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?
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
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)
To find the magnitude of the dryer's angular deceleration, we can use the following formula:
ω^2 = ω_0^2 + 2αθ
Where:
ω = final angular velocity
ω_0 = initial angular velocity
α = angular deceleration
θ = angle of rotation
In this scenario, the initial angular velocity is 75 rpm, and the final angular velocity is 0 rpm (since it comes to rest). The angle of rotation is 4 revolutions, which is equal to 4 * 2π radians.
Converting the initial and final angular velocities to radians per second (since the formula requires SI units), we have:
ω_0 = 75 rpm * (2π radians/1 revolution) * (1 minute/60 seconds) = 5π radians/second
ω = 0 radians/second
Substituting these values into the formula, we get:
(0)^2 = (5π)^2 + 2α(4 * 2π)
Simplifying the equation, we have:
0 = 25π^2 + 16π^2α
Now we can solve for α by rearranging the equation:
16π^2α = -25π^2
α = -25π^2 / (16π^2) (dividing both sides by 16π^2)
α = -25 / 16
Thus, the magnitude of the dryer's angular deceleration is 25/16 rad/s^2.