HW8-phy2048-Spring-2014-solution

# 150 ms 315 j 2 r 2 which implies that the work

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Unformatted text preview: bout the center of mass, m = 140 kg, and v = 0.150 m/s is the speed of its center of mass. Equation 11- 2 relates the angular speed to the speed of the center of mass: ω = v/R. Thus, ⎛ v2 ⎞ 1 1 2 K = mR 2 ⎜ 2 ⎟ + mv 2 = mv 2 = (140 kg ) ( 0.150 m/s ) = 3.15 J 2 ⎝R ⎠ 2 which implies that the work required is W = Δ K = 0 − 3.15 J = − 3.15 J . 1 PHY2048 Spring 2014 HW8 11P7. (a) We find its angular speed as it leaves the roof using conservation of energy. Its initial kinetic energy is Ki = 0 and its initial potential energy is Ui = Mgh where h = 6.0sin 30° = 3.0 m (we are using the edge of the roof as our reference level for computing U). Its final kinetic energy (as it leaves the roof) is (Eq. 11- 5) 1 1 K f = 2 Mv 2 +...
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