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7AF08HW10soln

# 7AF08HW10soln - Homework 10 solutions 10.59 a m 1 x cm 1 m...

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Unformatted text preview: Homework 10 solutions 10.59 a) m 1 x cm 1 + m 2 x cm 2 m 1 + m 2 = 7 × 0+1 . 4 × . 25 7+1 . 4 = 0 . 042 m = 4 . 2 cm b) I cm = I center of disc- M total x 2 cm = 1 2 M disc R 2 disc + mr 2- ( M disc + m ) x 2 cm = 1 2 7 × ( . 31) 2 + 1 . 4 × ( . 25) 2- (7 + 1 . 4) × ( . 42) 2 = 0 . 41 kg.m 2 10.69 Conservation of energy = ⇒ Mgy cm = 1 2 I end ω 2 = ⇒ Mg L 2 = 1 2 ML 2 3 v 2 L 2 = ⇒ v = √ 3 gL = 8 . 5 m / s 10.79 a) K tot = 1 2 M body v 2 cm +4 × 1 2 M wheels v 2 cm +4 × 1 2 I wheel ω 2 = 1 2 ( M total v 2 cm + 4 × 1 2 M wheel R 2 wheel v 2 cm R 2 wheel ) = 3 . 6 × 10 5 J ( v cm = ωR wheel because of pure rolling) b) K rot + K trans wheels K tot = 4 × 1 2 M wheels v 2 cm +4 × 1 2 I wheels ω 2 K tot = 0 . 20 c) F app- F fric = M total a cm , F fric R wheel = 4 × I wheel α , a cm = αR wheel . Some algebra gives a cm = F app M total +4 I wheel /R 2 wheel = 1 . 3 m / s 2 d) Ignoring rotational inertia of the wheels, a cm = F app /M total ....
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