Ch10Word - Chapter 10 1 CHAPTER 10 - Rotational Motion...

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Chapter 10 1 Page 1 CHAPTER 10 - Rotational Motion About a Fixed Axis 1. ( a ) 30° = (30°)( π rad/180°) = p/6 rad = 0.524 rad ; ( b ) 57° = (57°)(p rad/180°) = 19p/60 = 0.995 rad ; ( c ) 90° = (90°)(p rad/180°) = p/2 = 1.571 rad ; ( d ) 360° = (360°)(p rad/180°) = 2p = 6.283 rad ; ( e ) 420° = (420°)(p rad/180°) = 7p/3 = 7.330 rad . 2. The subtended angle in radians is the size of the object divided by the distance to the object: θ = 2 r Sun / r ; (0.5°)(p rad/180°) = 2 r Sun /(150 × 10 6 km), which gives r Sun ˜ 6.5 × 10 5 km . 3. We find the distance from = h / r ; (7.5°)(p rad/180°) = (300 m)/ r ; which gives r = 2.3 × 10 3 m . 4. From the definition of angular acceleration, we have α = ? ω /? t = [(20,000 rev/min)(2p rad/rev)/(60 s/min) – 0]/(5.0 min)(60 s/min) = 7.0 rad/s 2 . 5. From the definition of angular velocity, we have = ∆ / t , and we use the time for each hand to turn through a complete circle, 2p rad. ( a ) second = ∆ / t = (2p rad)/(60 s) = 0.105 rad/s . ( b ) minute = ∆ / t = (2p rad)/(60 min)(60 s/min) = 1.75 × 10 –3 rad/s . ( c ) hour = ∆ / t = (2p rad)/(12 h)(60 min/h)(60 s/min) = 1.45 × 10 –4 rad/s . ( d ) For each case, the angular velocity is constant, so the angular acceleration is zero . 6. ( a ) The Earth moves one revolution around the Sun in one year, so we have orbit = ∆ / t = (2p rad)/(1 yr)(3.16 × 10 7 s/yr) = 1.99 × 10 –7 rad/s . ( b ) The Earth rotates one revolution in one day, so we have rotation = ∆ / t = (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 × 10 –5 rad/s . 7. All points will have the angular speed of the Earth: = / t = (2p rad)/(1 day)(24 h/day)(3600 s/h) = 7.27 × 10 –5 rad/s. Their linear speed will depend on the distance from the rotation axis. ( a ) On the equator we have v = r Earth = (6.38 × 10 6 m)(7.27 × 10 –5 rad/s) = 464 m/s . ( b ) At a latitude of 66.5° the distance is r Earth cos 66.5°, so we have v = r Earth cos 66.5° = (6.38 × 10 6 m)(cos 66.5°)(7.27 × 10 –5 rad/s) = 185 m/s . ( c ) At a latitude of 40.0° the distance is r Earth cos 40.0°, so we have v = r Earth cos 40.0° = (6.38 × 10 6 m)(cos 40.0°)(7.27 × 10 –5 rad/s) = 355 m/s .
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Chapter 10 2 Page 2 8. The subtended angle in radians is the size of the object divided by the distance to the object. A pencil with a diameter of 6 mm will block out the Moon if it is held about 60 cm from the eye. For the angle subtended we have θ Moon = D pencil / r pencil ˜ (0.6 cm)/(60 cm) ˜ 0.01 rad . We estimate the diameter of the Moon from Moon = D Moon / r Moon ; 0.01 rad = D Moon /(3.8 × 10 5 km), which gives D Moon ˜ 4 × 10 3 km . 9. ( a ) ω = (2500 rev/min)(2p rad/rev)/(60 s/min) = 262 rad/s . ( b ) The linear speed of the point on the edge is the tangential speed: v = r = (0.175 m)(262 rad/s) = 46 m/s . Because the speed is constant, the tangential acceleration is zero. There will be a radial acceleration: a R = 2 R = (262 rad/s) 2 (0.175 m) = 1.2 × 10 4 m/s 2 radial . 10. ( a ) The angular speed of the merry-go-round is = (1 rev)(2p rad/rev)/(4.0 s) = 1.57 rad/s. The linear speed of the child is the tangential speed: v = r = (1.2 m)(1.57 rad/s) = 1.9 m/s . ( b ) The child will have a radial acceleration: a R = 2 R = (1.57 rad/s) 2 (1.2 m) = 3.0 m/s 2 radial . 11. In each revolution the ball rolls a distance equal to its circumference, so we have L = N (p D ); 3.5 m = (15.0)p D , which gives D = 0.074 m = 7.4 cm .
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Ch10Word - Chapter 10 1 CHAPTER 10 - Rotational Motion...

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