Worked examples › Example 6
Example 6Rolling disk driven through a spring
Problem
A homogeneous disk of mass $m$ and radius $R$ rolls without slipping. Its center O is connected to the wall by a spring $k$ and to a massless moving base B by a spring $4k$. The base moves as $x_B(t)=b\sin\omega t$. $x$ is the position of the center.
Use $m=4$ kg, $k=800$ N/m, $b=20$ mm.
Find: (a) the EOM in terms of $x$; (b) the particular solution; (c) the response amplitude at $\omega=15$ rad/s. Is it in phase or out of phase with the forcing?
Displacements in the drawing are scaled up so the motion is visible; the plot shows the true values.
Steady-state amplitude vs forcing frequency; the dot is your current ω.
Solution
Every solution follows the same five steps: coordinates, free-body diagrams, kinetics, kinematics, solve. Try the problem first, then open one step at a time to check your work.
Step 1 · Choose a coordinate system
Let $x$ be the position of the center O, positive to the right, with both springs unstretched when $x=0$ and $x_B=0$. Let $\phi$ be the disk's rotation, positive clockwise (so rolling right means $\phi$ increases).
Step 2 · Draw the free-body diagrams
Step 3 · Kinetics (Newton–Euler equations)
Translation: $\;m\ddot x=-kx+4k(x_B-x)+F_f$
Rotation about O ($I_O=\tfrac12mR^2$): $\;\tfrac12mR^2\,\ddot\phi=-F_fR$
Friction at the bottom acting in +x turns the disk counterclockwise, hence the minus sign.
Step 4 · Kinematics (constraints)
Rolling without slipping:
$$x=R\phi\qquad\Longrightarrow\qquad \ddot x=R\ddot\phi$$Step 5 · Solve
From rotation, $F_f=-\tfrac12m\ddot x$. Substitute into translation:
$$\tfrac32m\ddot x+5kx=4k\,x_B=4kb\sin\omega t,\qquad \omega_n=\sqrt{\frac{10k}{3m}}$$Particular solution: try $x_p=X\sin\omega t$:
$$X=\frac{4kb}{5k-\tfrac32m\omega^2}$$At ω = 15 rad/s ($m=4$ kg, $k=800$ N/m, $b=0.02$ m): $5k-\tfrac32m\omega^2=4000-1350=2650$ N/m and $4kb=64$ N, so
$$X=\frac{64}{2650}=0.0242\ \text{m}=24.2\ \text{mm}$$$X>0$, so the response is in phase with the forcing ($\omega<\omega_n=25.8$ rad/s).
Answer
(a) $\tfrac32m\ddot x+5kx=4kb\sin\omega t$ (b) $x_p=\dfrac{4kb}{5k-\tfrac32m\omega^2}\sin\omega t$
(c) $X=24.2$ mm, in phase with the forcing (since $\omega<\omega_n$).
Explore
- Raise ω above 25.8 rad/s. The stats switch to "180° out of phase": watch the disk move opposite to B.
- Untick "start on the particular solution" to see the free vibration that rides on top when you start from rest.
- Does the disk radius R affect anything? Why not?