Average Angular Acceleration of a Music CD over 75 minutes

[mcarloni]

A CD has a playing time of 74.2 minutes. When the music starts, the CD is rotating at an angular speed of 488 revolutions per minute (rpm). At the end of the music, the CD is rotating at 215 rpm. Find the magnitude of the average angular acceleration of the CD. Express your answer in rad/s^2.

knowns:
t = 74.2 min or 4344 sec
w0 = 488 rev/min * 2pi rad = 3066 rad/min / 60 sec/min = 51 rad/sec
w = 215 rev/min * 2pi rad = 1350 rad/min / 60 sec/min = 23 rad/sec

av ang acc = w - w0 / t = (23 - 51) / 4344 = -.0064 rad/sec^2

WileyPlus tells me my answer is wrong. Tried a positive answer, too.
I'm stumped!
UPDATE: SOLVED

Wiley needed more significant figures

 

[bob012345]

I'm answering this just for the record. Yes, the solution looks correct.

 

[Steve4Physics]

t = 74.2 min or 4344 sec

74.2mins = 4452s not 4344s. (Also note that the symbol for seconds is 's' not 'sec'.)

w0 = 488 rev/min * 2pi rad = 3066 rad/min / 60 sec/min = 51 rad/sec
w = 215 rev/min * 2pi rad = 1350 rad/min / 60 sec/min = 23 rad/sec

You should not round-off too much in intermediate steps - especially if you are going to find the difference between 2 rounded values.
$\omega_{initial}$ = 51.1032 rad/s (not 51 rad/s)
$\omega_{final}$ = 22.5147 rad/s (not 23 rad/s)

av ang acc = w - w0 / t = (23 - 515) / 4344 = -.0064 rad/sec^2

That should be $\frac {22.5147 rad/s- 51.1032 rad/s }{4452s} = -0.00642rad/s^2$.

You have made three mistake - one converting minutes to seconds and two rounding. By good luck your three mistakes have cancelled-out so that your final answer matches the correct answer to two significant figures!!!

 

[kuruman]

For even better accuracy, I would first take the difference then convert $$\begin{align}|\bar a| & =\frac{(488-215)~(\text{rev/min})}{74.2~(\text{min})}=\frac{273}{74.2}~(\text{rev/min}^2) \nonumber \\
& =\frac{273}{74.2}~(\text{rev/min}^2)\times(2\pi)~(\text{rad/rev})\times \left(\frac{1}{60}~(\text{min/s)}\right)^2.\nonumber
\end{align}$$

 

[Steve4Physics]

For even better accuracy, I would first take the difference then convert $$\begin{align}|\bar a| & =\frac{(488-215)~(\text{rev/min})}{72.4~(\text{min})}=\frac{273}{72.4}~(\text{rev/min}^2) \nonumber \\
& =\frac{273}{72.4}~(\text{rev/min}^2)\times(2\pi)~(\text{rad/rev})\times \left(\frac{1}{60}~(\text{min/s)}\right)^2.\nonumber
\end{align}$$

To avoid confusion note the typo' - the time is 74.2 minutes, not 72.4.

 

[kuruman]

To avoid confusion note the typo' - the time is 74.2 minutes, not 72.4.

There goes my purported accuracy. It's fixed now, thanks.

 

[bob012345]

Good catch @Steve4Physics ! I'm a bit embarrassed I did not catch the errors in the OP. I calculated with minutes and converted at the end and getting the same number I did not notice the minutes conversion in the OP was based on transposing the 2 and 4.

 

[kuruman]

Good catch @Steve4Physics ! I'm a bit embarrassed I did not catch the errors in the OP. I calculated with minutes and converted at the end and getting the same number I did not notice the minutes conversion in the OP was based on transposing the 2 and 4.

I am more than a bit embarrassed for making the same error in post #4.