A dentist's drill starts from rest. After 3.30 s of constant angular acceleration it turns at a rate of 2.1x10^4 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
(a) Divide the final angular velocity (in rad/s, NOT rpm) by the time interval.
(b) angle = (1/2)*(angular acceleration)*t^2
OR
(1/2)*(final angular velocity)*time
Why do you multiply by .5?
it is based off the equation:
displacement = (initial velocity)*(change in time) + 1/2(acceleration)*(change in time squared)
since that equation is linear, the equation recommended for this problem is:
"angle" = 0 + 1/2(acceleration)*(change in time squared)
that is how one gets the 1/2
To find the answers to the given questions, we can use the following steps:
Step 1: Convert the given angular velocity from rev/min to rad/s.
Step 2: Use the formula to find angular acceleration.
Step 3: Use the formula to find the angle through which the drill rotates.
Let's apply these steps to find the answers:
Step 1:
The conversion factor from rev/min to rad/s is given as:
1 rev/min = (2π rad) / (60 s)
Thus, the given angular velocity can be converted as follows:
2.1 × 10^4 rev/min = (2.1 × 10^4 × 2π) / 60 rad/s
= 220π rad/s
Step 2:
To find angular acceleration, we'll use the formula:
angular acceleration (α) = (final angular velocity - initial angular velocity) / time
Given: Initial angular velocity (ω0) = 0 rad/s
Final angular velocity (ω) = 220π rad/s
Time (t) = 3.30 s
Using the formula, we get:
α = (220π rad/s - 0 rad/s) / 3.30 s
= (220π rad/s) / 3.30 s
≈ 210.95 rad/s^2
Therefore, the angular acceleration is approximately 210.95 rad/s^2.
Step 3:
To find the angle (θ) through which the drill rotates, we'll use the formula:
θ = (initial angular velocity × time) + (0.5 × angular acceleration × time^2)
Given: Initial angular velocity (ω0) = 0 rad/s
Angular acceleration (α) ≈ 210.95 rad/s^2
Time (t) = 3.30 s
Using the formula, we get:
θ = (0 rad/s × 3.30 s) + (0.5 × 210.95 rad/s^2 × (3.30 s)^2)
= (0 rad/s) + (0.5 × 210.95 rad/s^2 × 10.89 s^2)
≈ 1147.15 rad
Therefore, the angle through which the drill rotates during this period is approximately 1147.15 radians.