Rotation w/constant angular acceleration

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The discussion revolves around a physics homework problem involving a flywheel that slows down with constant angular acceleration. The calculated time to come to rest is approximately 340 seconds, with an angular acceleration of -0.0048 rad/s². For the first 20 revolutions, the user computed the time to be about 99.67 seconds, while the textbook states it should be 98 seconds. The discrepancy is attributed to rounding errors during calculations, emphasizing the importance of maintaining precision throughout the process. The user seeks confirmation on the correctness of their method despite the slight difference in the final answer.
Ignitia

Homework Statement


A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration? (c) How much time is required for it to complete the first 20 of the 40 revolutions?

Homework Equations


Figured (a) and (b) no problem. (a) =335s (round to 340s) and (b) = -0.0048rad/s2

40revs = 80π
20revs = 40π

The Attempt at a Solution


For (c) I went:

θf - θi = wit+1/2αt2
40π = 1.5t - 0.0024t2
0=-0.0024t2 + 1.5t + 40π

Plug them into the quadratic formula: t = (-1.5) - √(1.5 2 - 4(-.0024*-40π)) / (2(-.0024))

Calculator spits out t = 99.67s. The book says it's 98. It's close enough I want to write it off as correct, but I'd prefer if it could be double checked.
 
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Ignitia said:
Calculator spits out t = 99.67s. The book says it's 98.
98 is nearer.
I expect your error comes from rounding off too soon. Always keep more digits than you intend to quote in the final answer. How many more is not always obvious. E.g. if you have a step that is equivalent to 100002-100001 then you need to keep 6 digits along the way to have just one digit at the end.
 
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I figured that might have been the reason, thanks. I just wanted to be sure my method was correct.
 
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