The tub of a washer (radius=45 cm) goes into its spin cycle,starting from rest a
ID: 1753371 • Letter: T
Question
The tub of a washer (radius=45 cm) goes into its spin cycle,starting from rest and gaining angular speed steadily for 8.00seconds, when its it turning at 5.00 rev/sec. At this point theperson doing the laundry opens the lid, and a safety switch turnsoff the washer. the tub smoothly slows to rest in 12.0seconds. a) Assuming it is constant, what is the angular acceleration,in radians/s ^2, of the tub for the first 8 seconds ( include anysigns)? _____________________ b) Assuming it is constant, what is the angular acceleration,in radians/s^2. of the tub for the last 12 seconds (include anysigns)? _____________________ c) Through how many total revolutions does the tub turns whileit is in motion? ____________________ d) An itsy bitsy spider was riding along on the rim ofthe tub as it spin. How far did this spider travel (in meters) bythe time the washer had come to rest? __________________ e) At the point where the washer is spinning its fastest, atwhat speed is the spider moving (in meters/sec)? _________________ f) At the point imeediately before the washer is opened, whatis the magnitude of the acceleration of the spider? ( hint:remeber, this includes finding two components of acceleration andthen finding the magnitude of the net acceleration... in youanswer, be sure to identify and label your values for both of thoseacceleration components as well as the total acceleration)? ____________________ ____________________ ____________________ The tub of a washer (radius=45 cm) goes into its spin cycle,starting from rest and gaining angular speed steadily for 8.00seconds, when its it turning at 5.00 rev/sec. At this point theperson doing the laundry opens the lid, and a safety switch turnsoff the washer. the tub smoothly slows to rest in 12.0seconds. a) Assuming it is constant, what is the angular acceleration,in radians/s ^2, of the tub for the first 8 seconds ( include anysigns)? _____________________ b) Assuming it is constant, what is the angular acceleration,in radians/s^2. of the tub for the last 12 seconds (include anysigns)? _____________________ c) Through how many total revolutions does the tub turns whileit is in motion? ____________________ d) An itsy bitsy spider was riding along on the rim ofthe tub as it spin. How far did this spider travel (in meters) bythe time the washer had come to rest? __________________ e) At the point where the washer is spinning its fastest, atwhat speed is the spider moving (in meters/sec)? _________________ f) At the point imeediately before the washer is opened, whatis the magnitude of the acceleration of the spider? ( hint:remeber, this includes finding two components of acceleration andthen finding the magnitude of the net acceleration... in youanswer, be sure to identify and label your values for both of thoseacceleration components as well as the total acceleration)? ____________________ ____________________ ____________________Explanation / Answer
Radius of tub r = 45 cm = 0.45 m initial angular speed w = 0 m / s time taken t = 8 s final angular velocity w ' = 5 rev / s = 31.415 rad / s (a). from the relation w ' = w + t angular accleration = ( w ' - w ) / t = 3.926 rad / s ^ 2 for last 12 s : ------------- Initial angular speed = 31.415 rad / s final angular speed ' = 0 rad / s time t ' = 12 s So, from the relation ' = + ' t ' from this angular accleration for last 12 s is '= ( ' - ) / t ' = -2.6179 rad / s ^ 2 let No.of revolutions in 1 st 8 s be then from the relation = w t + ( 1/ 2) t ^2 = 125.632 rad = ( 125.632 / 2 ) rev = 19.994 rev for last 12 s : Let the No.of revolutuins in last 12 be ' then from therelation '= t ' + ( 1/ 2) ' t '^ 2 = 376.98 - 188.4888 = 188.4912 rad = ( 188.4912 / 2 ) rev = 29.999 rev So, total revolutions = + ' = 49.9933 rev ~ 50 rev = 376.98 - 188.4888 = 188.4912 rad = ( 188.4912 / 2 ) rev = 29.999 rev So, total revolutions = + ' = 49.9933 rev ~ 50 rev w '^ 2 - w ^ 2= 2Related Questions
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