As this is a fairly short chapter, and I wanted to leave things for other people to blog about, I’m taking a leaf out of Heather’s book and making my second post my solution to Your Turn 11F.

a) Torque, τ is force, f × (cross product) distance from the pivot to the point the force is applied, r. The force described in equation 11.17 is for the force acting perpendicular to the long axis of the rod, so the angle between the force vector and the position vector is 90o. So the cross product can be treated as a regular product. I will assume the force applied to the rod is uniform along the rod, so the average position of the point where the force is being applied is just the halfway down the rod. Thus, τ= f ×r≈3.0ηLvr=1.5ηL^2v.

The velocity of a point rotating at ω rad/s is the angular velocity times the distance that point is from the pivot. The average distance a point on the rod is from the pivot is again just halfway down the rod. So the torque becomes τ≈0.75ηL^3ω.

For a rod of length 1μm and an angular velocity of 6 rev/s=12π/s, the torque is 2.83x10^-20Nm.

b) The work done by torque is just the torque times the angle moved. In this case, the angle moved is one third of a revolution, which is ⅔π. So the work done by the F1 motor every third of a revolution is 5.92x10^-20J. Thermal energy at room temperature is 4.1x10^-21J, which is less than the value calculated, which is a good sign.

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