A thin rigid massless rod goes through the center of

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5. A thin, rigid, massless rod goes through the center of a 20 kg rigid hollow sphere of 5.0 cm radius. One end of the rod, a distance of 8.0 cm from the center of the sphere is welded to a thin metal bar, perpendicular to the rod. The bar functions as an axis about which the whole array rotates at 7.0 rad/s. axis 8cm What is the kinetic energy of this rotating array? ( Hint: Use the parallel axis theorem). (a) 4.0 J (b) 0.82 J (c) 3.6 J (d) 3.1 J (e) 1.6 J We need to find the rotational kinetic energy of this configuration given the angular velocity. We know that KE rot 1 2 2 , and we are given ω about the metal bar, thus we need only to find I about the metal bar to calculate the rotational kinetic energy. Our strategy will be to find the moment of inertia about the center of mass of the sphere, then use the parallel axis theorem to find the moment of inertia about the metal bar. The moment of inertia about the center of mass of a hollow metal sphere is I c.m. 2 3 MR 2 . Using the parallel axis theorem we have that I bar I c.m. MD 2 I bar 2 3 MR 2 MD 2 I bar 2 3 20 . 05 2 20 . 08 2 I bar 0 . 0161 . 8
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Now inserting this into our equation for the rotational kinetic energy we get KE rot 1 2 2 KE rot 1 2 0 . 0161 7 2 KE rot 4 J. This is choice a. 6. A 300.0 kg solid disk of 10 cm radius is free to rotate about an axis through its center and perpendicular to the disk. A force is applied so that it spins-up from rest to a speed of 33 revolutions per minute. The work done by the force in this process is We learned last chapter that the net work done on an object is the change in kinetic energy of the object. In this problem, the only source of work is the force that is spinning the disc, so the change in kinetic energy of the disc must be the work that the force has done in spinning the disc from rest, where it has no kinetic energy. So the change in kinetic energy is Δ KE KE f KE i 1 2 I c.m. ω 2 f 0 . The moment of inertia for a solid disc is 1 2 MR 2 , and the angular velocity is 33 fracrevmin 3 . 46 rad sec , inserting these into our expression we get W Δ KE 1 2 1 2 MR 2 ω 2 f W 1 4 300 . 1 2 3 . 46 2 W 8 . 98 J. 9
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This is choice a. 7. The linear density of a 2.0 m long, non-uniform thin rod is given by λ x 0 . 3 x 2 , where x is in meters and λ is in kg/m, and the rod extends from x 0 to x 2 . 0 m. How far form the x 0 end is the center of mass? We want to find the center of mass of the nonuniform rod, the formula for the x coordinate of the center of mass of a continuous density is x c.m. 1 M xλ x dx . Where M is mass of the entire rod. To calculate M , we simply integrate λ over the length of the rod M λ x dx M 2 0 . 3 x 2 dx M . 1 x 3 2 0 M . 1 2 3 0 M . 8 Now we can simply compute the x position of the center of mass using the formula above. 10
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x c.m. 1 M L 0 xλ x dx x c.m. 1 . 8 2 0 x. 3 x 2 dx x c.m. 3 8 2 0 x 3 dx x c.m. 3 8 1 4 x 4 2 0 x c.m. 3 32 2 4 0 4 x c.m. 3 32 16 3 2 Which is choice a. Good luck! 11
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