cm}(\hat{r})+r_{cm}\alpha (\hat{\theta})=\frac{4mgr_{cm}\cos(\theta)}{I_O} r_{cm} (\hat{r})+\frac{r^2_{cm} mg\sin(\theta)}{I_O}(\hat{\theta})##.
##I_{O}=I_{disk_O}+I_{rod_O}=\frac{1}{2}mr^2+\frac{1}{3}mr^2[3+\frac{3l}{r}+\frac{l^2}{r^2}]=mr^2[\frac{3}{2}+\frac{l}{r}+\frac{l^2}{3r^2}].##
##\vec{r}_{cm}=\frac{1}{2m}((r+\frac{l}{2})\sin(\theta), (r+\frac{l}{2})\cos(\theta))=\frac{2r+l}{4}(\sin(\theta),\cos(\theta))## so ##r_{cm}=\frac{2r+l}{4}.##
Now, the problem is that I don't understand how to find ##\omega=\frac{d\theta}{dt}## and ##\alpha=\frac{d^2\theta}{dt^2}## as a function of ##\theta##. What I have done is: ##\tau_O=I_O\alpha \Leftrightarrow  r_{cm}2mg\sin(\theta)=mr^2[\frac{3}{2}+\frac{l}{r}+\frac{l^2}{3r^2}] \frac{d^2\theta}{dt^2}## so ##\frac{d^2\theta}{dt^2}=\frac{2r+l}{4r^2[\frac{3}{2}+\frac{l}{r}+\frac{l^2}{3r^2}]}g\sin(\theta)## but I don't see how to go from this to ##\frac{d\theta}{dt}## (even knowing the initial conditions ##\theta(0)=\frac{\pi}{2}, \omega(0)=0##), since there is a ##\sin(\theta)## involved in the integration, where ##\theta=\theta(t)##.
So, I would be grateful if someone could help me understand how to find these two quantities, thanks.

By knowing ##\omega## and ##\alpha## then I can find ##\vec{a}_{cm}=\omega^2 r_{cm}(\hat{r})+r_{cm}\alpha (\hat{\theta})=\frac{4mgr_{cm}\cos(\theta)}{I_O} r_{cm} (\hat{r})+\frac{r^2_{cm} mg\sin(\theta)}{I_O}(\hat{\theta})##.
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