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