ps10sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 2010 Problem Set 10: Two Dimensional Rotations Solutions Problem 1: A spaceship is sent to investigate a planet of mass m p and radius r p . While hanging motionless in space at a distance 5 p r from the center of the planet, the ship fires an instrument package with speed v 0 . The package has mass i m which is much smaller than the mass of the spacecraft. The package is launched at an angle ! with respect to a radial line between the center of the planet and the spacecraft. For what angle will the package just graze the surface of the planet? Solution: The gravitational force by the planet on the spaceship G pm F ! always points towards the center of the planet. The torque about the center of the planet (point labeled O ) due to the gravitational force is given by the expression , G O O m pm r F = " ! ! ! . (1) The vector , O m r ! points from the center of the planet to the spaceship so , O m r ! and G pm F ! are anti-parallel hence the torque O ! is zero. Therefore the angular momentum of the spaceship about the center of the planet is constant. In the figure below the several positions of the spaceship with the associated velocities are shown. (Think of this as an angular momentum diagram.) In the figure denote 0 1,0 v v ! Then the initial angular momentum of the spaceship about the center of the planet is
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2 , , 1 1,0 1 0 1 0 1 0 ˆ ˆ ˆ ˆ 5 ( cos sin ) 5 sin O i O i L r m v Rr m v r m v Rm v k ! !! = " = " # + = ! ! ! . (2) The angular momentum about the center of the planet when the spaceship just grazes the planet is , , 1 1, 1 1, ˆ O f O f f f L r m v Rm v k = ! = ! ! ! . (3) Because angular momentum about the center of the planet is constant, , , O i O f L L = ! ! . (4) Substituting Eqs. (2) and (3) into Eq. (4) and taking the z-component on both sides yields 5 Rm 1 v 0 sin = Rm 1 v 1, f . (5) Thus we can solve for the final speed v 1, f = 5 v 0 sin . (6) There is no non-conservative work done on the spacecraft so the mechanical energy is constant, E i = E f . (7) Choose infinity as the zero point for potential energy then the energy equation becomes 1 2 m 1 v 0 2 ! Gm 1 m p 5 R = 1 2 m 1 v 1, f 2 ! Gm 1 m p R . (8) Substitute in Eq. (6) for the final speed yielding 1 2 m 1 v 0 2 ! Gm 1 m p 5 R = 1 2 m 1 (5 v 0 sin " ) 2 ! Gm 1 m p R . (9) Collecting terms we can rewrite Eq. (9) as 4 Gm 1 m p 5 R = 1 2 m 1 v 0 2 (25sin 2 " 1) (10) We can now solve for the angle :
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3 ! = sin " 1 1 5 8 Gm p 5 Rv 0 2 + 1 # $ % % & ' ( ( . (11)
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4 Problem 2 A drum of mass A m and radius a rotates freely with initial angular velocity ,0 A ! . A second drum with mass B m and radius b a > is mounted on the same axle and is at rest, although it is free to rotate. A thin layer of sand with mass S m is distributed on the inner surface of the smaller drum. At 0 t = , small perforations in the inner drum are opened. The sand starts to fly out at a constant rate kg " s -1 and sticks to the outer drum.
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ps10sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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