- #1
Hak
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- Homework Statement
- In 1912, Smoluchowski proposed a Brownian ratchet device. In 1962, Feynman conducted its first quantitative analysis. This device consists of two parts: a small paddle immersed in an ideal gas at temperature ##T_1## and a ratchet immersed in an ideal gas at temperature ##T_2##, connected to the paddle by a linkage. A pulley shaft is fixed in the middle of the linkage, with an object suspended below it. The ratchet is asymmetric in shape and has a small spring piece called a pawl that catches on the ratchet teeth. All components of the system are lightweight. Smoluchowski pointed out that when liquid molecules undergo random motion and collide with the paddle wheel, they will set it in motion.
The pawl restricts the ratchet to rotate in only one direction. Therefore, even if ##T_1 = T_2##, the system will continue to rotate in the same direction, thus extracting mechanical work from the internal energy of a thermally equilibrated system, violating the second law of thermodynamics.
1. When ##T_1 = T_2##, how will the system rotate?
(a) Completely stationary
(b) Rotating clockwise
(c) Both clockwise and counterclockwise rotation, with a net clock-
wise rotation on average
(d) Both clockwise and counterclockwise rotation, with no net di-
rection
2. When ##T_1 > T_2##:
(a) Estimate the average angular velocity of the system. Assume that each tooth of the ratchet corresponds to a central angle of ##\theta##. The radius of the pulley shaft in the linkage is
##R##, and the mass of the suspended object is ##m##, with all other components being lightweight.
(b) Determine the energy ##E## transferred from the gas on one side of the small paddle wheel to the small paddle wheel during the rotation of one tooth of the ratchet.
- Relevant Equations
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1. (d) Both clockwise and counterclockwise rotation, with no net direction. Because the pawl, which is also at the same temperature as the paddle and the ratchet, will undergo Brownian motion and bounce up and down randomly. Sometimes, it will fail to catch the ratchet teeth and allow the ratchet to slip backward. The net effect of many such random collisions and failures will be that the system will move back and forth with no preferred direction.
2. When ##T_1 > T_2##:
(a)
I assume that the system is in a steady state, meaning that the angular velocity of the paddle, the ratchet, and the pulley shaft are equal. I also assumed that the system is in equilibrium, meaning that the net torque on the system is zero.
I would calculate the average force ##F## exerted by the gas molecules on the paddle by using the ideal gas law, which relates the pressure, volume, temperature, and number of molecules of an ideal gas; then, the viscous drag coefficient ##\gamma## of the paddle by using Stokes’ law for a sphere, which relates the drag force, viscosity, and radius of a spherical object moving in a fluid. Then, I would equate the torque due to ##F## and the torque due to ##\gamma##, which are opposite in direction, to find the angular velocity ##\omega## of the paddle. Then, I will multiply ##\omega## by the ratio of ##R## and ##r##, which are the radii of the pulley shaft and the ratchet, respectively, to find the angular velocity of the ratchet. Dividing the angular velocity of the ratchet by ##2\pi##, which is the angle of one full rotation, I'll find the number of rotations per unit time: multiplying the number of rotations per unit time by ##\theta##, which is the angle of one tooth of the ratchet, I'll find the average angular displacement per unit time. I get:$$\omega=\frac{F}{\gamma}\frac{R}{r}\frac{\theta}{2 \pi}$$
The average force ##F## can be approximated by using the ideal gas law:$$F=\frac{N_1k_BT_1}{A}−\frac{N_2k_BT_2}{A}$$where ##N_1## and ##N_2## are the number of gas molecules on each side of the paddle, ##k_B## is the Boltzmann constant, ##T_1## and ##T_2## are the temperatures of the gases, and ##A## is the area of the paddle.
The viscous drag coefficient ##\gamma## can be calculated by using Stokes’ law for a sphere:$$\gamma=6\pi \eta r_p$$,where ##\eta## is the viscosity of the gas and ##r_p## is the radius of the paddle.Substituting these expressions into the formula for ##\omega##, we get:$$\omega=\frac{\frac{(N_1T_1 -N_2T_2) k_B}{A}}{6\pi \eta r_p}\frac{R}{r}\frac{\theta}{2 \pi}$$(b) The energy ##E## transferred from the gas on one side of the small paddle wheel to the small paddle wheel during the rotation of one tooth of the ratchet can be obtained by multiplying the average torque ##\tau## by the angle ##\theta##:
$$E = \tau \theta$$The average torque ##\tau## can be found by multiplying the average force ##F## by the radius ##r_p## of the paddle:$$\tau=Fr_p$$Using the expression for F from part (a), I get:$$\tau=\frac{(N_1T_1−N_2T_2)k_B}{A}r_p$$Therefore,$$E = \frac{(N_1T_1−N_2T_2)k_B}{A}r_p \theta$$Where do I go wrong?
2. When ##T_1 > T_2##:
(a)
I assume that the system is in a steady state, meaning that the angular velocity of the paddle, the ratchet, and the pulley shaft are equal. I also assumed that the system is in equilibrium, meaning that the net torque on the system is zero.
I would calculate the average force ##F## exerted by the gas molecules on the paddle by using the ideal gas law, which relates the pressure, volume, temperature, and number of molecules of an ideal gas; then, the viscous drag coefficient ##\gamma## of the paddle by using Stokes’ law for a sphere, which relates the drag force, viscosity, and radius of a spherical object moving in a fluid. Then, I would equate the torque due to ##F## and the torque due to ##\gamma##, which are opposite in direction, to find the angular velocity ##\omega## of the paddle. Then, I will multiply ##\omega## by the ratio of ##R## and ##r##, which are the radii of the pulley shaft and the ratchet, respectively, to find the angular velocity of the ratchet. Dividing the angular velocity of the ratchet by ##2\pi##, which is the angle of one full rotation, I'll find the number of rotations per unit time: multiplying the number of rotations per unit time by ##\theta##, which is the angle of one tooth of the ratchet, I'll find the average angular displacement per unit time. I get:$$\omega=\frac{F}{\gamma}\frac{R}{r}\frac{\theta}{2 \pi}$$
The average force ##F## can be approximated by using the ideal gas law:$$F=\frac{N_1k_BT_1}{A}−\frac{N_2k_BT_2}{A}$$where ##N_1## and ##N_2## are the number of gas molecules on each side of the paddle, ##k_B## is the Boltzmann constant, ##T_1## and ##T_2## are the temperatures of the gases, and ##A## is the area of the paddle.
The viscous drag coefficient ##\gamma## can be calculated by using Stokes’ law for a sphere:$$\gamma=6\pi \eta r_p$$,where ##\eta## is the viscosity of the gas and ##r_p## is the radius of the paddle.Substituting these expressions into the formula for ##\omega##, we get:$$\omega=\frac{\frac{(N_1T_1 -N_2T_2) k_B}{A}}{6\pi \eta r_p}\frac{R}{r}\frac{\theta}{2 \pi}$$(b) The energy ##E## transferred from the gas on one side of the small paddle wheel to the small paddle wheel during the rotation of one tooth of the ratchet can be obtained by multiplying the average torque ##\tau## by the angle ##\theta##:
$$E = \tau \theta$$The average torque ##\tau## can be found by multiplying the average force ##F## by the radius ##r_p## of the paddle:$$\tau=Fr_p$$Using the expression for F from part (a), I get:$$\tau=\frac{(N_1T_1−N_2T_2)k_B}{A}r_p$$Therefore,$$E = \frac{(N_1T_1−N_2T_2)k_B}{A}r_p \theta$$Where do I go wrong?