Solution: Sliding and Rotating Disc
The position of the lamp is given by the vector sum of the center of mass translation and the rotation of the disc. Let the time interval be $\tau = 1.0$ s. The displacement between consecutive pulses is:
$$\Delta \vec{r} = \vec{v}_c \tau + \Delta \vec{r}_{rot}$$Since $\vec{v}_c$ is constant, the vector $\vec{v}_c \tau$ is a constant vector. The term $\Delta \vec{r}_{rot}$ represents the chord of the rotation. If we plot the displacement vectors $\Delta \vec{r}$ from a common origin, their tips must lie on a circle centered at the tip of the vector $\vec{v}_c \tau$.
From the problem statement and graph, the coordinates of the pulses are:
- Red ($R$): $(4, 10)$
- Green ($G$): $(1, 5)$
- Yellow ($Y$): $(8, 0)$
- Blue ($B$): $(15, 5)$
We calculate the displacement vectors for each interval:
- $\vec{d}_1 = \vec{G} – \vec{R} = (1-4, 5-10) = (-3, -5)$ cm
- $\vec{d}_2 = \vec{Y} – \vec{G} = (8-1, 0-5) = (7, -5)$ cm
- $\vec{d}_3 = \vec{B} – \vec{Y} = (15-8, 5-0) = (7, 5)$ cm
The vectors $\vec{d}_1(-3, -5)$, $\vec{d}_2(7, -5)$, and $\vec{d}_3(7, 5)$ must lie on a circle. We find the center of this circle (which corresponds to $\vec{v}_c \tau$).
- Intersection of perpendicular bisector of $\vec{d}_1, \vec{d}_2$:
Midpoint of x-coordinates is $(-3+7)/2 = 2$.
Since y-components are equal ($-5$), the perpendicular bisector is the vertical line $x = 2$. - Intersection of perpendicular bisector of $\vec{d}_2, \vec{d}_3$:
Midpoint of y-coordinates is $(-5+5)/2 = 0$.
Since x-components are equal ($7$), the perpendicular bisector is the horizontal line $y = 0$.
The intersection is $(2, 0)$. Thus:
$$\vec{v}_c \tau = (2, 0) \implies \vec{v}_c = 2 \hat{i} \text{ cm/s}$$Radius of Disc ($R$):
The magnitude of the rotational step $|\Delta \vec{r}_{rot}|$ is the distance from the center $(2,0)$ to any tip, e.g., $(7,5)$.
This chord length corresponds to rotation $\theta$. By analyzing the relative vectors ($(-5,-5)$, $(5,-5)$, $(5,5)$), we see they are $90^\circ$ apart. Using chord formula for $\theta = 90^\circ$:
$$5\sqrt{2} = 2R \sin(45^\circ) = 2R \left(\frac{1}{\sqrt{2}}\right) = R\sqrt{2} \implies R = 5 \text{ cm}$$Angular Velocity ($\omega$):
$$\omega = \frac{\theta}{\tau} = \frac{\pi/2}{1} = 0.5\pi \text{ rad/s}$$
Velocity $v_c = 2 \text{ cm/s}$
Angular velocity $\omega = 0.5\pi \text{ rad/s}$