Solution to Question 32
Consider the plank moving down the incline with a steady velocity $v$. As the plank moves, it continuously encounters stationary rollers and sets them into rotational motion.
Since the velocity is steady, the system has reached a dynamic equilibrium: the rate at which the plank loses gravitational potential energy must exactly equal the total rate of energy dissipation and transfer (Kinetic Energy gained by the rollers + Thermal Energy dissipated due to slipping friction).
Let us track the system dynamics over a displacement interval $x = l$ (the periodic structural spacing of the rolling mill). Within this window, the plank passes over exactly one spatial gap and completely spins up one new stationary roller to operational velocity.
The total work done by gravity on the plank over this distance is:
$$ W_{\text{gravity}} = (M g \sin\theta) \cdot l $$When the system reaches steady state, slipping ceases at the boundary after an initial transient spin-up, and the peripheral tangental velocity of the active cylinders perfectly matches the linear speed $v$ of the plank.
Angular velocity of the roller: $\omega = \frac{v}{r}$
Moment of inertia of a uniform solid cylinder: $I = \frac{1}{2}m r^2$
The clean rotational Kinetic Energy ($K_{\text{roller}}$) locked into one fully accelerated roller is:
$$ K_{\text{roller}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{1}{2} m r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{4} m v^2 $$The target cylinder is accelerated entirely by the localized kinetic friction force exerted by the moving plank. While the cylinder’s surface velocity scales from $0$ to $v$, the driver maintains a rigid velocity $v$.
The absolute work tracking equation ($W_{\text{plank}}$) transmitted across this slipping interface is evaluated using the linear force path integral:
$$ W_{\text{plank}} = \int f_{\text{friction}} \cdot v \, dt = v \int f_{\text{friction}} \, dt $$Applying the angular impulse-momentum theorem directly to the roller ($\int \tau \, dt = \int f_{\text{friction}} r \, dt = I\omega$), we can cleanly isolate the total friction impulse: $\int f_{\text{friction}} \, dt = \frac{I\omega}{r}$. Substituting this back into our tracking equation yields:
$$ W_{\text{plank}} = v \left( \frac{I \omega}{r} \right) = \left(\frac{v}{r}\right) (I \omega) = I \omega^2 $$Substituting structural parameters ($I = \frac{1}{2}mr^2$ and $\omega = \frac{v}{r}$):
$$ W_{\text{plank}} = \left( \frac{1}{2} m r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{2} m v^2 $$Observation: Out of this total work supplied ($\frac{1}{2}mv^2$), exactly half ($\frac{1}{4}mv^2$) transitions into structured rotational kinetic energy, while the remaining half ($\frac{1}{4}mv^2$) is irrevocably lost to thermal energy across the sliding interface.
Equating total potential energy spent by gravity to the total input work demanded by the tracking state over displacement interval $l$ ($W_{\text{gravity}} = W_{\text{plank}}$):
$$ M g l \sin\theta = \frac{1}{2} m v^2 $$Isolating the steady-state velocity vector:
$$ v^2 = \frac{2 M g l \sin\theta}{m} $$