HW10_solns

HW10_solns - Solutions HW 10 S07 P112 1 See textbook Figure...

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Solutions HW 10 S07 P112 1) See textbook Figure 10.41. Let the z axis be along the axis of rotation of the wheel with the z + direction pointing out of the page. N 50 . 7 1 = F . N 30 . 5 2 = F . m 330 . 0 = R . m N 726 . 0 ) m 330 . 0 )( N 50 . 7 ( ) m 330 . 0 )( N 30 . 5 ( 1 2 = = = R F R F z τ . 2) a) a rad = 0, a tan = α r = 2 ( ) 0.300 m ( ) = 2 and so a = 2 . b) θ = π 3 rad, so a rad = ω 2 r = 2 0.600 rad s 2 ( ) 3 rad ( ) 0.300 m ( ) = 2 . The tangential acceleration is still 0.180 m s 2 , and so a = 2 ( ) 2 + 2 ( ) 2 = 0.418 m s 2 . c) For an angle of 120 ° a rad = 2 , and a = 2 , since a tan is still 0.180 m/s 2 . 3) The moment of inertia can be calculated using the Parallel Axis Theorem: I = I COM + Md 2 , where I COM is the moment of inertia of the plate rotated about its center of mass, and d is the distance from the COM to the new turning point. d 2 is given by 2(R/2) 2 , and so I = 1 6 MR 2 + 1 2 MR 2 = 2 3 MR 2 . 4) The magnitude of vector C is |C|=|A||B| |sin(140)|=2.57. The direction is in the positive z direction. 6) The angular momentum of a rotating body is given by L=I ω , where I is the moment of inertia of a car tire (approximately a solid cylinder of mass 7kg and radius r= 0.3m, so

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I = 0.5 (7 kg)(0.32 m) 2 = 0.3 kg m 2 ). ω is the angular velocity on the highway, and in the case of rolling without slipping: ω r =v. Let’s estimate the linear velocity to about 80 km/h = 22m/s. The angular momentum is then approximately L = (0.3 kg m 2 )(22 m/s)/(0.3 m) = 22 kg m 2 /s.
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HW10_solns - Solutions HW 10 S07 P112 1 See textbook Figure...

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