Solution
System Analysis:
Consider the plank moving down the incline with a steady velocity $v$. As the plank moves, it encounters stationary rollers and sets them into rotational motion. Since the velocity is steady (constant), the rate at which the plank loses Gravitational Potential Energy must equal the rate at which energy is transferred to the rollers (Kinetic Energy + Heat dissipated due to friction).
Step 1: Energy Balance for Displacement $l$
Let us analyze the energy exchange as the plank moves a distance $x = l$ (the center-to-center spacing between two adjacent rollers). In this distance, the plank effectively passes over one gap and engages one new roller, bringing it from rest to a steady angular velocity.
The work done by gravity on the plank for a displacement $l$ down the incline is:
$$ W_{\text{gravity}} = (M g \sin\theta) \cdot l $$Step 2: Energy Transferred to a Roller
As the plank moves at a steady speed $v$, it spins the rollers. Assuming pure rolling is achieved, the tangential speed of the roller’s surface matches the plank’s speed.
Angular velocity of the roller: $\omega = \frac{v}{r}$
Moment of inertia of a solid cylinder (roller): $I = \frac{1}{2}m r^2$
The final Kinetic Energy ($K_{\text{roller}}$) acquired by one roller is:
$$ K_{\text{roller}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{1}{2} m r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{4} m v^2 $$Step 3: Accounting for Work Done by Friction
The roller is brought from rest to angular velocity $\omega$ by the friction force exerted by the moving plank. This is a dynamic process where the plank moves at constant velocity $v$ while the roller surface accelerates from $0$ to $v$.
The total work done ($W_{\text{plank}}$) by the plank to spin up the roller is given by:
$$ W_{\text{plank}} = \int \tau d\theta_{\text{roller}} + \text{Heat Loss} $$Alternatively, using the work-energy theorem for a constant speed driver spinning a mass via friction:
$$ W_{\text{plank}} = \int f v dt = v \int f dt $$From the impulse-momentum relation for the roller ($\int f r dt = I \omega$), we have $\int f dt = \frac{I \omega}{r}$. Substituting this back:
$$ W_{\text{plank}} = v \left( \frac{I \omega}{r} \right) = \frac{v}{r} (I \omega) = \omega (I \omega) = I \omega^2 $$Substituting $I = \frac{1}{2}m r^2$ and $\omega = v/r$:
$$ W_{\text{plank}} = \left( \frac{1}{2} m r^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{2} m v^2 $$Note: This total work ($ \frac{1}{2} m v^2 $) is split exactly into kinetic energy of the roller ($\frac{1}{4} m v^2$) and heat dissipated by friction ($\frac{1}{4} m v^2$).
Step 4: Final Calculation
In the steady state, the work done by gravity over distance $l$ supplies the total energy required to spin up one roller (including the heat loss).
$$ W_{\text{gravity}} = W_{\text{plank}} $$ $$ M g l \sin\theta = \frac{1}{2} m v^2 $$Solving for $v$:
$$ v^2 = \frac{2 M g l \sin\theta}{m} $$