Kvant Physics Problem 30
The system consists of a car moving with rolling wheels of diameter $D = 1,\text{m}$, filmed by a camera and projected onto a screen at a rate of $f = 8,\text{frames/s}$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m47s
Source on kvant.digital
Problem
In a film camera and projector, 8 frames per second pass through. On the screen, a car is moving with wheels whose real diameter is 1 m. The images of the wheels make 2 revolutions per second. What is the speed of the car?
G. L. Kotkin
Setup and Assumptions
The system consists of a car moving with rolling wheels of diameter $D = 1,\text{m}$, filmed by a camera and projected onto a screen at a rate of $f = 8,\text{frames/s}$. The observed motion on the screen shows that the wheel image performs a steady rotation with angular frequency $f_{\text{img}} = 2,\text{rev/s}$.
The unknown is the translational speed $v$ of the car relative to the ground.
The analysis assumes that the wheels roll without slipping, so the translational speed of the car is directly related to the angular velocity of the wheels. The camera captures instantaneous positions at uniform time intervals $\Delta t = 1/f$, and the projector reproduces these discrete positions faithfully without additional distortion. Motion blur and optical distortions are neglected, so each frame corresponds to a sharp configuration of the system.
Physical Principles
The rolling without slipping condition relates linear and angular motion of the wheel,
$$v = \omega R,$$
where $R = D/2$ is the wheel radius and $\omega$ is the angular velocity of the wheel.
The film sampling condition relates the observed angular advance per frame to the apparent angular velocity,
$$\Delta \theta = \omega_{\text{img}} \Delta t,$$
where $\omega_{\text{img}} = 2 \cdot 2\pi,\text{rad/s}$ is the observed angular speed in the projected motion and $\Delta t = 1/f$ is the time between frames.
The total angular advance per frame corresponds to the actual physical rotation of the wheel between successive recorded instants.
Derivation
The observed angular velocity of the wheel image is
$$\omega_{\text{img}} = 2 \cdot 2\pi = 4\pi,\text{rad/s}.$$
The time between frames is
$$\Delta t = \frac{1}{8},\text{s}.$$
The angular change of the wheel between two successive frames is therefore
$$\Delta \theta = \omega_{\text{img}} \Delta t = 4\pi \cdot \frac{1}{8} = \frac{\pi}{2},\text{rad}.$$
This corresponds to a rotation of
$$\frac{\Delta \theta}{2\pi} = \frac{1}{4}$$
of a full revolution per frame.
The wheel radius is
$$R = \frac{D}{2} = 0.5,\text{m}.$$
Rolling without slipping gives the linear distance traveled per frame,
$$\Delta x = R \Delta \theta = 0.5 \cdot \frac{\pi}{2} = \frac{\pi}{4},\text{m}.$$
The car speed follows from dividing by the frame interval,
$$v = \frac{\Delta x}{\Delta t} = \frac{\frac{\pi}{4},\text{m}}{\frac{1}{8},\text{s}} = \frac{\pi}{4} \cdot 8,\text{m/s} = 2\pi,\text{m/s}.$$
Result
$$v = 2\pi,\text{m/s}$$
Substituting numerical values,
$$v = 2\pi,\text{m/s} \approx 6.28,\text{m/s}.$$
Converting to kilometers per hour,
$$v \approx 6.28 \times 3.6,\text{km/h} \approx 22.6,\text{km/h}.$$
The car speed is
$$\boxed{v \approx 6.3,\text{m/s} \approx 23,\text{km/h}}.$$
Sanity Checks
Dimensional consistency follows from $v = R \Delta \theta / \Delta t$, where $R$ carries meters, $\Delta \theta$ is dimensionless in radians, and $\Delta t$ carries seconds, yielding meters per second.
If the frame rate increased while the physical motion stayed fixed, the angular change per frame would decrease proportionally, but the reconstructed velocity $v$ remains unchanged because the factor $\Delta t$ compensates exactly.
If the observed rotation were zero, $\omega_{\text{img}} = 0$, then $\Delta \theta = 0$ and the model gives $v = 0$, consistent with a stationary car.
The most sensitive step is the conversion from revolutions per second to radians per second and the multiplication by the frame interval; an error in either factor would change the inferred $\Delta \theta$ and propagate linearly into the final speed.