- #1

raxp

- 11

- 3

Imagine we have a cuboid of mass ##M## and height ##2 r## that slides without friction on a horizontal surface. It is accelerated by a line or rod that is connected to the axle of a wheel of mass ##m## (homogeneously distributed) and radius ##r##. A rope or something similar is also connected to the wheel axle and a constant force ##f## is applied through it perfectly parallel to the ground. The entire system accelerates as a result and at any given moment, its velocity is ##v##. See the figure below.

The kinetic energy of the system is $$E = \frac 1 2 M v^2 + \frac 1 2 m v^2 + \frac 1 2 (I + m r^2) \omega^2$$ where ##I = \frac 1 2 m r^2## is the moment of inertia of the wheel around its center and ##\omega## is its angular velocity; the extra term ##m r^2## comes from the parallel axis theorem since the wheel rotates around its lower perimeter and not its center. Now, since the wheel rolls without slipping, ##v = r \omega## and the kinetic energy reduces to a function of the single variable ##v## alone: $$E = \frac 1 2 (M + \frac 5 4 m) v^2$$

The person pulling the system performs work ##f d## where ##d = \int v \mathrm d t## is the total displacement. The power of this work, work per unit time, is hence ##f v## and all of this is converted into kinetic energy of the system, hence the equation $$\frac {\mathrm d E} {\mathrm d t} = (M + \frac 5 4 m) v a = f v$$ where ##a## is the linear acceleration. Assuming ##v \neq 0##, we may solve this equation for ##a## to obtain Newton's second law with an effective mass: $$a = \frac f {M + \frac 5 4 m}$$.

Now consider the case where the attached rope is connected not to the wheel axle but to a point on the wheel perimeter. At a given moment, the situation is as depicted below.

The work performed per unit time is now greater, since the top of the wheel is moving at a speed ##2 v##, meaning that the power produced is now ##2 f v##. Carrying out the same derivation as before, we obtain $$a = \frac {2 f} {M + \frac 5 4 m}$$ but this expression is independent of ##r##, so that in the limit ##r \to 0## we have two different expressions for the induced acceleration, even though the point of application of the force is the same in both cases!

Where lies the error?