A phonograph record accelerates from rest to 34.0 rpm in 5.53 s. (a) What is its angular acceleration in rad/s2? rad/s2 (b) How many revolutions does it go through in the process? rev
Solution) wo = 0 ( rest )
w = 34 rpm
1 revolution = 2(pi) radians
1 minute = 60 seconds
w = (34)(2(pi))/(60) rad/s
w = 3.56 rad/s
t = 5.53 s
(a) Angular acceleration = ?
w = wo + (angular acceleration)(t)
3.56 = 0 + (angular acceleration)(5.53)
Angular acceleration = (3.56)/(5.53)
Angular acceleration = 0.64 rad/s^2
(b) theeta = ? ( number of revolutions )
w^2 = wo^2 + 2(angular acceleration)(theeta)
(3.56^2) = 0 + 2(0.64)(theeta)
Theeta = (3.56^2)/(2×0.64) = 9.9 rad
Theeta = 9.9 rad
1 rev = 2(pi) rad
1 rad = 1/(2(pi)) rev
Theeta = (9.9)/(2(pi))
Theeta = 1.57 revolutions
A phonograph record accelerates from rest to 34.0 rpm in 5.53 s. (a) What is its...
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