A k r k b k f b k r k f k b c k r k b k f d k r k b k

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(a) K r K b K f (b) K r K f K b (c) K r K b K f (d) K r K b K f (e) K b K f K r We know that the rotational kinetic energy of an object is given by the equation KE rot 1 2 I c.m. ω 2 . The moment of inertia about the center of mass of a disc is 1 2 MR 2 . We have two wheels of mass M and a bicycle/rider combination which is also of mass M . We also know that because the wheels roll without slipping the angular velocity of each wheel can be related to the velocity of the center of mass of the bike by the equation v c.m. . If we insert this into the equation for the rotational kinetic energy we can calculate the rotational kinetic energy for each wheel in terms of the velocity of the center of mass and the radius of the wheel. KE 1 ,rot 1 2 I c.m. ω 2 KE 1 ,rot 1 2 1 2 MR 2 1 v c.m. R 1 2 KE 1 ,rot 1 4 Mv 2 c.m. . 5
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So we see that the radius of the wheel does not effect the rotational kinetic energy. If we were to calculate the rotational kinetic energy of wheel 2, we would get the same expression KE 2 ,rot 1 4 Mv 2 c.m. . The total kinetic energy for each wheel is the sum of rotational and trans- lational kinetic energies. Because each wheel has the same center of mass velocity and mass, each wheel has a total kinetic energy KE wheel 1 4 Mv 2 c.m. 1 2 Mv 2 c.m. 3 4 Mv 2 c.m. . The rider and bike have only translational kinetic energy, since they are of mass M and move with velocity v c.m. , we can instantly write down the kinetic energy for the rider/bike combination as KE r 1 2 Mv 2 c.m. . So we see from comparing our terms that choice a is correct. 6
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4. The figure shows a bar that is free to rotate about an axis through point P perpendicular to the plane of the figure (that is, through the page). P ~ F 2 ~ F 1 ~ F 3 Three forces, ~ F 1 , ~ F 2 and ~ F 3 , as indicated are applied a distance of 3.0 cm from the axis of rotation producing torques τ 1 , τ 2 and τ 3 , respectively. The forces have magnitude ~ F 1 ~ F 2 1 . 0 N and ~ F 3 2 . 0 N. The angle between ~ F 1 and ~ F 2 is 30 . Rank order the torques according to their magnitudes, smallest to largest. (a) τ 3 τ 2 τ 1 . (b) τ 1 τ 2 τ 3 . (c) τ 3 τ 1 τ 2 . (d) τ 3 τ 1 τ 2 . (e) τ 1 τ 2 τ 3 . We know right away that ~ τ 3 0 because the force is in the same direction as the vector from the pivot point to where the force is being applied. Now we know that ~ τ ~ r ~ F sin θ . The forces ~ F 1 and ~ F 2 are applied at the same point and have the same magnitude. Thus the only difference between the torques they cause comes from the angle θ . ~ F 1 is applied at 90 degrees with respect to ~ r , while ~ F 2 is applied at 60 degrees. Because sin 90 sin 60 0, we can then say that ~ τ 1 ~ τ 2 ~ τ 3 , which is choice a. 7
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