Ch 7 Physics for Scientists and Engineers

# Ch 7 Physics for Scientists and Engineers - 7.1. Solve: (a)...

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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, θπ =− − + () 246 rad in one sec, or ωπ 6 rad s. 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 =−− = 0 2 2 rad and =+ rad s. (b)

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7.2. Solve: Since ωθ = () dd t we have θ f = i + area under the ω -versus- t graph between t i and t f From t = 0 s to t = 2 s, the area is 1 2 20 2 20 rad / s s rad ( ) = . From t = 2 s to t = 4 s, the area is 20 2 40 rad / s s rad ( ) = . Thus, the area under the -versus- t graph during the total time interval of 4 s is 60 rad or (60 rad) × (1 rev/2 π rad) = 9.55 revolutions.
7.3. Model: Treat the record on a turntable as a particle rotating at 45 rpm. Solve: (a) The angular velocity is ω π ×= 45 1 60 2 rpm s rad 1 rev 1.5 rad / s min (b) The period is T == = 2 rad 1.5 rad / s 1.33 s

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7.4. Model: The airplane is to be treated as a particle. Visualize: Solve: (a) The angle you turn through is θθ π 21 180 71 6 −== = = × ° s r 5000 miles 4000 miles 1.25 rad 1.25 rad rad . (b) The plane’s angular velocity is ω = == 0 139 tt 1.25 rad 9 hr rad / hr . = × 0 139 3 86 10 5 .. rad hr 1 hr 3600 s rad / s Assess: An angular displacement of approximately one-fifth of a complete rotation is reasonable because the separation between Kampala and Singapore is approximately one-fifth of the earth’s circumference.
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: v r T == × () ×× 2 π 2 1.5 10 m 365 days 24 hr 1 day 3600 s 1 hr 11 = 3.0 × 10 4 m/s (b) Since v = r ω , the angular acceleration is × × v r 30 10 15 10 20 10 4 11 7 . . . ±m/s m rad / s (c) The centripetal acceleration is a v r r × × 2 4 2 11 3 60 10 . . . m 2 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.

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7.6. Solve: Speed, radial velocity, radial acceleration, tangential acceleration, and the magnitude of the net force are constant. Furthermore, the radial velocity and tangential acceleration are zero.
7.7. Solve: The pebble’s angular velocity ωπ = () ( ) = 30 2 . rev / s rad rev 18.85 rad / s. The speed of the pebble as it moves around a circle of radius r = 30 cm = 0.30 m is vr == = ω 18 85 0 30 5 65 .. . rad s m m / s The radial acceleration is a v r r = 2 06 5.65 m / s 0.30 m 1m / s 2 2

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7.8. Solve: Let R E be the radius of the earth at the equator. This means R E + 300 m is the radius to the top of the tower. Letting T be the period of rotation, we have vv R T R T top bottom E E m 300 m 24 hrs m 24 3600 s ±m/s −= + () = 2 300 2 2 600 218 10 2 π .
7.9. Model: The rider is assumed to be a particle. Solve: Since we have avr r = 2 /, va r v r 2 == () ()⇒= 98 m / s 12 m 34.3 m / s 2 Assess: 34.3 m/s 70 mph is a large yet understandable speed.

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7.10. Solve: The plane must fly as fast as the earth’s surface moves, but in the opposite direction. That is, the plane must fly from east to west. The speed is vr == × () × = ω π 2 6 4 10 1680 3 rad 24 hr km km hr 1680 km hr 1 mile 1.609 km 1040 mph .
7.11. Model:

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## This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.

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Ch 7 Physics for Scientists and Engineers - 7.1. Solve: (a)...

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