A dentist's drill starts from rest. After 3.20 s of constant angular acceleration it turns at a rate of 2.7 104 rev/min.
(a) Find the drill's angular acceleration.
(b) Determine the angle (in radians) through which the drill rotates during this period.
I have a question. 2.7104 was that suppose to actually be 2.7*10^4?
i've a question, where did u get 6.28 rad/rev?
2pie equals 6.28
(a) Well, it seems like the dentist's drill had a pretty impressive acceleration! To find the angular acceleration, we can start by converting the final angular velocity to radians per second. Since 1 revolution is equal to 2π radians, we'll have:
Final angular velocity = 2.7 x 10^4 rev/min
= (2.7 x 10^4) x (2π rad/1 rev) x (1 min/60 s)
= 900π rad/s
Now, we can use the formula for angular acceleration:
ωf = ωi + αt
Where ωf is the final angular velocity (900π rad/s), ωi is the initial angular velocity (0 rad/s because the drill starts from rest), α is the angular acceleration, and t is the time (3.20 s).
Plugging in the known values, we can solve for α:
900π rad/s = 0 rad/s + α(3.20 s)
α = (900π rad/s) / (3.20 s)
α ≈ 879.63 rad/s²
So, the drill's angular acceleration is approximately 879.63 rad/s². That's some pretty fast spinning!
(b) To find the angle through which the drill rotates, we can use the formula for angular displacement:
θ = ωit + (1/2)αt²
Where θ is the angular displacement, ωi is the initial angular velocity (0 rad/s), α is the angular acceleration (879.63 rad/s²), and t is the time (3.20 s).
Plugging in the known values:
θ = (0 rad/s)(3.20 s) + (1/2)(879.63 rad/s²)(3.20 s)²
θ ≈ (1/2)(879.63 rad/s²)(10.24 s²)
θ ≈ 4,511.03 rad
So, the drill rotates approximately 4,511.03 radians during this period. That's quite a twist!
To find the drill's angular acceleration, we can use the following equation of motion for rotational motion:
ω = ω_0 + αt
where
ω = final angular velocity
ω_0 = initial angular velocity
α = angular acceleration
t = time
Given that the initial angular velocity (ω_0) is zero and the final angular velocity (ω) is 2.7 × 10^4 rev/min, we first need to convert ω to radians/second.
1 revolution = 2π radians.
So, 1 rev/min = (2π radians/1 revolution) × (1/60 min)
= π/30 radians/sec
Therefore, 2.7 × 10^4 rev/min = (2.7 × 10^4) × (π/30) radians/sec
Now, substituting the values into the equation, we have:
2.7 × 10^4 rev/min = 0 + α × 3.20 s
Solving for α, we get:
α = (2.7 × 10^4 rev/min) / (3.20 s)
To find the angle through which the drill rotates, we can use another equation of motion:
θ = ω_0t + (1/2) αt^2
However, since the initial angular velocity (ω_0) is zero, the equation simplifies to:
θ = (1/2) αt^2
Substituting the values, we get:
θ = (1/2) α × (3.20 s)^2
Now, we can calculate both the angular acceleration (α) and the angle (θ) using the given values.
A dentist's drill starts from rest it truns at a rate of 2.22*10^5 rpm at 3.5s.
1)find the angular acceleration of the drill.
2)through how many revolution the drill rotates during this period.
V = 2.7104 rev / min.
V = 2.7104 rev/min * 1 / 60 min/s = 0.04517 rev/s.
a. a = 0.04517 rev/s / 3.2 s =
0.01412 rev/s^2.
b. A = 0.04517 rev/s * 3.2 s * 6.28 rad/rev = 0.91 Radians.