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hw8_sol_s11

# hw8_sol_s11 - Phys 2101 Homework 8 Solution Spring `11...

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Phys 2101 Homework 8 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. We note that the component of v perpendicular to r has magnitude v sin where = 30°. A similar observation applies to F . (a) Eq. 11-20 leads to   2 3.0 m 2.0 kg 4.0 m/s sin30 12 kg m s. rmv   (b) Using the right-hand rule for vector products, we find r p points out of the page, or along the + z axis, perpendicular to the plane of the figure. (c) Similarly, Eq. 10-38 leads to  2 sin 3.0 m 2.0 N sin 30 3.0N m. rF  (d) Using the right-hand rule for vector products, we find r F is also out of the page, or along the + z axis, perpendicular to the plane of the figure. 2. (a) The acceleration vector is obtained by dividing the force vector by the (scalar) mass: a = F / m = (3.00 m/s 2 )i ^ (4.00 m/s 2 )j ^ + (2.00 m/s 2 )k ^ . (b) Use of Eq. 11-18 leads directly to L = (42.0 kg . m 2 /s)i ^ + (24.0 kg . m 2 /s)j ^ + (60.0 kg . m 2 /s)k ^ . (c) Similarly, the torque is r F = ( 8.00 N m )i ^ (26.0 N m )j ^ (40.0 N m )k ^ . (d) We note (using the Pythagorean theorem) that the magnitude of the velocity vector is 7.35 m/s and that of the force is 10.8 N. The dot product of these two vectors is v . F = 48 (in SI units) . Thus, Eq. 3-20 yields = cos 1 [ 48.0/(7.35 10.8)] = 127 . 3. For the 3.1 kg particle, Eq. 11-21 yields

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  2 1 1 1 2.8 m 3.1kg 3.6 m/s 31.2 kg m s. r mv Using the right-hand rule for vector products, we find this ) ( 1 1 p r is out of the page, or along the + z axis, perpendicular to the plane of Fig. 11-41. And for the 6.5 kg particle, we find   2 2 2 2 1.5 m 6.5 kg 2.2 m/s 21.4 kg m s. r mv And we use the right-hand rule again, finding that this ) ( 2 2 p r is into the page, or in the z direction.
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