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09prob_11

# 09prob_11 - Physics 9A-C Assignment#9 Cole[9.82]a).E=E 2 2...

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Assigment #9, 5/27/11 Page 1 of 4 UC Davis Physics 9A-C Assignment #9 Cole [9.82] a) No nonconservative forces do work so conserve energy. = E õ mgh = 1 2 mv 2 + 1 2 Â ω 2 where ∑  = v/r. Solve for Â: Â = 2 gh v 2 v 2 mr 2 = 2 9.8 m/s 2 ( ) 2 m ( ) 5 m/s ( ) 2 5 m/s ( ) 2 8 kg ( ) 0.37 m ( ) 2 = 0.622 kg·m 2 . b) Moving mass toward the axis of rotation would decrease the moment of inertia. The best we can do is move approximately all of the mass to the radius. The mass of the wheel is M = w g = 280 N 9.8 m/s 2 = 28.6 kg. Doing this gives Â = Mr 2 = (28.6 kg)(0.37 m) 2 = 3.92 kg·m 2 . Anything larger is impossible. [9.83] a) Gravity acts at the center of mass. The work done by gravity is - Î U cm . The potential energy of the meter stick will be least when it is vertical, at which point, the center of mass will have lowered by 0.5 m õ Î U = 0 - mgh =- (0.16 kg)(9.8 m/s 2 )(0.5 m) = -0.784 J b) The change in potential energy gets converted into kinetic energy Î K +ÎU = 0 õ 1 2 Â p 2 = - ÎU õ ω = 2 Δ U Â p = 2 mgh ( ) 1 3 mL 2 = 5.42 rad/s where we have used the moment of inertia about the pivot. c) v = r ∑   = L ∑   = (1 m)(5.42/s) = 5.42 m/s d) v = 2 gh = 2 9.8 m/s 2 ( ) 1 m ( ) = 4.43 m/s. õ v ( c ) = 3/2 v ( d ) , that is, the average linear acceleration of the end of the stick is larger than g ! [9.85] The energies we must account for are the work done by friction, KE of the 2 masses, potential energy of mass B , and the rotational energy of the pulley. Take the zero point for potential energy to be the starting height for block (B). As (B) falls, its potential energy will decrease. Recall, work done by friction changes the energy of the system: W f = E - õ k m A gd = ( 1 2 m A v 2 + 1 2 m B v 2 2 -m B gd) - 0 We need one more relation connecting the angular speed of the pulley. If the string does not slip then v = R . õ m B gd μ m A gd = 1 2 m A + m B + Â R 2 v 2 . Solving for v : v = 2 m B gd μ m A gd ( ) m A + m B + Â R 2 . [9.89] a) The total moment of inertia is the sum Â = Â 1 + Â 2 = 1 2 M 1 R 1 2 + 1 2 M 2 R 2 2 = 1 2 0.8 kg ( ) 2.5 × 10 2 m ( ) 2 + 1 2 1.6 kg ( ) 5.0 × 10 2 m ( ) 2 = 2.25 ª 10 -3 kg·m 2 .

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