A dentist's drill starts from rest. After 3.30 s of constant angular acceleration it turns at a rate of 2.1 104 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
18.7
To solve this problem, we can use the equations of rotational motion.
(a) We are given that the drill starts from rest and reaches an angular velocity of 2.1 x 10^4 rev/min after 3.30 s of constant angular acceleration. We need to find the drill's angular acceleration.
First, we need to convert the angular velocity to radians per second (rad/s). We know that 1 revolution is equal to 2π radians. Therefore,
2.1 x 10^4 rev/min = (2π rad/rev) * (2.1 x 10^4 rev/min) = 4.2 x 10^4π rad/min
To convert to rad/s, we divide by 60:
4.2 x 10^4π rad/min = (4.2 x 10^4π rad/min) / (60 min/s) ≈ 2199.9 rad/s
Now we can use the equation for angular acceleration:
ω = ω0 + αt
where ω is the final angular velocity, ω0 is the initial angular velocity (which is zero in this case), α is the angular acceleration, and t is the time.
Plugging in the values:
2199.9 rad/s = 0 + α(3.30 s)
Rearranging the equation to solve for α:
α = 2199.9 rad/s / 3.30 s ≈ 666.6 rad/s^2
Therefore, the drill's angular acceleration is approximately 666.6 rad/s^2.
(b) To find the angle through which the drill rotates during this period, we can use the equation:
θ = ω0t + (1/2)αt^2
where θ is the angle rotated, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.
Plugging in the values:
θ = 0 + (1/2)(666.6 rad/s^2)(3.30 s)^2
Simplifying the equation:
θ = (1/2)(666.6 rad/s^2)(10.89 s^2) ≈ 3827.3 rad
Therefore, the angle through which the drill rotates during this period is approximately 3827.3 radians.