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Unformatted text preview: Physics 200 Problem Set 6 Solution 1. (i) What is the moment of inertia I CM of a propeller with three blades (treated as rods) of mass m , length L , at 120 ◦ relative to each other? (ii) If a torque τ acts on this how long will it take to reach an angular velocity ω ? (iii) How many revolutions will it have made before reaching this ω ? (iv) Get the numerical answers if L = 1 . 25 m, m = 12 kg, τ = 3000 N · m, ω = 2000 rad / s. Answer: (i) We know (if not, look on p. 296) that the moment of inertia of a single rod rotating around its end is 1 3 mL 2 . It’s not hard to convince oneself that if there are three of them rotating around the same axis and in the same plane, the moment of inertia is just three times this, I CM = mL 2 . (ii) Since ω = αt and τ = I CM α , t = I CM ω τ = mL 2 ω τ . (iii) From our knowledge of constant acceleration problems, ω 2 = 2 αθ = ⇒ θ = ω 2 2 α = I CM ω 2 2 τ = mL 2 ω 2 2 τ . The number of revolutions it made is N = θ 2 π = mL 2 ω 2 4 πτ . (iv) I CM = (12 kg)(1 . 25 m) 2 = 19 kg · m 2 , t = (12 kg)(1 . 25 m) 2 (2000 s 1 ) 2(3000 N · m) = 13 s , N = (12 kg)(1 . 25 m) 2 (2000 s 1 ) 2 4 π (3000 N · m) = 2000 . 2. Consider I for a rectangle of sides a (along the yaxis) and b (along the xaxis) about the two symmetry axes. (Rotate the rectangle about one of these axes and think of it as composed of rods.) Show that about the axis parallel to x , I = 1 12 Ma 2 . Going back to the very definition of I , show that if this rectangle is spun around an axis through its CM and perpendicular to its area the moment of inertia will be I = 1 12 M ( a 2 + b 2 ). Answer: To find I x , the moment of inertia about the symmetry axis parallel to the xaxis, we think of dividing the rectangle into many thin rods: 1 x y a b Since each “rod” has length a , it should be obvious that the sum of these contributions is simply 1 12 Ma 2 , i.e., the same as if there was one rod of mass M rotating around the axis, but let’s be more explicit. If we divide up the rectangle into n rods (where n is large so they really are rods, though in the end it doesn’t matter) so that each rod has mass M/n , then with each rod contributing 1 12 ( M n ) a 2 and there being n rods, we get I x = 1 12 Ma 2 as expected. The same logic of course applies to the symmetry axis parallel to the yaxis. From the definition of the moment of inertia, I = X i r 2 i Δ m i . Notice from the picture that r 2 i = ( x i b/ 2) 2 + ( y i x/ 2) 2 , so we can write x y a b (b/2,a/2) (x i ,y i ) r I = X i x i b 2 ¶ 2 Δ m i + X i ‡ y i a 2 · 2 Δ m i = I y + I x = I x + I y = 1 12 M ( a 2 + b 2 ) . It’s easy to see that this kind of argument will work for any flat object, as long as you choose your axes to be perpendicular to each other....
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 Fall '08
 RAMAMURTISHANKAR
 Physics, Force, Inertia, Mass, Sin, kg, Ax Bz

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