A Pure Rolling Disk Consider the system shown in Figure 5.60a, in which a uniform disk of mass m and radius r rolls on a horizontal surface. A translational spring of stiffness k is attached to the disk. Assuming that there is no slipping between the disk and the surface, derive the differential

Question

A Pure Rolling Disk

Consider the system shown in Figure 5.60a, in which a uniform disk of mass [latex]m[/latex] and radius [latex]r[/latex] rolls on a horizontal surface. A translational spring of stiffness [latex]k[/latex] is attached to the disk. Assuming that there is no slipping between the disk and the surface, derive the differential equation of motion.

Accepted Answer

The free-body diagram of the system is shown in Figure [latex]5.60 \mathrm{~b}[/latex], in which the normal force [latex]N[/latex] and the friction force [latex]f[/latex] are reaction forces at the contact point. When the disk is at the static equilibrium position, we have [latex]f_{\mathrm{k}}=k \delta_{\mathrm{st}}[/latex], where [latex]\delta_{\mathrm{st}}[/latex] is the static deformation of the spring. Then,

[latex] \begin{gathered} +\curvearrowright: \sum M_{\mathrm{IC}}=0, \ r \cdot k \delta_{\mathrm{st}}=0, \end{gathered} [/latex]

or [latex]\delta_{\mathrm{st}}=0[/latex]. We choose the static equilibrium position as the origin. When the disk rolls, the spring force is [latex]f_{\mathrm{k}}=k\left(x+\delta_{\mathrm{st}} ight)=k x[/latex]. Because of no slipping, the contact point is the IC, and [latex]\mathbf{r}_{\mathrm{C} / \mathrm{IC}} \| \mathbf{a}_{\mathrm{IC}}[/latex]. Applying Equation 5.37

[latex]\sum M_{\mathrm{{IC}}}=I_{\mathrm{{IC}}} \alpha.[/latex] (5.37)

gives

[latex] \begin{gathered} +\curvearrowright: \sum M_{\mathrm{IC}}=l_{\mathrm{IC}} \alpha, \ r \cdot k x=l_{\mathrm{IC}} \alpha, \end{gathered} [/latex]

where

[latex]I_{\mathrm{IC}}=I_{\mathrm{C}}+m d^{2}=\frac{1}{2} m r^{2}+m r^{2}=\frac{3}{2} m r^{2}.[/latex]

Introducing the assumption of no slipping, [latex]x=r heta[/latex], we obtain the differential equation of motion

[latex]\frac{3}{2} m r^{2} \ddot{ heta}+k r^{2} heta=0 .[/latex]

Strictly speaking, the disk in Figure 5.59 or Example 5.12 is not a purely rotational system, and the rolling motion involves both translation and rotation. Such a system can also be modeled using both Newton's second law and the moment equation. We will present the corresponding discussion in the next section.

Question 5.12a: A Pure Rolling Disk Consider the system shown in Figure 5.60...

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