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The two disks are brought slowly together and the

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Unformatted text preview: 1 = = 4.5rev/s. I2 1.45 (a) The angular momenta ωo ωo ωo 2r are r 2r ωo r L1 = I1ω o and L2 = I2ω o, but they are in opposite directions. Since there are no external torques angular momentum is conserved. Question 4: Two disks, with identical masses but The initial angular momentum is different radii (r and 2r) are spinning on frictionless L1 + ( −L2 ) = I1ω o − I2ω o = (I1 − I2 )ω o, and the final angular momentum is bearings with the same angular speed ω o but in opposite (I1 + I2 )ω . directions. The two disks are brought slowly together and the resulting frictional force between their surfaces eventually brings them to a common angular velocity. (a) What is the magnitude of the final angular speed in terms of ω o ? (b) What is the change in the rotational kinetic energy? 1 1 But I1 = M(2 r )2 = 2Mr 2 and I2 = Mr 2. 2 2 1 1 ∴ 2 − Mr 2ω o = 2 + Mr 2ω , 2 2 i.e., 32 5 Mr ω o = Mr 2ω 2 2 3 ∴ ω = ω o. 5 The total initial rotational kinetic energy: 1 1 1 Ki = I1ω o2 + I2ω o2 = (I1 + I2 )ω o2 2 2 2 The total final rotational kinetic energy: 1 K f = (I1 + I2 )ω 2. 2 ∴ ∆K K f − Ki ω 2 − ω o2 = = , K Ki ω o2 3 but ω = ω o, so (ω 2 − ω o2 ) < 0, i.e., ∆K is always < 0. 5 ∆K ∴ = −0.64, K i.e., 64% of the kinetic energy is “lost”. • What’s happened to it? Note: this is the analog of the inelastic collision we studied in chapter 8. Question 5: (a) Assuming the Earth to be a homogeneous sphere of radius R and mass M, show that the period of rotation about its axis is 4 πM 2 T= R, L where L is the angular momentum of the Earth about its axis. (b) Suppose the radius changes by a small amount ∆R . Show that the fractional change in period is ∆T ∆R ≈2 . T R (c) By how many kilometers would the Earth have to expand for the period to change by 0.25 days/year so that leap years would no longer be necessary? (a) If the angular velocity is ω we have • angular momentum L = Iω , 2 • moment of inertia I = MR 2. 5 2π The period of rotation is T = ω ω = 2 π 2 πI 4 π M 2 R. = = L L 5L I QED. DISCUSSION PROBLEM [10.2]: A student sits on a stool that is free to rotate. She is handed a wheel that is spinning in the horizontal plane as shown below. (b) Let R ⇒ R + ∆R so T ⇒ T + ∆T , then 4 πM ∆T dT T ≈ = 2 R = 2 2 R . 5L R ∆R dR ∴ ∆T ∆R ≈2 . T R Assuming she (and the stool) are stationary to begin (c) ∆R ≈ R ∆T...
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