A clothes dryer is spinning at 75 rpm. A student opens its door and it comes to rest after rotating through four revolutions. Assume constant deceleration. What was the magnitude of the dryer's angular deceleration?

Answer 1

Vi = 75 * 2 pi radians/min *1 min/60 sec
= 5 pi/2 radians/second

v = Vi - a t
0 = 5 pi/2 - a t
so t = 5 pi/2a

4*2pi = Vi t + (1/2) a t^2

8 pi = ( 5 pi/2)(5 pi/2a) + (1/2)a (5 pi/2a)^2

etc

Answer 2

Vi = 75 * 2 pi radians/min *1 min/60 sec
= 5 pi/2 radians/second

v = Vi - a t
0 = 5 pi/2 - a t
so t = 5 pi/2a
+++++++++ a is assumed negative +++
4*2pi = Vi t - (1/2) a t^2

8 pi = ( 5 pi/2)(5 pi/2a) - (1/2)a (5 pi/2a)^2

8 pi a = (5 pi/2)^2 = (1/2)(5 pi/2)^2

8 pi a = (1/2)(5 pi/2)^2
16 pi a = 25 pi^2/4
a = (25/64) pi
remember it is negative (deacceleration)