HW4 - PHYS125 Homework Assignment #4 Grading Rubric Stage 1...

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PHYS125 Homework Assignment #4 Grading Rubric Stage 1 2 points each Problem 7.54, 4, 8.24, 8.35, 8 Stage2: Penny on a Turntable ~3 points for Explanation ~3 point sfor Free Body Diagram ~2 points for Angular Velocity ~2 points for correct solution Stage 3: Penguins ~4 points FBD on penguins (1 point per penguin) ~2 points acceleration constant ~3 points correct algebra ~1 point correct solution
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Stage 0 Solutions: 7.1. Solve: (a) From t = 0 s to t = 1 s the particle rotates clockwise from the angular position + 4 π rad to 2 rad. Therefore, () 246 θ ππ Δ= − −+ = rad in one sec, or 6 r a ds ωπ =− . From t = 1 s to t = 2 s, ω = 0 rad/s. From t = 2 s to t = 4 s the particle rotates counterclockwise from the angular position 2 rad to 0 rad. Thus and 022 r a θπ Δ=−− = d rad s =+ . (b) 7.5. Model: The earth is a particle orbiting around the sun. Solve: (a) The magnitude of the earth’s velocity is displacement divided by time: ( ) 11 21 . 51 0 m 2 24 hr 3600 s 365 days 1 day 1 hr r v T × == ×× = 3.0 × 10 4 m/s (b) Since v = r , the angular acceleration is 4 7 11 3.0 10 m/s 2.0 10 rad/s 1.5 10 m v r × = × × (c) The centripetal acceleration is ( ) 2 4 2 32 11 6.0 10 m/s r v a r × × Assess: A tangential velocity of 3.0 × 10 4 m/s or 30 km/s is large, but needed for the earth to go through a displacement of 2 (1.5 × 10 11 m) 9.4 × 10 8 km in 1 year. 7.9. Model: The rider is assumed to be a particle. Solve: 2 Since / , we have r av r = ( )( ) 22 98 m/s 12 m 34.3 m/s r va r v = Assess: 34.3 m/s 70 mph is a large yet understandable speed.
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7.13. Model: The vehicle is to be treated as a particle in uniform circular motion. Visualize: On a banked road, the normal force on a vehicle has a horizontal component that provides the necessary centripetal acceleration. The vertical component of the normal force balances the weight. Solve: From the physical representation of the forces in the r-z plane, Newton’s second law can be written 2 sin r mv Fn r θ == cos 0 cos z F nm gn θθ =− = ⇒= m g Dividing the two equations and making the conversion 90 km hr 25 m/s = yields: () 2 2 2 25 m/s tan 0.128 7.27 9.8 m/s 500 m v rg = ⇒= ° Assess: Such a banking angle for a speed of approximately 55 mph is clearly reasonable and within our experience as well. 7.17. Model: The satellite is considered to be a particle in uniform circular motion around the moon. Visualize: Solve: The radius of moon is 1.738 × 10 6 m and the satellite’s distance from the center of the moon is the same quantity. The angular velocity of the satellite is 4 22 r a d 1 m i n 9.52 10 rad/s 110 min 60 s T ππ ω × = × and the centripetal acceleration is ( )( ) 2 26 4 1.738 10 m 9.52 10 rad/s 1.58 m/s r ar × × = 2 The acceleration of a body in orbit is the local “g” experienced by that body.
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7.19. Model: Model the passenger in a roller coaster car as a particle in uniform circular motion. Visualize: Note that the normal force of the seat pushing on the passenger is the passenger’s apparent weight. n r Solve: Since the passengers feel 50% heavier than their true weight, 1.50 nw = . Thus, from Newton’s second law: 22 1.50 0.50 r mv mv Fnw ww m g rr =− = − = = () ( ) ( ) 2 0.50 0.50 30 m 9.8 m/s 12.1 m/s vg r ⇒= = = 7.21.
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This note was uploaded on 06/20/2008 for the course PHYS 125 taught by Professor Mutchler during the Spring '08 term at Rice.

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HW4 - PHYS125 Homework Assignment #4 Grading Rubric Stage 1...

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