Question 9.10: Attwood’s machine Two blocks of masses m and 2m are connecte...

The system is shown in Figure 9.8. We suppose that the string does not slip on the pulley and that the pulley is smoothly pivoted about its axis of symmetry.
Let z be the upward displacement of the mass m (from some reference configuration) and $v(=\dot{z})$ its upward velocity at time t. Then, since the string is inextensible, the mass 2m must have the same displacement and velocity, but measured downwards.
The angular velocity ω of the pulley is determined from the condition that the string does not slip. In this case, the velocity of the rim of the pulley and the velocity of the string must be the same at each point where they are in contact, that is, aω = v.
Hence ω = v/a. Also, from the table in the Appendix, the moment of inertia of a uniform circular disk of mass M and radius a about its axis of symmetry is $\frac{1}{2} M a^{2}$.
Hence, the total kinetic energy of the system is

$T=\frac{1}{2} m v^{2}+\frac{1}{2}(2 m) v^{2}+\frac{1}{2}\left(\frac{1}{2}(4 m) a^{2}\right)\left(\frac{v}{a}\right)^{2}=\frac{5}{2} m v^{2} .$

The gravitational potential energy of the system (relative to the reference configuration) is

$V=m g z-(2 m) g z=-m g z.$

We must now dispose of the constraint forces. (i) At the smooth pivot that supports the pulley, the reactions are perpendicular to the velocities of the particles on which they act. Hence these reactions do no work. (ii) Since there is no slippage between the string and the three material bodies of the system, the total work done by the string on the bodies must be equal and opposite to the total work done by the bodies on the string.∗ (iii) The internal forces that keep the pulley rigid do no work in total. Hence the constraint forces do no work in total.

Energy conservation therefore applies in the form

$\frac{5}{2} m v^{2}-m g z=E,$

where E is the total energy. If we now differentiate this equation with respect to t (and cancel by mv), we obtain

$\frac{d v}{d t}=\frac{1}{5} g$

which is the equation of motion of the system. Thus the upward acceleration of the mass m is g/5. (If the pulley were massless, the result would be g/3.)