Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 6.00 s, it rotates 28.0 rad. During that time, what are the magnitudes of (a) the angular acceleration and (b) the average angular velocity? (c) What is the instantaneous angular velocity of the disk at the end of the 6.00 s? (d) With the angular acceleration unchanged, through what additional angle (rad) will the disk turn during the next 6.00 s?

To answer the given questions, we need to use the kinematic equations for rotational motion.

(a) The magnitude of the angular acceleration can be found using the formula:
angular acceleration (α) = (final angular velocity - initial angular velocity) / time

From the given information, the initial angular velocity is 0 (starting from rest), the final angular velocity is 28.0 rad (given in the problem), and the time is 6.00 s.

Therefore, the angular acceleration is:
α = (28.0 rad - 0) / 6.00 s

Calculating the value:
α = 4.67 rad/s^2

(b) The average angular velocity can be found using the formula:
average angular velocity (ω_avg) = (final angular velocity + initial angular velocity) / 2

As the initial angular velocity is 0 (starting from rest), and the final angular velocity is 28.0 rad, we can substitute these values into the formula:

ω_avg = (28.0 rad + 0) / 2

Calculating the value:
ω_avg = 14.0 rad/s

(c) The instantaneous angular velocity at the end of the 6.00 s can be calculated by using the formula:
final angular velocity (ω_final) = initial angular velocity + (angular acceleration * time)

Using the given values:
initial angular velocity = 0 (starting from rest)
angular acceleration = 4.67 rad/s^2
time = 6.00 s

Substituting these values into the formula:
ω_final = 0 + (4.67 rad/s^2 * 6.00 s)

Calculating the value:
ω_final = 28.0 rad/s

(d) To find the additional angle (θ) the disk will turn during the next 6.00 s, we can use the equation:
θ = initial angular velocity * time + (1/2) * angular acceleration * time^2

Using the given values:
initial angular velocity = 28.0 rad (the final angular velocity of the previous 6.00 s)
angular acceleration = 4.67 rad/s^2
time = 6.00 s

Substituting these values into the formula:
θ = 28.0 rad * 6.00 s + 0.5 * 4.67 rad/s^2 * (6.00 s)^2

Calculating the value:
θ = 168 rad

Therefore, the disk will turn an additional 168 rad during the next 6.00 s.

To find the answers to these questions, we can use the kinematic equations for rotational motion.

(a) To find the magnitude of the angular acceleration (α), we can use the equation:

θ = ω₀t + (1/2)αt²

Given that the disk rotates 28.0 rad in 6.00 s (θ = 28.0 rad and t = 6.00 s), and starting from rest (ω₀ = 0), we can rearrange the equation to solve for α:

28.0 rad = (1/2)α(6.00 s)²

First, let's calculate (1/2) × (6.00 s)²:

(1/2) × (6.00 s)² = 18.0 s²

Now, we can rearrange the equation:

28.0 rad = 18.0 s² × α

Divide both sides by 18.0 s²:

α = 28.0 rad / 18.0 s²

Calculating this gives us the magnitude of the angular acceleration (α).

(b) To find the average angular velocity (ω_avg), we can use the equation:

ω_avg = (ω₀ + ω) / 2

Given that the disk starts from rest (ω₀ = 0) and rotates 28.0 rad in 6.00 s (t = 6.00 s), we can plug in these values:

ω_avg = (0 + 28.0 rad) / 2 × (6.00 s)

Simplifying this gives us the average angular velocity (ω_avg).

(c) To find the instantaneous angular velocity (ω) at the end of the 6.00 s, we can use the equation:

ω = ω₀ + αt

Given that the disk starts from rest (ω₀ = 0), the time (t) is 6.00 s, and we have already calculated the angular acceleration (α), we can plug these values into the equation to find ω.

(d) To find the additional angle (θ') the disk will turn during the next 6.00 s, we can use the same equation as in part (a):

θ = ω₀t + (1/2)αt²

Given that the initial angular velocity (ω₀) is the instantaneous angular velocity at the end of 6.00 s (ω) and the time (t) is 6.00 s, we can plug in these values to calculate θ'.

a) θ = ½ α t² => α = 2 θ / t² = 2 * 20 / 8² = 0.625 rad/s²

b) ωa = ∆θ / ∆t = 20 / 8 = 2.5 rad/s

c) ωi = α t = 0.625 * 8 = 5 rad/s

d) θ = ωo t + ½ α t² = 5 * 6 + ½ 0.625 * 6² = 41.25 rad