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widelec
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Homework Statement
I am providing a solution up to the point when I'm having a little issue with defining the generalized force.
An eccentrically hollow cylinder of radius [itex]r[/itex] rolls down a plane of inclination angle [itex]\alpha[/itex]. Inside the cylinder, there is a cylinder-shaped hole of radius [itex]\frac{r}{2}[/itex] and center located [itex]\frac{r}{2}[/itex] from the cylinder's center. Using the Lagrange method, formulate the differential equations of the system, assuming the density of the cylinder is [itex]\rho[/itex] and its length is equal to [itex]1 \ m[/itex].
Assume the potential energy is zero and there is an external force acting on the system.
Variables and data:
[itex]l,r[/itex] - length (= 1 m) and radius of the cylinder
[itex]d[/itex] - distance between the cylinder's center of rotation and its center of mass (calculated below)
[itex]m=l \rho \pi r^{2}[/itex] - mass of the full cylinder
[itex]m_{s}=l \rho \pi \left [ r^{2} - \left ( \frac{r}{2} \right )^{2} \right ] = \frac{3}{4}m[/itex] - mass of the hollowed cylinder
[itex]g[/itex] - gravitational acceleration
[itex]x_{1},y_{1}[/itex] - coordinates of the center of rotation
[itex]x_{s},y_{s}[/itex] - coordinates of the center of mass
[itex]\varphi_{1}[/itex] - rotation angle of the cylinder
Homework Equations
The constraints are:
1. [itex]y_{1}=0[/itex]
2. [itex]x_{1}=C_{1}-r \varphi_{1}[/itex] (C1 being the initial position)
3. [itex]x_{s}=x_{1}+d \cos(\varphi_1), \ y_{s}=d \sin(\varphi_1)[/itex]
[itex]\varphi_{1}[/itex] is chosen as the generalized coordinate.
The Attempt at a Solution
The differentials are:
[itex]\dot{x}_{1}=-r \dot{\varphi}_{1}[/itex]
[itex]\dot{x}_{s}=-r \dot{\varphi}_{1}-d \dot{\varphi}_{1} \sin(\varphi_{1})[/itex]
[itex]\dot{y}_{s}=d \dot{\varphi}_{1} \cos(\varphi_{1})[/itex]
The distance between centers of rotation and mass is calculated:
[itex]d=\frac{1}{m_{s}}\sum _{i=1}^{2} m_{i}d_{i}=\frac{4}{3m} \left (\frac{1}{2} m \cdot 0 +\frac{1}{4}m \cdot \frac{1}{2}r\right )=\frac{1}{6}r[/itex]
Moment of inertia of the hollow cylinder with regard to the center of rotation (Steiner's theorem):
[itex]J_{0}=\frac{1}{2}mr^2 - \left [ \frac{1}{2} \left ( \frac{1}{4}m\right ) \left ( \frac{r}{2}\right )^2 + \left ( \frac{1}{4}m\right ) \left ( \frac{r}{2}\right )^2 \right ] = \frac{13}{32}mr^2[/itex]
and in regard to the center of mass:
[itex]J_{s}=J_{o}-m_{s}d^2=\frac{37}{96}mr^2[/itex]
Thus, the kinetic energy (and the Lagrangian, since potential is zero) of the system is equal to:
[itex]\mathcal{L}=T=\frac{1}{2}J_s \dot{\varphi}_1^2 + \frac{1}{2}m_s \left ( \dot{x}_s^2+\dot{y}_s^2\right )[/itex]
Skipping the calculations, I've come to the following form:
[itex]\mathcal{L}=\frac{37}{64}mr^2\dot{\varphi}_1^2+\frac{1}{8}mr^2 \dot{\varphi}_1^2 \sin\varphi_1[/itex]
So the respective differentials are:
[itex]\frac{\partial \mathcal{L}}{\partial \dot{\varphi}_1}=\frac{37}{32}mr^2\dot{\varphi}_1+\frac{1}{4}mr^2 \dot{\varphi}_1\sin\varphi_1[/itex]
[itex]\frac{d}{dt} \left [ \frac{\partial \mathcal{L}}{\partial \dot{\varphi}_1} \right ]=\frac{37}{32}mr^2\ddot{\varphi}_1+\frac{1}{4}mr^2\left ( \ddot{\varphi}_1 \sin{\varphi_1} + \dot{\varphi}_1^2\cos{\varphi_1}\right )[/itex]
[itex]\frac{\partial \mathcal{L}}{\partial \varphi}_1=\frac{1}{8}mr^2\dot{\varphi}_1^2\cos\varphi_1[/itex]
Now, we need to formulate the Lagrange equation:
[itex]\frac{d}{dt} \left [ \frac{\partial \mathcal{L}}{\partial \dot{\varphi}_1} \right ]-\frac{\partial \mathcal{L}}{\partial \varphi}_1=\tilde{P}_{\varphi_1}[/itex]
where [itex]\tilde{P}_{\varphi_1}[/itex] is the generalized force acting in the direction of [itex]\varphi_1[/itex].
This is the point where I'm having trouble defining the generalized force.
Since we assume the cylinder does not slip along the plane, we should consider only the frictional force that causes the cylinder to roll.
Because frictional force acts in the same direction as the X axis, I write it out as:
[itex]F_{x_1}=-m_sg \sin{\alpha}=-\frac{3}{4}mg\sin{\alpha}[/itex]
Now, to convert it to act along [itex]\varphi_1[/itex], I believe it needs to be done this way:
[itex]\tilde{P}_{\varphi_1}=\frac{\partial x_1}{\partial \varphi_1}F_{x_1}=(-r) \cdot (-\frac{3}{4}mg\sin{\alpha})=\frac{3}{4}mgr\sin{\alpha}[/itex]
Is this going to be the only external force acting on the system? Does hollowing the cylinder eccentrically make any difference in terms of forces when compared to a "standard" full cylinder rolling down a plane?
Or perhaps I'm making a mistake somewhere? Any help would be greatly appreciated.
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