A dentist’s drill starts from rest. After 1.28 s
of constant angular acceleration it turns at a
rate of 38840 rev/min.
Find the drill’s angular acceleration.
Answer in units of rad/s^2
Throughout what angle does the drill rotate
during this period?
Answer in units of rad.
To find the drill's angular acceleration, we need to use the formula:
ω = ω0 + αt
where
ω is the final angular velocity (in rad/s),
ω0 is the initial angular velocity (in rad/s),
α is the angular acceleration (in rad/s^2), and
t is the time (in seconds).
Given:
ω = 38840 rev/min
t = 1.28 s
First, let's convert the final angular velocity from revolutions per minute (rpm) to radians per second (rad/s):
ω = (38840 rev/min) * (2π rad/rev) * (1 min/60 s)
Since there are 2π radians in one revolution, and 60 seconds in one minute, we can simplify:
ω = 4052π rad/s
Since the drill starts from rest (ω0 = 0), we can rearrange the formula to solve for α:
α = (ω - ω0) / t
α = (4052π rad/s - 0 rad/s) / 1.28 s
Now, let's calculate α:
α = 4052π rad/s / 1.28 s ≈ 3171.33 rad/s^2
Therefore, the drill's angular acceleration is approximately 3171.33 rad/s^2.
To calculate the angle through which the drill rotates during this period, we can use the following formula:
θ = ω0t + 0.5αt^2
Given:
ω0 = 0 (since the drill starts from rest)
t = 1.28 s
α ≈ 3171.33 rad/s^2
Substituting the values, we get:
θ = 0.5 * (3171.33 rad/s^2) * (1.28 s)^2
θ ≈ 2587.80 rad
Therefore, the drill rotates approximately 2587.80 radians during this period.
To find the angular acceleration of the drill, we can use the formula:
ω = ω₀ + αt
where ω is the final angular velocity (in rad/s), ω₀ is the initial angular velocity (in rad/s), α is the angular acceleration (in rad/s^2), and t is the time (in seconds).
Given:
ω = 38840 rev/min
t = 1.28 s
First, let's convert the final angular velocity to rad/s:
ω = 38840 rev/min * (2π rad/rev) * (1 min/60 s) ≈ 4078.087 rad/s
Since the drill starts from rest, the initial angular velocity ω₀ is zero.
Using the formula ω = ω₀ + αt, we can rearrange it to solve for α:
α = (ω - ω₀) / t
Substituting the values, we have:
α = (4078.087 rad/s - 0 rad/s) / 1.28 s ≈ 3184.066 rad/s^2
Therefore, the drill's angular acceleration is approximately 3184.066 rad/s^2.
Now, let's find the angle through which the drill rotates during this period. We can use the formula:
θ = ω₀t + 0.5αt²
where θ is the angular displacement (in radians).
Since the drill starts from rest, the initial angular velocity ω₀ is zero.
θ = 0.5αt²
Substituting the values, we have:
θ = 0.5 * 3184.066 rad/s^2 * (1.28 s)² ≈ 2582.706 rad
Therefore, the drill rotates approximately 2582.706 radians during this period.
each revoution= 2PI radians
avg speed=38840/2 rev
angletotal=2PI*avgspeed*1.28/60 radians