This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 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 Z R drr Z π dθr sin θ + 2 Z R drr Z 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 n x CM = Rθ- | ¯ y CM | sin θ = Rθ- (4 R/ 9 π ) sin θ y CM = R- | ¯ y CM | cos θ = R- (4 R/ 9 π ) cos θ. ⇓ n ˙ 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. Thusis the rotational kinetic energy about the center of mass....
View Full Document