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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
b) 954 J
c) About 0 kgm/s
d) 112 kgm 2/s
e) 6.66 kgm 2
f) 16.9 rad/s
g) 951 J
h) 51.8 kgm 2
i) 2.16 rad/s
j) 121 J
k) Force is towards their center of mass, so when they extend arms they are moving parallel to the force, but in the opposite direction. This constitutes negative work, which means their KE decreases

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.
MgLsinθ =1/2Mv 2 + 1/2I w 2
MgLsinθ = 1/2Mv 2 + ½(1/2)MR 2(v/R) 2
g(2πR)sin = 1/2 v 2 + 1/4v 2
v 2 = 4/3(9.8m/s2)(1.07m)sin15
v = 1.9 m/s

b) w= v/R
= 190cm/s/17cm
=11.2 rad/s
=1.8 rev/s
v top = 2v cm = 3.8 m/s

c) Work = Change in K.E
Fd = 1/2mv2
τθ = 1/2I w 2
τ = Fr perp
F = 1/2I w 2/θr perp
=1/2(1/2)MR 2 w 2/ θ2R, θ = π/4
= 5.4N

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
Straight w = 0.5 rev/0.65s = 4.8 rad/s

c) L = I t w t = I s w s
L = I t19.3 = I s 4.8
Ratio Tuck: Straight = 1:4

d) ∆L/∆t = τ
L f = I w = τ∆t
τ = 1/3Ml 2 w/∆t
= 1/3(50kg)(1.64m^2)(4.8 rad/s)/0.2s
=1076 Nm

e) τ = rFsinθ
F = 1076/(0.8m*sin70)
= 1431 N


Rotation Problem Sets


Posted on 8/1/05

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