Let D denote the coin that is a sharp-edged circular disk that rolls on a plane support S fixed in a reference frame R (Figure 7.19). Line n1 is the tangent to the periphery of D at the point of contact C between D and S. Let n1^1, n2^1 and n3^1 be three mutually perpendicular unit vectors, as

Question

Let D denote the coin that is a sharp-edged circular disk that rolls on a plane support S fixed in a reference frame R (Figure 7.19). Line $\bar{n}_1$ is the tangent to the periphery of D at the point of contact C between D and S. Let $\bar{n}_1^1, \bar{n}_2^1$ and $\bar{n}_3^1$ be three mutually perpendicular unit vectors, as shown, $\bar{n}_1^1$ is along $\bar{n}_1$ and $\bar{n}_2^1$ passes through the center G of D. Finally, $\bar{n}_x, \bar{n}_y$ and $\bar{n}_z$ are fixed in R and are mutually perpendicular.

(a) Find the governing equations of motion for the system

(b) Solve for the constraint forces at point C.

(a) Find the governing equations of motion for the system

(b) Solve for the constraint forces at point C.

Accepted Answer

Let the rolling coin be body [latex]B_1[/latex] and the following are the kinematics needed in the analysis. As usual we derive all kinematical quantities in an explicit form, where we define the angular velocity of body B, with respect to R as

[latex]\bar{\omega}=\{\dot{\phi}\}^T\left\{\bar{n}^1 ight\}=\{\dot{\phi}\}^T\left[S^{1 R} ight]\left\{\bar{n}^R ight\}[/latex] (7.296)

Where

[latex]\begin{aligned}\{\dot{\phi}\}^T &=\left[\begin{array}{lll}\dot{\phi}_1 & \dot{\phi}_2 & \dot{\phi}_3\end{array} ight] \\left\{\bar{n}^R ight\}^T &=\left[\begin{array}{lll}\bar{n}_x & \bar{n}_y & \bar{n}_z\end{array} ight]\end{aligned}[/latex]

First we note that

[latex]\left\{\bar{n}^1 ight\}=\left[\begin{array}{ccc}1 & 0 & 0 \0 & \cos \left(\phi_1-90^{\circ} ight) & \sin \left(\phi_1-90^{\circ} ight) \0 & -\sin \left(\phi_1-90^{\circ} ight) & -\cos \left(\phi_1-90^{\circ} ight)\end{array} ight]\{\bar{n}\}[/latex]

expressing [latex]\{\bar{n}\}[/latex] with respect to R we get

[latex]\left\{\bar{n}^1 ight\}=\left[\begin{array}{ccc}1 & 0 & 0 \0 & \sin \phi_1 & -\cos \phi_1 \0 & \cos \phi_1 & \sin \phi_1\end{array} ight]\left[\begin{array}{ccc}\cos heta & \sin heta & 0 \-\sin heta & \cos heta & 0 \0 & 0 & 1\end{array} ight]\left\{\bar{n}^R ight\}[/latex]

Therefore, the shifter is simply

[latex]\left[S^{1 R} ight]=\left[\begin{array}{ccc}\cos heta & \sin heta & 0 \-\sin \phi_1 \sin heta & \sin \phi_1 \sin heta & -\cos \phi_1 \-\cos \phi_1 \sin heta & \cos \phi_1 \cos heta & \sin \phi_1\end{array} ight] (7.297)[/latex]

with

[latex]\dot{\phi}_3 \sin \phi_1=\dot{ heta}[/latex] (7.298)

Hence the angular velocity components are

[latex]\{\omega\}=\left[\begin{array}{c}\dot{\phi}_1 \cos heta-\dot{\phi}_2 \sin \phi_1 \sin heta-\dot{\phi}_3 \cos \phi_1 \sin heta \\dot{\phi}_1 \sin heta+\dot{\phi}_2 \sin \phi_1 \sin heta+\dot{\phi}_3 \cos \phi_1 \cos heta \-\dot{\phi}_2 \cos \phi_1+\dot{\phi}_3 \sin \phi_1\end{array} ight] (7.299)[/latex]

The coin has six possible degrees of freedom defined by

[latex]x^T=\left[\begin{array}{llllll}\phi_1 & \phi_2 & \phi_3 & x & y & z\end{array} ight][/latex] (7.300)

where [latex]\phi_1, \phi_2[/latex] and [latex]\phi_3[/latex] are defined as the rotations shown along the unit vectors given in Figure 7.19, and x, y and z represent the position of the center mass of the coin in R.

In a matrix form the angular velocity can be expressed as

[latex]\bar{\omega}=\{\dot{x}\}^T[\omega]\left\{\bar{n}^R ight\}[/latex] (7.301)

where [ω] is the partial angular velocity array defined by

[latex][\omega]=\left[\begin{array}{c}{\left[S^{1 R} ight]} \0_{3 imes 3}\end{array} ight][/latex] (7.302)

which is found through the simple differentiation

[latex]\frac{\partial \bar{\omega}}{\partial \dot{x}_{\ell}}=[\omega]\left\{\bar{n}^R ight\}[/latex] (7.303)

The angular acceleration is

[latex]\bar{\alpha}=\frac{d}{d t} \bar{\omega}=\left(\{\ddot{x}\}^T[\omega]+\{\dot{x}\}^T[\dot{\omega}] ight)\left\{\bar{n}^R ight\}[/latex] (7.304)

and

[latex][\dot{\omega}]=\left[\begin{array}{c}{\left[\dot{S}^{1 R} ight]} \0_{3 imes 3}\end{array} ight][/latex] (7.305)

Differentiation of [latex]\left[S^{1 R} ight][/latex] can be explicitly found

[latex]\left[\dot{S}^{1 R} ight]=\left[\begin{array}{ccc}\dot{\phi}_3 s \phi_1 s heta & \dot{\phi}_3 s \phi_1 c heta & 0 \\dot{\phi}_1 c \phi_1 s heta-\dot{\phi}_3 s \phi_1 s \phi_1 c heta & \dot{\phi}_1 c \phi_1 c heta-\dot{\phi}_3 s \phi_1 c heta & \dot{\phi}_1 s \phi_1 \-\dot{\phi}_2 s \dot{\phi}_1 s heta-\dot{\phi}_3 s \phi_1 c \phi_1 c heta & -\dot{\phi}_2 s \phi_1 c heta-\dot{\phi}_3 s \dot{\phi}_1 c \phi_1 s \phi & \dot{\phi}_1 c \phi_1\end{array} ight][/latex] (7.306)

The velocity of the center mass of the rolling coin is defined by

[latex]\bar{v}=\{\dot{x}\}^T[V]\left\{\bar{n}^R ight\}=\dot{x} \bar{n}_x+\dot{y} \bar{n}_y+\dot{z} \bar{n}_z[/latex] (7.307)

where [V] is the partial velocity array which is seen to take a simple form

[latex][V]=\left[\begin{array}{l}0 \I\end{array} ight][/latex] (7.308)

where I denotes the identity matrix and 0 a zero matrix. The partial velocity is

[latex]\frac{\partial \bar{v}}{\partial \dot{x}_{\ell}}=[V]\left\{\bar{n}^R ight\}[/latex] (7.309)

The acceleration of the center mass is found through differentiation of equation (7.307).

[latex]\bar{a}=\{\ddot{x}\}^T[V]\left\{\bar{n}^R ight\}=\ddot{x} \bar{n}_x+\ddot{y} \bar{n}_y+\ddot{z} \bar{n}_z[/latex] (7.310)

The equations of motion are found through a, b, c formulation given in Chapter 5 which states

[latex]\{f\}=a\{\ddot{x}\}+b\{\dot{x}\}+c\{\dot{x}\}[/latex] (7.311)

where, as before, the coefficient of the equations above are

[latex]\begin{aligned}&{[a]=m[V][V]^T+[\omega]\left[I_{1 R} ight][\omega]^T} \&{[b]=[\omega]\left[I_{1 R} ight][\dot{\omega}]^T} \&{[c]=[\omega]\left[\Omega_x^{R 1} ight]\left[I_{1 R} ight][\omega]^T}\end{aligned}[/latex]

where [latex]\left[I_{1 R} ight][/latex] denotes the moment of inertia of the disk with respect to [latex]\left\{\bar{n}^R ight\}[/latex] in R and is given by

[latex]\left[I_{1 R} ight]=\left[S^{R 1} ight]\left[I_1 ight]\left[S^{1 R} ight][/latex] (7.312)

where [latex]I_1[/latex] is the moment of inertia with respect to the center axis.

[latex]\left[I_1 ight]=\left[\begin{array}{lll}I_{11} & & 0 \& I_{12} & \0 & & I_{13}\end{array} ight][/latex] (7.313)

The values [latex]I_{11}, I_{12}[/latex] and [latex]I_{13}[/latex] are the moments of inertia with respect to [latex]\bar{n}_1^1, \bar{n}_2^1[/latex] and [latex]\bar{n}_3^1[/latex], respectively. In this problem, [latex]I_{11}=I_{13}[/latex] are assumed to be [latex]I_0[/latex], then [latex]I_{12}=2 I_0[/latex]; hence

[latex]\left[I_1 ight]=I_0\left[\begin{array}{lll}1 & 0 & 0 \0 & 2 & 0 \0 & 0 & 1\end{array} ight] (7.314)[/latex]

From equations (7.312) and (7.297), we obtain the moment of inertia of the disk with respect to R.

[latex]\left[I_{1 R} ight]=I_0\left[\begin{array}{ccc}1+s^2 \phi_1 s^2 heta & & ext { symmetric } \-s^2 \phi_1 s heta c heta & 1+s^2 \phi_1 c^2 heta & \c \phi_1 s heta_1 s heta & -c \phi_1 s \phi_1 c heta & 1+c^2 \phi_1\end{array} ight] (7.315)[/latex]

The expression [latex]\left[\Omega_x^{R 1} ight][/latex] is a skew matrix composed of [latex]\{\dot{x}\}^T[\omega][/latex].

Note that

[latex]\begin{aligned}{\left[\Omega^{R 1} ight] } &=\left[\dot{S}^{R 1} ight]\left[S^{1 R} ight]=\left[\begin{array}{ccc}0 & -\dot{ heta} s \phi_1 & -\dot{ heta} c \phi_1 \\dot{ heta} s \phi_1 & 0 & -\dot{\phi}_1 \\dot{ heta} c \phi_1 & \dot{\phi}_1 & 0\end{array} ight] (7.316) \&=\left[\begin{array}{ccc}0 & -\dot{\phi}_3 s^2 \phi_1 & -\dot{\phi}^3 s \phi_1 c \phi_1 \\phi_3 s^2 \phi_1 & 0 & -\dot{\phi}_1 \\dot{\phi}_3 s \phi_1 c \phi_1 & \dot{\phi}_1 & 0\end{array} ight]\end{aligned}[/latex]

To obtain the equations we now can evaluate the coefficients [a], [b] and [c]. The first term written explicitly after substitution yields

[latex]\begin{aligned}{[a] } &=m\left[\frac{0_{3 imes 3}}{I} ight]\left[0_{3 imes 3} \mid I ight]+\left[\frac{\left[S^{1 R} ight]}{0_{3 imes 3}} ight]\left[S^{R1} ight]\left[I_1 ight]\left[S^{1 R} ight]\left[S^{R 1}\overset{\mid }{\mid } {0_{3 imes 3}} ight] (7.317) \&=\left[\begin{array}{cc}I_1 & 0_{3 imes 3} \0_{3 imes 3} & m I\end{array} ight]_{6 imes 6}\end{aligned}[/latex]

Let us combine the other two terms in the equations of motion

[latex]\left[b^* ight]=[b]+[c][/latex] (7.318)

Substituting the terms of [b] and [c] we get

[latex]b^*=\left[\frac{\left[S^{1 R} ight]}{0} ight]\left[S^{R 1} ight]\left[I_1 ight]\left[S^{1 R} ight]\left[S^{R 1}\overset{\mid }{\mid } 0 ight]+\left[\frac{\left[S^{1 R} ight]}{0} ight]\left[\Omega_x^{R 1} ight]\left[S^{R 1} ight]\left[I_1 ight]\left[S^{R 1} ight]\left[S^{R 1}\overset{\mid }{\mid } 0 ight][/latex] (7.319)

Since

[latex]\left[S^{1 R} ight]\left[\dot{S}^{R 1} ight]=\left[\dot{S}^{R 1} ight]\left[S^{1 R} ight]=\left[\Omega^{R 1} ight][/latex] (7.320)

Equation (7.319) reduces to

[latex]\left[b^* ight]=\left[\begin{array}{cc}b_{11} & 0 \0 & 0\end{array} ight]_{6 imes 6}[/latex] (7.321)

where

[latex]\left[b_{11} ight]=\left[S^{1 R} ight]\left[\Omega_x^{R 1} ight]\left[S^{R 1} ight]\left[I_1 ight]+\left[I_1 ight]\left[\Omega^{R 1} ight][/latex]

which is shown to take the following form

[latex]\begin{aligned}\left[b_{11} ight]&=I_0\left[\begin{array}{ccc}0 & -2\left(\omega_3 s \phi_1+\omega_2 c \phi_1 c heta-\omega_1 s heta c \phi_i ight) \ & -\omega_3 c \phi_1+\omega_2 c heta s \phi_1 -\omega_1 s heta s \phi_1 &\ \omega_3 s \phi_1+\omega_2 c \phi_1 c heta & 0 & \omega_3 s heta-\omega_1 c heta \ -\omega_1 s heta c \phi_1 & & 0 \ \omega_3 c \phi_1-\omega_2 c heta_s \phi_1 & 2\left(\omega_2 s heta+\omega c heta ight) & 0 \ +\omega_1 s heta s \phi_1 & \end{array} ight]\&+I_0\left[\begin{array}{ccc}0 & -\dot{\phi}_3 s^2 \phi_1 & -\dot{\phi}^3 s \dot{\phi}_1 c \phi_1 \ 2 \dot{\phi}_3 s^2 \dot{\phi}_1 & 0 & -2 \dot{\phi}_1 \ \dot{\phi}_3 s \phi_1 c \phi_1 & \dot{\phi}_1 & 0\end{array} ight]\&=I_0\left[\begin{array}{ccc}0 & -2 \dot{\phi}_3-\dot{\phi}_2&+\dot{\phi}_3 c \phi_1 \& & s \phi_1 \\dot{\phi}_3 & 0 & -\dot{\phi}_1 \-\dot{\phi}_2-\dot{\phi}_3 & 2 \dot{\phi}_1 & 0 \c \phi_1 s \dot{\phi}_1\end{array} ight]+I_0\left[\begin{array}{ccc}0 & -\dot{\phi}_3^2 s \phi_1 & -\dot{\phi}^3 s \phi_1\&&c \phi_1 \2 \dot{\phi}_3 s^2 \phi_1 & 0 & -2 \dot{\phi}_1 \\dot{\phi}_3 s \phi_1 c \phi_1 & \dot{\phi}_1 & 0\end{array} ight] \&=I_0\left[\begin{array}{ccc}0 & -\dot{\phi}_3\left(2+s^2 \phi_1 ight) & \dot{\phi}_2 \\dot{\phi}_3\left(1+2 s^2 \phi_1 ight) & 0 & -3 \dot{\phi}_1 \-\dot{\phi}_2 & 3 \dot{\phi}_1 & 0\end{array} ight] (7.322)\end{aligned}[/latex]

Note how equations (7.317) and (7.321) combined with equation (7.322) form the equations of motion of the rolling disk. What remains is the evaluation of the generalized active forces and any constraint conditions imposed on the system.

The vector of the center mass of the disk is [latex]\overline{O D}[/latex], as shown in Figure 7.19, which is equal to

[latex]\overline{O D}=\overline{O C}+\overline{C D}[/latex] (7.323)

Since we are assuming no slip, differentiation of the equation above yields

[latex]\overset{.}{\overline{OD}}=\overset{.}{\overline{OC}}+\overset{.}{\overline{CD}}=\overset{.}{\overline{CD}}[/latex]

The velocity of point C is

[latex]\bar{v}_c=\overset{.}{\overline{OD}}=\overset{.}{\overline{CD}}=\bar{\omega} imes r \bar{n}_3^1[/latex]

From equation (7.297) and (7.298), we get the velocity of contact C as

[latex]\begin{aligned}\bar{v}_c=& r\left[\begin{array}{ccc}\bar{n}_x & \bar{n}_y & \bar{n}_z \\omega_1 & \omega_2 & \omega_3 \-\cos \phi_1 \sin heta & \cos \dot{\phi}_1 \cos heta & \sin \dot{\phi}_1\end{array} ight] \=&\left(\omega_2 \sin \phi_1-\omega_3 \cos \phi_1 \cos heta ight) \bar{n}_x-\left(\omega_3 \cos \phi_1 \sin heta+\omega_1 \sin \phi_1 ight) \bar{n}_y \&\left.+\left(\omega_2 \cos \phi_1 \cos heta-\omega_2 \cos \phi_1 \sin heta ight) \bar{n}_z ight] r \=& r\left(\dot{\phi}_1 s heta s \phi_1+\dot{\phi}_2 c heta ight) \bar{n}_x-r\left(\dot{\phi}_1 c \phi_1-\dot{\phi}_2 s heta ight) \bar{n}_y+r \dot{\phi}_1 c \phi_1 \bar{n}_z (7.324)\end{aligned}[/latex]

Expressing the center mass velocity in a matrix form, we get

[latex]\left[\begin{array}{c}v_{c x} \v_{c y} \v_{c z}\end{array} ight]=r\left[\begin{array}{c}\dot{\phi}_1 c heta s \phi_1+\dot{\phi}_2 s heta \-\dot{\phi}_1 c heta s \phi_1+\dot{\phi}_2 s \phi \\dot{\phi}_1 c \phi_1\end{array} ight][/latex]

We need to impose the condition that [latex]v_c[/latex] is zero which can be expressed as

[latex][B]\{\dot{x}\}=0[/latex] (7.325)

where

[latex][B]=\left[\begin{array}{cccccc}s heta s \phi_1 & c heta & 0 & -1 & 0 & 0 \-c heta s \phi_1 & s heta & 0 & 0 & -1 & 0 \c \phi_1 & 0 & 0 & 0 & 0 & 1\end{array} ight][/latex] (7.326)

The generalized active forces follow the derivations stated in Chapter 4 , where using the definition of those forces given by

[latex]\begin{aligned}\{f\} &=\bar{F} \cdot \frac{\partial \bar{v}}{\partial x_{\ell}}+\bar{M} \cdot \frac{\partial \bar{\omega}}{\partial \dot{x}_{\ell}} \&=[V]\{F\}+[\omega]\{M\} (7.327)\end{aligned}[/latex]

where [latex]\bar{F}[/latex] denotes the resultant external force in R and [latex]\bar{M}[/latex] the corresponding external moment in R acting on the body. Due to gravity, we have

[latex]\bar{F}=-m g \bar{n}_Z[/latex] (7.328)

which can be written further as

[latex]\{F\}=\left[\begin{array}{lll}0 & 0 & -m g\end{array} ight]^T[/latex] (7.329)

Similarly,

[latex]\begin{aligned}\bar{M}=-mgrs \phi_1\left[\begin{array}{lll}1 & 0 & 0\end{array} ight]\left\{n^{-1} ight\} &=-mgrs \phi_1\left[\begin{array}{lll}1 & 0 & 0\end{array} ight]\left[S^{1 R} ight]\left\{\bar{n}^R ight\} (7.330)\\{M\} &=-mgrs \phi_1\left[S^R 1 ight]\left\{\begin{array}{l}1 \0 \0\end{array} ight\} (7.331)\end{aligned}[/latex]

Therefore, the generalized active forces array becomes

[latex]\begin{aligned}\{f\} &=\left[\frac{0}{I} ight]\{F\}+\left[\frac{S^{1 R}}{I} ight]\{M\} \&=-m g\left[\begin{array}{c}r s \phi_1 \0 \0 \0 \0 \1\end{array} ight] (7.332)\end{aligned}[/latex]

Reaction Force N at Point c

[latex]\bar{N}=m g+m \ddot{z} \quad\left(N ext { is in } \bar{n}_Z ext { direction } ight)[/latex] (7.333)

The friction force at c has no contribution to the generalized force because [latex]\bar{v}_c=0[/latex].

The equations of motion of the rolling disk or rolling coin are found making use of Kane's equations.

[latex]f+f^*=f^c[/latex] (7.334)

where the generalized external force array f is given by equation (7.332), the generalized inertia force array [latex]f^*[/latex] is given by (7.319), and the constraint force array [latex]f^c[/latex] is

[latex]f^c=[B]^T\{\lambda\}[/latex] (7.335)

where {λ} is the array of Lagrange undetermined multipliers

[latex]\{\lambda\}=\left[\begin{array}{l}\lambda_1 \\lambda_2 \\lambda_3\end{array} ight][/latex] (7.336)

Combining equation (7.298), (7.317), (7.321) and (7.322), we obtain the equations of motion of the system:

[latex]\begin{aligned}{[a]\{\ddot{x}\}+[b *]\{\dot{x}\}+[B]^T\{\lambda\} } &=\{f\} (7.337) \\dot{ heta} &=\dot{\phi}_3 \sin \phi_1 (7.338)\end{aligned}[/latex]

or in a compact form by an augmented matrix form of the equations above. That is, taking θ as a generalized coordinate and equation (7.338) as constraint equation and embedding it into constraint matrix [B], we arrive at

[latex]\left[\begin{array}{llllll}I_0 & & & & && 0 \& 2 I_0 & & & & \& & I_0 & & & \& & & m & & \& & & &m & & \& & & && m & \0 & & & && & 0\end{array} ight]\left[\begin{array}{c}\ddot{\phi}_1 \\ddot{\phi}_2 \\ddot{\phi}_3 \\ddot{x} \\ddot{y} \\ddot{z} \\ddot{ heta}\end{array} ight][/latex]

[latex]\begin{array}{r}+I_0\left[\begin{array}{cccc}0 & -\dot{\phi}_3\left(2+s^2 \phi_1 ight) & \dot{\phi}_2 & \\dot{\phi}_3\left(1+2 s^2 \phi_1 ight) & 0 & -3 \dot{\phi}_1 & 0_{3 imes 4} \\dot{\phi}_2 & 3 \dot{\phi}_1 & 0 & \ & 0_{4 imes 3} && 0_{4 imes 4}\end{array} ight]\left[\begin{array}{c}\dot{\phi}_1 \\dot{\phi}_2 \\dot{\phi}_3 \\dot{x} \\dot{y} \\dot{z} \\dot{ heta}\end{array} ight] \+\left[\begin{array}{cccc}s heta s \phi_1 & -c heta s \phi_1 & c \phi_1 & 0 \c heta & s heta & 0 & 0 \0 & 0 & 0 & s \phi_1 \-1 & 0 & 0 & 0 \0 & -1 & 0 & 0 \0 & 0 & -1 & 0 \0 & 0 & 0 & -1\end{array} ight]\left[\begin{array}{l}\lambda_1 \\lambda_2 \\lambda_3 \\lambda_4\end{array} ight]=-m g\left[\begin{array}{c}r s \phi_1 \0 \0 \0 \0 \1 \0\end{array} ight] (7.339)\end{array}[/latex]

We can easily find an orthogonal matrix T numerically in time to make [latex][T][B]^T=0[/latex]; then we reduce the equations of motion to

[latex][T][a]\{\ddot{x}\}+[T]\left[b^* ight]\{\dot{x}\}=[T]\{f\}[/latex] (7.340)

Numerical Simulation

The problem above was simulated to investigate its motion for 1.6 s. The following data was used.

Input Data

[latex]\begin{gathered}m=1, \quad r=1 \I_0=I_{11}=I_{13}=\frac{m r^3}{4}=0.25, \quad I_{12}=\frac{m r_2}{2}=0.5 \{\left[I_1 ight]=\left[\begin{array}{ccc}0.25 & 0 & 0 \0 & 0.5 & 0 \0 & 0 & 0.25\end{array} ight]}\end{gathered}[/latex]

The following initial conditions were used

[latex]\left[\begin{array}{c}\phi_1 \\phi_2 \\phi_3 \x \y \z \ heta\end{array} ight]_{t=0}=\left[\begin{array}{c}\frac{\pi}{2} \0 \0 \0 \0 \1 \\frac{\pi}{2}\end{array} ight]\left[\begin{array}{c}\dot{\phi}_1 \\dot{\phi}_2 \\dot{\phi}_3 \\dot{x} \\dot{y} \\dot{z} \\dot{ heta}\end{array} ight]_{t=0}=\left[\begin{array}{c}-0.1 \-1 \0.1 \-\epsilon \0.1 \0 \0.1\end{array} ight][/latex]

where ∈ is a very small positive quantity. The time step Δt = 0.1; the total number of time steps NTT = 17. The following results were found.

Discussion

Mass center motion is shown in Figure 7.20.
The value of θ increases as t increases; this means that the disk is dropping downward while Φ increases and ψ decreases (i.e., the disk moves rightward against the x − z plane, as shown in Figure 7.21).
As [latex]\ddot{z}[/latex] decreases, the reaction force at c, N, decreases.

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