- #1

lorenz0

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- 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)^3-r^3]=\frac{\lambda r^3}{3}[(1+\frac{l}{r})^3-1]=\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)##

So, I would be grateful if someone could help me understand how to find these two quantities, thanks.

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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.

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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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