Question

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

Answer 1

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