Plate-tilt dynamics

Rolling-without-slipping dynamics

Newton-Euler derivation of the rolling-without-slipping equation of motion for a uniform solid sphere on a tilted plate, yielding the in-plane acceleration $\tfrac{5}{7}\, g \sin\theta_y$.

01 — Setup

Setup

A planar plate is tilted by angle $\theta_y$ about its $\hat y$ axis. A uniform solid sphere of mass $m$ and radius $\rho$ rests on the plate and is free to translate along the slope. Gravity acts vertically.

$\theta_y = 0$ corresponds to the plate horizontal (no motion). $\theta_y > 0$ corresponds to the right side of the plate below the left, with the ball rolling along the slope.

horizon θy tilted plate rotated by θy from horizontal solid sphere · mass m, radius ρ

02 — Reference frames

Reference frames

Two frames are used. $\ddot x$ denotes the ball's acceleration along the plate's $\hat x_P$ axis (i.e.\ along the slope).

  • World frame {W}
    $\hat x_W, \hat y_W, \hat z_W$. Fixed; gravity points along $-\hat z_W$.
  • Plate frame {P}
    $\hat x_P, \hat y_P, \hat z_P$. Attached to the plate; rotates with it.
  • Tilt $\theta_y$
    Rotation of {P} about the world's $\hat y$ axis.

All velocities and accelerations referenced in this derivation ($\ddot x$, $v$, $\omega$) are expressed in $\{P\}$.

x̂W ẑW {W} x̂P ẑP $\{P\}$

03 — Free-body diagram

Free-body diagram

The ball is acted upon by three forces.

  • Gravity $m g$
    Vertical, along $-\hat z_W$.
  • Normal $N$
    Along $+\hat z_P$, perpendicular to the plate surface.
  • Static friction $f$
    Tangent to the plate, opposing the tendency of the contact point to slip; points up the slope ($-\hat x_P$).

With $f = 0$ the sphere would slide down the slope without rotating; $f$ is the force that produces the torque about the ball's centre and so couples translation to rotation.

f friction N normal mg gravity

04 — Translation along the slope

Translation equation

Projecting Newton's second law along $\hat x_P$, the gravity component along the slope is $m g \sin\theta_y$; the normal force is perpendicular to $\hat x_P$ and does not contribute.

along $\hat x_P$ $$m\, \ddot x \;=\; m\, g\, \sin\theta_y \;-\; f$$

05 — Torque about the ball's centre

Rotational equation

Friction acts at the contact point at distance $\rho$ from the ball's centre, producing a torque $f\rho$ about the centre. Gravity acts through the centre and the normal force is perpendicular to the moment arm, so neither contributes.

torque about ball centre $$I\, \alpha \;=\; f \cdot \rho$$

For a uniform solid sphere:

moment of inertia $$I \;=\; \tfrac{2}{5}\, m\, \rho^{2}$$

Here $\alpha$ is the angular acceleration of the ball about its centre.

06 — Rolling-without-slipping constraint

Kinematic constraint

The rolling-without-slipping condition requires that the contact point has zero velocity relative to the plate at the instant of contact, which is equivalent to:

$$v \;=\; \rho\, \omega \quad\Longrightarrow\quad \ddot x \;=\; \rho\, \alpha$$

Translational speed equals radius times angular rate. Differentiating once: translational acceleration equals radius times angular acceleration.

ω v = ρω ρ contact (zero rel. vel.)

07 — Eliminate $f$ and $\alpha$

Eliminate $f$ and $\alpha$

Three equations are available: Newton's second law along $\hat x_P$, the torque equation about the ball's centre, and the rolling constraint. Substituting $\alpha = \ddot x / \rho$ into the torque equation removes $\alpha$; substituting the resulting expression for $f$ into Newton's second law removes $f$, leaving a single equation in $\ddot x$.

substitute $\alpha = \ddot x / \rho$ $$I \cdot \frac{\ddot x}{\rho} \;=\; f \cdot \rho \;\;\Longrightarrow\;\; f \;=\; \frac{I}{\rho^{2}}\,\ddot x \;=\; \tfrac{2}{5}\, m\, \ddot x$$
plug into translation $$m\, \ddot x \;=\; m\, g\, \sin\theta_y \;-\; \tfrac{2}{5}\, m\, \ddot x$$
collect $\ddot x$ $$\bigl(1 + \tfrac{2}{5}\bigr)\, m\, \ddot x \;=\; m\, g\, \sin\theta_y \;\;\Longrightarrow\;\; \tfrac{7}{5}\, m\, \ddot x \;=\; m\, g\, \sin\theta_y$$

08 — Result

Rolling equation

exact $$\ddot x \;=\; \tfrac{5}{7}\, g\, \sin\theta_y$$

For small tilts, $\sin\theta_y \approx \theta_y$, so:

small-angle, used in the controller $$\ddot x \;\approx\; \tfrac{5}{7}\, g\, \theta_y$$

09 — Extension to the other axis

The $\hat y_P$ equation

The derivation above was carried out along $\hat x_P$, using a tilt $\theta_y$ about $\hat y_W$. The argument is symmetric: along $\hat y_P$ the slope is produced by a tilt $\theta_x$ about $\hat x_W$, and an identical chain of Newton, torque and rolling-constraint equations yields the same factor $\tfrac{5}{7}$ with a single sign change driven by the wiring convention.

exact, both axes $$\ddot x \;=\; \phantom{-}\tfrac{5}{7}\, g\, \sin\theta_y, \qquad \ddot y \;=\; -\tfrac{5}{7}\, g\, \sin\theta_x$$

Under the small-angle approximation $\sin\theta \approx \theta$, the controller uses:

small-angle, both axes $$\ddot x \;\approx\; \phantom{-}\tfrac{5}{7}\, g\, \theta_y, \qquad \ddot y \;\approx\; -\tfrac{5}{7}\, g\, \theta_x$$

The minus sign in the $\hat y_P$ equation comes from the wiring convention $\theta_{\text{tar}} = u_y\, \hat x_P^{\,W}(q) - u_x\, \hat y_P^{\,W}(q)$ fixed on the control page: a positive tilt about the world's $\hat x_W$ axis accelerates the ball in the $-\hat y_P$ direction.

10 — Interpretation of $\tfrac{5}{7}$

Effective inertia

The factor $\tfrac{5}{7}$ is the ratio $1 / \!\bigl(1 + I / (m\rho^{2})\bigr)$; with $I = \tfrac{2}{5}m\rho^{2}$ for a uniform solid sphere, this is $\tfrac{5}{7}$. Equivalently, $\tfrac{5}{7}$ of the gravitational power along the slope drives translation and $\tfrac{2}{7}$ drives rotation.

$$\frac{1}{1 + I / (m\rho^{2})} \;=\; \frac{1}{1 + 2/5} \;=\; \frac{5}{7}$$

Sliding (no rotation)

$\ddot x = g\,\sin\theta$
all gravity goes to translation

Rolling sphere

$\ddot x = \tfrac{5}{7}\, g\,\sin\theta$
$\tfrac{5}{7}$ translation · $\tfrac{2}{7}$ rotation

11 — Connection to the wiring stage

Sign convention used in the wiring stage

The result $\ddot x \propto +\theta_y$ fixes the sign of the tilt command relative to the desired ball acceleration: a ball with positive $x$-error must be driven toward $-\hat x$, so the corresponding tilt must satisfy $\theta_y < 0$, i.e.\ a rotation about $-\hat y_W$. This is the sign that appears in the wiring formula on the Control page:

$$\theta_{\text{tar}} \;=\; u_y\, \hat x_P^{\,W}(q) \;-\; u_x\, \hat y_P^{\,W}(q)$$
Control pipeline →

References

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  3. Hauser, J., Sastry, S., Kokotović, P. (1992). Nonlinear control via approximate input-output linearization: the ball and beam example. IEEE Transactions on Automatic Control, 37(3), 392–398. eecs.berkeley.edu/~sastry
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