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In the figure below, the disc rolls without slipping on an inclined plane tilted by an angle $\epsilon$ relative to the horizontal plane. The axes $X_1,Y_1,Z_1$ represent the space frame and $X,Y,Z$ represent the body frame. ($X$ and $X_1$ point out of the plane and are not shown in the figure.)

The Euler angles in the body frame are $\Psi$, $\theta$, and $\phi$ (only the latter two shown in the figure). $\Psi$ is the angle of $Z_1$ about $Z$, and, $\Psi =0 $ when the disk is at its lowest point (equilibrium position.)

The mass of the system (rod+disk) is $m$ and the distance from $O$ to the center of mass is $l$. The principal moments of inertia of the system relative to $O$ are $I_{xx} = I_{yy}$ and $I_{zz}$. enter image description here

I am supposed to obtain the relationship between $\Psi$ and $\phi$ based on the constraint that the disc rolls without slipping. Also, the kinetic and potential energies are required. The entire system's degree of freedom is reported as 1 and the Lagrangian equation of motion is to be expressed in terms of $\Psi$. Now, the potential energy is easy: $$ U = mgl \sin \epsilon ( \cos \Psi_0 - \cos \Psi ) = mgl \sin \epsilon (1- \cos \Psi ). $$

Moreover, $\theta$ is fixed and can be computed as below:

$$\theta = \cos^{-1} \frac{r_2}{r_3} = \sin^{-1} \frac{r_1}{r_3}.$$ How can I incorporate the rolling without slipping constraint in the solution? I'd appreciate any help/hint on how to approach the two remaining tasks. Thanks.

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    $\begingroup$ Isn't $\theta$ fixed? $\endgroup$
    – JAlex
    May 4 '21 at 13:13
  • $\begingroup$ Yes, it is. I just edited the post. $\endgroup$
    – Neutrino
    May 4 '21 at 13:18
  • $\begingroup$ What book is this question from? @Neutrino $\endgroup$
    – Buraian
    May 9 '21 at 11:10
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Get the kinematic relationship of the velocity vector at the contact point and set the components to zero as no slip is allowed.

If A is the contact point, and O is the fixed pivot then

$$ \vec{v}_A = \vec{\omega} \times (\vec{r}_A - \vec{r}_O) $$

where $\vec{\omega}$ depends on the joint speeds $\dot{\phi}$ and $\dot{\psi}$.

Specifically the no interpenetration constraint is

$$\hat{Z}_1 \cdot \vec{v}_A = 0 $$

and the no-slip constraint is

$$\hat{X}_1 \cdot \vec{v}_A = 0 $$

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    $\begingroup$ i don’t think that your first rolling condition is correct the plane is $X_1,Y1$ $\endgroup$
    – Eli
    May 4 '21 at 18:41
  • $\begingroup$ @Eli On the $X_1Y_1$ plane the normal is $Z_1$ and hence the dot product with velocity along the normal should be zero. No velocity components normal to the plane. $\endgroup$
    – JAlex
    May 4 '21 at 18:46
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    $\begingroup$ but this is not the rolling condition. $\endgroup$
    – Eli
    May 4 '21 at 18:55
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    $\begingroup$ you have 2 rolling non holonomic conditions and 1 holonomic condition. $\endgroup$
    – Eli
    May 4 '21 at 18:56
  • $\begingroup$ @Eli I have omitted the constraint along $Y_1$ since it is trivially met since $\theta$ is assumed to be constant. $\endgroup$
    – JAlex
    May 5 '21 at 16:58
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The rolling without sping constrain is $a=\alpha r$ where a is the linear acceleration, $\alpha$ the angular acceleration, and r the radius of the disc. This means that if you integrate both sides you get $$ \int_0^{t^\prime}\frac{dv}{dt}dt=r\int_0^{t^\prime}\frac{d\omega}{dt}dt\\ v=r\omega $$ where v is the linear velocity and $\omega$ the angular one. That way you can compute the kinetic energy. As for the expression in terms of the Euler angles, you can calculate the acceleration in terms of the inclination $\epsilon$. The acceleration will be constant so the angular velocity will be easily calculated. I hope this is enough of a hint for you. Let me know if you still need more help.

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enter image description here

The velocity at the contact point A is zero

$$\vec v_A=\vec v-S_x(\alpha)\,\left(\vec\omega\times\vec\rho\right)$$

where $~S_x$ is the transformation matrix between "$\omega$" system and "1" system

constraint equations

the rolling conditions

$$\vec v_A\cdot \vec{\hat{X}}_1=0\\ \vec v_A\cdot \vec{\hat{Y}}_1=0$$

and

$$v_z=0\\\omega_x=0$$

you have now 4 constraint equations and 6 degrees of freedom so you get 2 generalized coordinates $v_x~,\omega_z$

$$ \left[ \begin {array}{c} v_{{x}}\\ v_{{y}} \\v_{{z}}\\ \omega_{{x}} \\ \omega_{{y}}\\ \omega_{{z}} \end {array} \right] = \left[ \begin {array}{c} v_{{x}} \\ 0\\ 0\\ 0 \\ {\frac {v_{{x}}}{\rho_{{z}}}}+{\frac {\omega_{{z} }\rho_{{y}}}{\rho_{{z}}}}\\ \omega_{{z}}\end {array} \right] $$

with :

$$v_x=-\cos \left( \psi \right) l\cos \left( \alpha \right) \dot\psi \\ \omega_z=\cos(\alpha)\dot\psi$$

you obtain 1 generalized coordinate which is $\psi$

$$ \left[ \begin {array}{c} v_{{x}}\\ v_{{y}} \\ v_{{z}}\\ \omega_{{x}} \\ \omega_{{y}}\\ \omega_{{z}} \end {array} \right] = \left[ \begin {array}{c} -\cos \left( \psi \right) l\cos \left( \alpha \right)\dot\psi \\ 0 \\ 0\\ 0\\ -{ \frac {\cos \left( \alpha \right) \left( \cos \left( \psi \right) l- \rho_{{y}} \right)\dot\psi }{\rho_{{z}}}}\\ \dot\psi \, \cos \left( \alpha \right) \end {array} \right] $$

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    $\begingroup$ This answer might violate the homework policy. Please do not supply complete answers to homework like questions. $\endgroup$
    – JAlex
    May 4 '21 at 17:44

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