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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 π rad. Therefore, θ π π π = − − + ( ) = − 2 4 6 rad in one sec, or ω π = − 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 π rad to 0 rad. Thus θ π π = − − ( ) = 0 2 2 rad and ω π = + rad s . (b)
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7.2. Solve: Since ω θ = ( ) d dt we have θ f = θ 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 ω -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.
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7.3. Model: Treat the record on a turntable as a particle rotating at 45 rpm. Solve: (a) The angular velocity is ω π π = × × = 45 1 60 2 rpm s rad 1 rev 1.5 rad / s min (b) The period is T = = = 2 rad 2 rad 1.5 rad / s 1.33 s π ω π π
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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 θ θ π 2 1 180 71 6 = = = = × ° = ° s r 5000 miles 4000 miles 1.25 rad 1.25 rad rad . (b) The plane’s angular velocity is ω θ θ = = = 2 1 2 1 0 139 t t 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.
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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 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 ω = = × × = × v r 3 0 10 1 5 10 2 0 10 4 11 7 . . . m / s m rad / s (c) The centripetal acceleration is a v r r = = × ( ) × = × 2 4 2 11 3 3 0 10 1 5 10 6 0 10 . . . m / s m m / s 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 π (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.
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7.7. Solve: The pebble’s angular velocity ω π = ( )( ) = 3 0 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 v r = = ( ) ( ) = 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 1 m / 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 v v R T R T top bottom E E m 300 m 24 hrs m 24 3600 s m / s = + ( ) = ( ) = ( ) = × 2 300 2 2 600 2 18 10 2 π π π π .
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7.9. Model: The rider is assumed to be a particle.
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