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

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Page 1 Chapter  10 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 θ    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   = ? ϖ /? 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  ϖ   =∆ θ / , and  we use the time for each hand  to turn  through  a complete circle, 2p rad. ( a ) ϖ second =∆ θ / = (2p rad)/(60 s) =       0.105 rad/s . ( b ) ϖ minute =∆ θ / = (2p rad)/(60 min)(60 s/min)  =       1.75  × 10 –3  rad/s . ( c ) ϖ hour =∆ θ / = (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 ϖ orbit =∆ θ / = (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 ϖ rotation =∆ θ / = (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: ϖ  =  θ /  = (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 r Earth ϖ   = (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 r Earth  cos 66.5 °   ϖ   = (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 r Earth  cos 40.0 °   ϖ   = (6.38 × 10 6  m)(cos 40.0 ° )(7.27  × 10 –5  rad/s)  =          355 m/s .
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Page 2 Chapter  10 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 .
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