A dentist's drill starts from rest. After 3.00 s of constant angular acceleratio

Question Description

A dentist's drill starts from rest. After 3.00 s of constant angular acceleration it turns at a rate of 2.0  104 rev/min.

(a) Find the drill's angular acceleration.

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

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Final Answer

α = (ω - ωo)/t 
α = (2 e4 rev/min)(2π rad/rev)(min / 60 s) / (3s) 
α = 698.13 rad/s^2 

θ = αt^2 / 2 
θ = (698.13rad/s^2)(3 s)^2 / 2 
θ = 3141.59 rad = 3.142 * 10^3 rad

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Gopal_R (1109)
Duke University

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