Angular Acceleration/ Find angle in period - PLEASE help

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The discussion focuses on calculating angular acceleration and the angle through which a dentist's drill rotates during a period of acceleration. The user successfully calculates the angular acceleration as approximately 605.047 rad/s² but struggles with finding the angle of rotation. Initially, they incorrectly assume constant speed for the angle calculation, leading to an incorrect result of 14825 radians. After clarification, they realize the correct formula involves using angular acceleration and time, resulting in an angle of approximately 7412.58 radians. The conversation highlights the importance of using the right formulas in rotational motion problems.
hoseA
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Angular Acceleration/ Find angle in period -- PLEASE help!

I'm having trouble with part "B" please help.

A.)A dentist's drill starts from rest. After 4.95 s
of constant angular acceleration, it turns at a
rate of 28600 rev/min.
Find the drill's angular acceleration. An-
swer in units of rad/s^2.

B.)Find the angle through which the drill rotates
during this period. Answer in units of rad.

A.) [(28600*2pi)/60]/4.95 = 605.047 rad/s^2

B.) [(28600*2pi)/60] * 4.95 = 14825 rad. (but this is not the right answer?)

Am I using the wrong formula? Any help appreciated.
 
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hoseA said:
B.) [(28600*2pi)/60] * 4.95 = 14825 rad. (but this is not the right answer?)
This assumes that the speed is constant at (28600*2pi)/60 rad/s. But the speed is not constant; the drill starts from rest and accelerates.
 
Thanks so much. Didn't realize that. So it's theta = .5 (605.047)(4.95^2)

= 7412.58 radians.

Thanks again. :)
 
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