Work done by friction between two bodies with different velocities

Question

The question titles arose when solving the following problem:

Figure 5 is a schematic of a centrifuge that rotates with angular velocity $\Omega$. Solid material is first fed into the centrifuge at a radius $r_0$ where it is accelerated up to the speed $r_0 \Omega$ by friction with the rough inner wall.

Determine the work done by the centrifuge to accelerate a point mass $m$ to the speed $r_0 \Omega$. The mass is initially stationary. The cylindrical section of the centrifuge has a rough inner surface with a coefficient of friction $\mu$.

And this is the beginning of the solution provided to us (where $mr_0\dot\theta^2$ and $mr_0\ddot\theta$ are d'Alembert forces):

The work in the solution is given by: $$Work\ done = \int_0^T F\Omega r_0 dt$$ However, the force $F$ acts on the mass $m$ which moves with velocity $\dot\theta r_0$. Hence, the work equation I would expect would be: $$Work\ Done = \int_0^T F\dot\theta r_0 dt$$

Where $\dot\theta r_0$ is the distance that the force moves.

Why does this approach yield a different answer than when considering the force as acting on the centrifuge (and ignoring that the point of application moves relative to the centrifuge)?

Answer 1

In the first expression for the work done, we have the final speed $(r_0\Omega)$ in the integral. But the speed of the mass isn't constant as it starts at zero.

The second expression looks correct, except that you've called $\dot{\theta}r_0$ the 'distance the force moves'. But it's the speed of the mass in the centrifuge. The force also depends on $\dot{\theta}$. To evaluate the integral we have to know how $\dot{\theta}$ varies, and we don't know that--unless I'm missing something.

This is what I'm getting:

$$\begin{align} W &= \int_0^t Fv \text{ } dt = \int_0^t \mu mr_0 \dot{\theta}^2 \cdot r_0\dot{\theta} \text{ } dt\\ W &= \mu m r_0^2 \int_0^t \dot{\theta}^3 \text{ } dt\end{align}$$

How does $\dot{\theta}^3$ depend on $t$? Unless I'm making a mistake, I don't see how we can know this based on what's given.

I agree with my2cts. Since work is equal to $\Delta KE$, in the end this should evaluate to $\frac{1}{2}m(r_0\Omega)^2$.

Answer 2

The work done is equal to the change in kinetic energy.

Answer 3

The same force acts both on the mass and centrifuge. The work done by the centrifuge would be the friction force time the distance travelled along the centrifuge, as shown in the answer, i.e.: $$Work\ done\ by\ centrifuge = \int_0^T F \Omega r_0 dt$$

The other expression suggested, i.e. $\int_0^T F \dot\theta r_0 dt$, is the part of the work transmitted to the mass, i.e. the same friction force, but timed by the actual displacement of the solid lump in the centrifuge. As noted in the question, this is a different quantity, that is less than the work done by the centrifuge. $$Work\ done\ on\ mass = \int_0^T F \dot\theta r_0 dt$$

The difference between the two comes from the fact that some of the work done by the centrifuge is dissipated by the friction force and lost as heat. The difference between the two expressions is the integral $\int_0^T F (\Omega - \dot\theta) r_0 dt$, which is exactly the integral of the friction force times sliding velocity - the work dissipated as heat.

Answer 4

The Work according to friction force is \begin{align*} W &=\int F_\mu \,ds=\int_{0}^{t} F_\mu \frac{ds}{dt}\,dt\\ \end{align*} The equation of motions with $g=0$ and $\left[\frac{ds}{dt}\right]^2=0$ ist: \begin{align*} m\,\frac{ds^2}{dt^2} & =F_\mu \\ F_\mu&=\mu\,F_N\\ F_N&=-m\,r_0\,\Omega^2\\ \\ & \boxed{\,\frac{ds^2}{dt^2}+ \mu\,r_0\,\Omega^2=0} \end{align*}

Solution with Intial conditions $s(0)=0\,,D(s)(0)=0 $

\begin{align*} s(t)&=-\frac{1}{2 }\,\mu\,r_{{0}}{\Omega}^{2}{t}^{2}\\ \frac{ds}{dt}&=-\mu\,r_{{0}}{\Omega}^{2}t\\ \end{align*} Work: \begin{align*} &W=\int_{0}^{t} F_\mu \frac{ds}{dt}\,dt\\\\ &\boxed {W=-\int_{0}^{t} \left[\mu\,m\,r_0\,\Omega^2\right]\, \frac{ds}{dt}\,dt=-\frac{m}{2}\,{\mu}^{2}\,{r_{{0}}}^{2}{\Omega}^{4}{t}^{2}} \end{align*}