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D 1 cos 2 also mechanical energy is conserved ie 096m

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Unformatted text preview: T 0.25 day 0.25 day = = and we need . 2T T 1 year 365 day ∴ ∆R = 6.37 × 103 × 0.25 = 2.18 km. 2 × 365 with, what happens if she inverts the wheel (through 180o) ? We have a perfectly inelastic collision and no external torques. So, angular momentum is conserved at the collision. Question 6: A uniform rod of length 1.20m and mass 0.80kg is pivoted at its top. The rod, initially at rest, is struck by a lump of putty of mass 0.30kg at a point x cos θ d θ x x x(1 − cos θ) 0.96m below the pivot point. If the putty sticks to the d rod and the maximum deflection of the rod is 60o from θ 2 + d× cm the vertical, what was the speed of the putty at impact? d (1 − cos θ) 2 Also, mechanical energy is conserved, i.e., 0.96m 1.20m 0.30kg v M = 0.80kg Ki + Ui = K f + Uf . But K f = Ui = 0, so Ki = Uf . The rotational kinetic 1 energy of the rod and putty at impact is Ki = Iω 2, 2 where I is the moment of inertia of the rod plus putty. An increase in potential energy occurs for both the rod and the putty. The increase in potential energy at the top of the swing is d Uf = Mg (1 − cos θ) + mgx (1 − cos θ) , 2 where M is the mass of the rod and m is the mass of putty. Conservation of mechanical energy gives 12 d Iω = g M + mx (1 − cos θ). 2 2 Also, at impact angular momentum is conserved, i.e., the angular momentum of the putty before the impact equals the angular momentum of the rod and putty after the impact. mvx 0.288 v ∴ mvx = Iω ⇒ ω = = , I I 1 where I = Md 2 + mx 2 3 0.80 × (1.20)2 = + 0.30 × (0.96)2 = 0.660 kg ⋅ m2. 3 ∴ω = 0.288 v = 0.436 v , 0.660 which gives ω in terms of v, the velocity of the putty. Hence, substituting for ω gives 1 × 0.660 × (0.436 v )2 2 0.80 × 1.20 = 9.81 + 0.30 × 0.96 (1 − cos 60o ). 2 ∴ 0.0627 v 2 = 9.81(0.480 + 0.288) × Hence v = 1 = 3.767. 2 3.767 = 7.75m/s. 0.0627 Physics of the Gyroscope The Gyroscope The axle of a wheel is supported by a pivot at O and held A spinning wheel exhibits a remarkable property: horizontal by a hand, say. The weight of the wheel r produces a torque τ in the horizontal plane, so, if the end r τ r τ d rr ∆L = τ∆t r L ω O r F ω r mg r L ∆φ gyro.mov To download a copy of this movie go to the “Useful notes ...” link on the homepage and look under Chapter 10. r ∆L r L of the axle is released the wheel falls. If the wheel is given a very large angular velocity ω before the end is released, it does not fall; rather, the axle rotates slowly in the...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.

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