Question

As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. A) If the CD rotates clockwise (let's take clockwise rotation as positive) at 500rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60s with constant angular acceleration, what is alpha_vec, the angular acceleration of the CD, as it spins to a stop? B)How many revolutions does the CD make as it spins to a stop?

Answer 1

writing + Δ Math- & Chemistry + x'x. α β γ δ ε e λ μ π ρ σ τ.co.:Γ #"ΣΔΩ Ω Add/Edit Custom Equation dddt Cutom iagm Gine New Step Answered By "Mohsen a) wi-2nf/60 2 3.1416500/60- 52.36 rad/s a (wf-wi)/t (0-52.36)/2.6-20.1 rad/s 2 02-52.36 52.36 /(2*(-20.138)) 68.069 rad 68.069/(23.1416) 10.8 revolutions

Answer 2

(500*2*pi/60)/2.6 = 20.138 rad/s^2

Answer 3

Well so I am going to use radians/seconds instead of revolutions/minutes.
So 500rev/min*2π/60s=52.36 rad/s

Vi is velocity intial and Vf is velocity final.

Vi=52.36 rad/s

Vf=o rad/s

a=?

t= 2.60s

SO I am going to use the rotational kinematic slope equation to solve the acceleration. The equation is

a=(Vf-Vi)/t

So plug stuff in: a=(0-52.36)/2.60

So the acceleration is negative since it is slowing down. So the acceleration is -20.14 rad/s^2.

BTW is I missunderstood the time being seconds instead of mins then just do the same as above but instead the acceleration should equal -500/2.60. It should come out to be -192.31 rev/min^2.

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