# Angular Acceleration Problem

Problem(physics class 201/Portland Community College)
During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an angular displacement of 0.250 rev. What is the angular acceleration of the CD?

This is my formula from my Kinetics formula in my book where ∂ = angular acceleration

(1)Kinetics formula
V^2 = V(initial)^2 + 2ax

(2)so I converted to:
ω^2 = ω(initial)^2 + 2∂θ
(477 rev/mins)^2 = (0 rad/s)^2 + 2(∂)(0.250 rev)
[(477 rev/mins)^2 - (0 rad/s)^2)]/ (2 (0.250 rev))= ∂
[(477 rev/mins)^2 - 0] / (.50 rev) = ∂
(477 rev/mins)^2 / .50 rev = ∂
(227529 rev^2/mins^2) / .50 rev = ∂
455,058 rev/mins^2 = ∂

**** I can not get the answer! the Answer is "e" ****

## Answers and Replies

TSny
Homework Helper
Gold Member
Hi, gcombina.

Watch the units.

Note the units in your answer compared to the units in the choices of answer.

Last edited:
so 477 rev/min = 477 rad/60 s??? meaning 7.095 rad/s??

TSny
Homework Helper
Gold Member
so 477 rev/min = 477 rad/60 s??? meaning 7.095 rad/s??

How many radians in a revolution?

Note: I believe your original answer is correct in rev/min2. So, you could just convert it to rad/s2. However, I think it would be worthwhile for you to also work the problem by first converting the given data to SI units.

I got it!
thanks!
ω^2 = ω(initial)^2 + 2∂θ

converted
477 rev/min into rad/s ===> converted to 477 (2pi)rad/60s) because 1 revolution equals a 2pi radian
0.25 rev ====> converted to 1.57 because 1rev = 2pi therefore, 0.25 (2pi) = 0.25 (2(3.1415)) = 1.57a

after converting the revolutions to radians, I just plug in the numbers
(49.95 rad/s )^2 = (0 rad/s)^2 + 2(α)(1.57)
[(2495 rad/s) - (0 rad/s)^2] = 2 (α) (1.57)
2495 rad/s = 3.14 (α)
(2495 rad/s) / (3.14) = α
α = 795

))) thanks!

TSny
Homework Helper
Gold Member
Good work!