Answer 1: a) Assuming there is now friction and thus no external forces or torques, angular and linear momentum are conserved; total energy is conserved; kinetic energy may or may not be conserved
Answer 2: a) If the rotor shaft were connected directly to the generator, it would have to spin at 1200 rpm. At this rotational speed the tips of the blades would be moving at a speed of v= wr, where w is the angular velocity and r is the length of the blade. For this example, w is 126 rad/s and r is 23 m. So the speed of the blade tips is 2900 m/s or 6500 mph! This is not reasonable.b) Where the question mark is on the diagram above, place a gear box that connects the rotor shaft to the generator shaft. For example, a typical gear ratio is 50:1, which means that the generator rotates at 1200 rpm, but the turbine only rotates at 24 rpm. This gives a more reasonable velocity of 58 m/s.
Answer 3: a)Ravi’s full can will get to the ground first
Answer 4: a) The moment of inertia of a rod rotated around one end is , where M is the mass of the rod, and L is its length. The mass of each rod is M = r V = 4.4 kg, and there are three of them, so I = ML 2= 28 kg-m 2.
b) The angular velocity w corresponding to 25 rpm is 2.6 rad/s. Angular momentum L = I w = 73 kg-m 2/s.
c) We want to reduce the angular momentum to zero in 30 s, so:
Answer 5: a) The high speed shaft is moving at 1200 rpm, which corresponds to an angular velocity of 126 rad/s.
b) The angular velocity of the low speed shaft is 1/50 of that of the high speed shaft, or 2.5 rad/s. The tips of the blades will be moving at v = w r = 55 m/s.
c) 1500 rpm corresponds to 160 rad/s. The angular velocity of the turbine shaft is smaller by a factor of 50, or 3.2 rad/s. The tips of the blades will be moving at v = w r = 69 m/s.
Answer 6: a) Change in P.E.= Change in K.E.
b) w= v/R
c) Work = Change in K.E
Answer 7: a) The moment of inertia for tuck position is smaller then that of pike position. To get the torque needed to make the 2 ½ rotations, Jenny needs to lean more.
b) Tuck w = 2 rev./0.65s = 3 rev/s = 19.3 rad/s
c) L = I t w t = I s w s
d) ∆L/∆t = τ
e) τ = rFsinθ
Posted on 8/1/05
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