HW9_solutions

HW9_solutions - Physics 112 - HW #9 Solutions Spring 2006...

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- 1 - Physics 112 - HW #9 Solutions Spring 2006 8-65 [Railroad Handcar] (a) Since the mass is thrown sideways out of the car in the car's reference frame, it's momentum component along the car's direction of motion doesn't change. So the car's momentum and speed don't change. v car f = 5.00 m/s East (b) The mass's final velocity with respect to the track is 0. Use Cons. of Momentum: Mv car o = m(0) + (M - m)v car f v car f = Mv car o M - m = (200 kg)(5.00 m/s) 200 kg - 25 kg = 5.71 m/s East (c) Cons. of Momentum again: Mv car o - mv throw = (M + m)v car f v car f = Mv car o - mv throw M + m = (200 kg)(5.00 m/s) - (25 kg)(6.00 m/s) 200 kg + 25 kg = 3.78 m/s East 9-52 The center of mass (C.M.) for a circular hoop is located by symmetry at the center of the hoop. The moment of inertia of the hoop about an axis perpendicular to the plane of the hoop and passing through the center of mass, is I C.M. = MR 2 , where M and R are the mass and radius of the hoop. We may calculate the moment of inertia about another axis parallel to the first but passing through a point on the rim, a distance D = R, from the center of mass of the hoop, through the “Parallel Axis Theorem”, which states I = MD 2 + I C.M. = MD 2 + MR 2 = 2 MR 2 , since D = R for a point on the rim. 9-47 [Flywheel Energy Storage] The radial acceleration at the rim of the flywheel is related to the angular frequency ω by a=e 2 R, so the maximum angular frequency is ω = a/R . The rotational energy is then E= 1 2 I ω 2 = 1 2 M R 2 2 a R = 1 4 M a R= 1 4 (70.0 kg)(3500m/s 2 )(1.20 m) E=73500J 9-89 [Disks, Block, & String] (a) I = 1 2 M 1 R 1 2 + 1 2 M 2 R 2 2 = 1 2 (0.80 kg)(0.0250 m) 2 + 1 2 (1.60 kg)(0.0500 m) 2 = 2.25 × 10 -3 kg-m 2 (b) Cons. of Energy: K o + U go = K f + U gf 0 + mgH = 1 2 mv 2 + 1 2 I ω 2 + 0 ω = v R 1 mgH = 1 2 mv 2 + 1 2 Iv 2 R 1 2 [m = mass of block = 1.50 kg, H = 2.00 m]
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- 2 - v = 2gH 1 + I R 1 2 = 2(9.8 m/s 2 )(2.00 m) 1 + (2.25 × 10 -3 kg-m) (0.0250 m) 2 = 3.40 m/s (c) Same as (b), but with R 2 instead of R 1 v = 4.95 m/s v is greater here than in (b) because in (b) a greater fraction of the system's total KE is rotational KE. 9-41 [Wagon Wheel] (a) I = I rim + 8 I spoke = m rim R 2 + 8 1 3 m spoke R 2 = m rim + 8 3 m spoke R 2 = 1.40 kg + 8 3 (0.280 kg) (0.300 m) 2 = 0.193 kg-m 2 [CHECK: If written as I = CMR 2 , where M = the total mass, what is the value of C? Total mass is M = 1.40 kg + 8(0.280 kg) = 3.64 kg C = I MR 2 = 0.193 kg-m 2 (3.64 kg)(0.300 m) 2 = 0.59 This make sense since it lies between 1/3 = 0.33 for the spokes and 1 for the rim.] (b) Angular velocity is ω = v R = 2.0 m/s 0.300 m = 6.667 m/s Translational KE is K tra = ½ m v 2 = ½ (3.64 kg) (2.0 m/s) 2 = 7.28 J Rotational KE is K rot = ½ I ω 2 = 1/2 ( 0.193 kg-m 2 ) (6.667 m/s) 2 = 4.29 J Total KE is K total = K tra +K rot = 7.28 J + 4.29 J = 11.57 J K tra K total
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.

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HW9_solutions - Physics 112 - HW #9 Solutions Spring 2006...

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