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Eccentrically hollow cylinder - Lagrangian (external forces)

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data
    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

    cseVOX5s.jpg ([itex]h_s[/itex] is not considered in this approach - no potential energy)

    2. Relevant 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.

    3. 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.
     
    Last edited: Apr 8, 2014
  2. jcsd
  3. Apr 9, 2014 #2
    1) Assumption about no potential energy doesn't seem correct. Not only does the cylinder have potential energy but the potential energy changes as it rolls down the ramp.

    2) However, maybe if you assume that gravity is an external force in this situation then you can ignore the potential energy.

    3) When you introduced the constraint [itex]x_{1}=C_{1}-r \varphi_{1}[/itex] then I think this is effectively your friction force. So no need to add this as an external force.
     
  4. Apr 9, 2014 #3
    paisiello2, thank you very much for your reply.

    However, I did come to a conclusion after an extensive discussion with my professor. I am now 100% sure this is correct.

    We have to consider virtual work as follows (minus signs due to directions opposite to axes):

    [itex]\delta A=(-m_{s}g\sin{\alpha}) \delta x_s+(-m_{s}g\cos{\alpha})\delta y_s[/itex]

    where:

    [itex]\delta x_s=\frac{\partial x_s}{\partial \varphi_1}, \ \delta y_s=\frac{\partial y_s}{\partial \varphi_1}[/itex]

    thus:

    [itex]\delta A=\frac{3}{4}mgr\left[ \sin{\alpha}-\frac{1}{6}\cos(\alpha+\varphi_1)\right ]=\tilde{P}_{\varphi_1}[/itex]

    Substituting the above to the aforementioned Lagrange equation along with the previously calculated derivatives gives a result consistent with the one given in my coursebook where potential energy is considered (and no external forces acting, in turn).
     
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