Question

A dentist's drill starts from rest. After 2.60 s of constant angular acceleration it turns at a rate of 2.70 ✕ 10⁴ rev/min.

(a) Find the drill's angular acceleration.

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

Answer 1

Final RPM of the drill = N = 2.7 x 10⁴ rev/min

Time period = t = 2.6 sec

Initial angular velocity of the drill = ₁ = 0 rad/s (Starts from rest)

Final angular velocity of the drill = ₂

Angular acceleration of the drill =

Angle through which the drill rotated during the time period =

₂ = 2N/60

₂ = 2(2.7x10⁴)/60

₂ = 2827.43 rad/s

₂ = ₁ + t

2827.43 = 0 + (2.6)

= 1087.47 rad/s²

= ₁t + t²/2

= (0)(2.6) + (1087.47)(2.6²)/2

= 3675.65 rad

a) Angular acceleration of the drill = 1087.47 rad/s²

b) Angle through which the drill rotates during the period = 3675.65 radians