Science
A dentist's drill starts from rest. After 2.60 s of constant angular acceleratio

Question Description

A dentist's drill starts from rest. After 2.60 s of constant angular acceleration it turns at a rate of 2.7  104 rev/min.

(a) Find the drill's angular acceleration.
 rad/s2

(b) Determine the angle (in radians) through which the drill rotates during this period.

 rad

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Final Answer

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α = (ω - ωo)/t 
α = (2.7e4 rev/min)(2π rad/rev)(min / 60 s) / (2.60 s) 
α = 1087 rad/s^2 

(b) 
θ = αt^2 / 2 
θ = (1087rad/s^2)(2.60 s)^2 / 2 
θ = 418 rad

Please let me know if you need any clarification. I'm always happy to answer your questions.

ngunyag (125)
New York University

Anonymous
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