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**Unformatted text preview: **Physics 101 Classical Physics Spring 2010 Homework 8 Solutions DUE IN CLASS ON April 1 1. The world’s power consumption is about 13 . 5 × 10 12 W (power is energy used per unit time, so that 1 Watt=1 joule/second). If some clever pre-med figured out how to take energy from the Earth’s rotation and use it to power everything in the world for one year, how many seconds longer would one Earth day take? (note: you should not round off when calculating this number—use all the precision your calculator has and pick the number below that is closest to your answer) (a) 50 s, (b) 50 μ s, (c) 12 s, (d) 70 μ s , (e) 84600 s Treating the earth as a solid sphere of mass m and radius r , the energy in the earth’s rotation is E = 1 2 Iω 2 = 1 5 mr 2 ω 2 = 1 2 2 5 mr 2 ω 2 = (0 . 2)(5 . 9742 × 10 2 4 kg)(6 . 3781 × 10 6 m) 2 2 π 24 · 3600 s 2 = 2 . 57 × 10 29 J If we want to take an amount of energy Δ E from this to power the earth for a year, then using the earth’s power consumption this would be Δ E = Pt = (13 . 5 × 10 1 2 W)(3600 · 24 · 365 s) = 4 . 26 × 10 20 J If E and ω are the energy in the earth’s rotation and angular velocity after this energy is extracted, then we have E = 1 2 Iω 2 ⇒ E ω 2 = I 2 E = 1 2 Iω 2 ⇒ E ω 2 = I 2 The right hand side of both expressions are the same since the moment of inertia of the earth does not change, so E ω 2 = E ω 2 ⇒ 1 ω = 1 ω r E E If we let T = 24 hr be the time it takes the earth to complete 1 revolution, then T = T +Δ T is the time it will take the earth to complete a rotation after the energy is extracted for some small Δ T . Then, we have ω = 2 π/T and ω = 2 π/ ( T + Δ T ). Now, if we let E = E- Δ E for some small Δ E which was calculated above, the above expression becomes T + Δ T = T r E E- Δ E = T 1- Δ E E- 1 / 2 1 Solving for Δ T , Δ T = T 1- Δ E E- 1 / 2- 1 ! You could try plugging in the numbers above at this point, but your calculator may not have the precision to handle this. To simplify this further, since Δ E/E is small, we can keep only the first term in the Taylor expansion of (1- Δ E/E )- 1 / 2 . Δ T ≈ T 1 + 1 2 Δ E E- 1 = 1 2 Δ E E T Plugging in the number of seconds in a day for T and the values calculated above for Δ E and E , we find Δ T = 72 μ s 2. A mine elevator (a ‘cage’) is supported by a steel cable 2 cm in diameter. The mass of the elevator and its contents are 800 kg. By how much is the cable stretched when the elevator is 250 m below sea level (in other words, when the cable is 250 m long)? Young’s moduls for steel is 210,000 MN/m 2 . (a) 0.97 cm, (b) 5.94 cm, (c) 2.97 cm (d) 0 cm, (e) 250 m Young’s modulus is the proportionality constant for relating the tensile stress to the tensile strain, so F A = Y Δ L L where F is the force applied, A is the cross-sectional area, L is the original length of the cable, and Δ L is the amount the cable stretches. Solving for Δ L , and plugging in the numbers Δ L = LF Y A =...

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