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s11f1029

# s11f1029 - Solution for 29 Dening as the mass density we...

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Solution for # 29 Defining ρ as the mass density, we use polar coordinates x = r cos θ , y = r sin θ . Given a coordinate system ( x , y ) which rotates with the disk (see the figure), the location of the CM is ¯ x CM = 0 and ¯ y CM = ρ M R 0 drr π 0 dθr sin θ + 2 R 0 drr 2 π π dθr sin θ = ρ M R 3 3 2 - R 3 3 4 = - 2 3 ρR 3 M = - 4 R 9 π . Where in the last line we have used M = ρ πR 2 2 + 2 ρ πR 2 2 = 3 2 ρπR 2 . Again referencing the figure, we see that the relationships between the lab coordinates and the coordinates of the center of mass are x CM = - | ¯ y CM | sin θ = - (4 R/ 9 π ) sin θ y CM = R - | ¯ y CM | cos θ = R - (4 R/ 9 π ) cos θ. ˙ x CM = R ˙ θ - (4 R/ 9 π ) cos θ ˙ θ ˙ y CM = (4 R/ 9 π ) sin θ ˙ θ. The kinetic energy of the disk is made of 2 terms: T = T trans + T rot , where T trans is the translational kinetic energy of the mass M with coordinates ( x CM , y CM ) and velocity ( ˙ x CM , ˙ y CM ), and T rot is the rotational kinetic energy about the center of mass. Thus T = 1 2 M ( ˙ x 2 CM + ˙ y 2 CM ) + 1 2 I 3 ˙ θ 2 . Here I 3 is the moment of inertia about the CM, with respect to the z-axis, perpendicular to the plane of the figure. To find I 3 we will calculate it with respect to the center of disk (since it is easier!) and use the parallel axis theorem, (32.12) of Landau and Lifshitz, to obtain it with respect

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