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

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Unformatted text preview: Page 1 Chapter 10 1 CHAPTER 10 - Rotational Motion About a Fixed Axis 1. ( a ) 30 = (30)( 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: = 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 = ? /? 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 ....
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Ch10Word - Page 1 Chapter 10 1 CHAPTER 10 - Rotational...

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