Solution to Question 32
The plank moves down the incline at a steady speed $v$. This implies that the loss in gravitational potential energy is exactly consumed by the work done to rotate the rollers and the associated heat loss due to friction.
Consider the plank moving forward by a distance equal to the separation between rollers, $l$. During this interval:
- The plank descends a vertical height $h = l \sin \theta$.
- The plank encounters one new stationary roller and accelerates it to an angular velocity $\omega$.
The gravitational potential energy lost by the plank as it travels distance $l$ is: $$ \Delta PE = M g h = M g l \sin \theta $$
The energy is consumed in two forms: kinetic energy of the roller and heat dissipated by friction during the spin-up process.
1. Kinetic Energy of the Roller:
The roller is a solid cylinder of mass $m$ and radius $r$. Its moment of inertia is $I = \frac{1}{2} m r^2$.
When the plank moves at speed $v$, the roller surface eventually acquires speed $v$ (assuming no slip in steady state), so angular velocity $\omega = v/r$.
$$ KE_{roller} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{4} m v^2 $$
2. Work against Friction (Heat Loss):
When a surface slides over a roller to accelerate it from rest to speed $v$ via a constant friction force, the work done by the driving force is split equally between the kinetic energy gained and the heat dissipated. This is because the process is similar to a perfectly inelastic collision.
$$ \text{Heat Loss} = KE_{roller} = \frac{1}{4} m v^2 $$
Therefore, the total energy consumed per roller (per distance $l$) is:
$$ E_{total} = KE_{roller} + \text{Heat Loss} = \frac{1}{4} m v^2 + \frac{1}{4} m v^2 = \frac{1}{2} m v^2 $$
Equating the energy supplied by gravity to the energy consumed: $$ M g l \sin \theta = \frac{1}{2} m v^2 $$
Rearranging to solve for $v$: $$ v^2 = \frac{2 M g l \sin \theta}{m} $$ $$ v = \sqrt{\frac{2 M g l \sin \theta}{m}} $$