A disk rotates with constant angular acceleration. The initial angular speed of the disk is 2π rad/s. After the disk rotates through 17π radians, the angular speed is 18π rad/s.

(a) What is the magnitude of the angular acceleration?
answer in rad/s2

(b) How much time did it take for the disk to rotate through 17π radians?
answer in s

(c) What is the tangential acceleration of a point located at a distance of 3 cm from the center of the disk?
answer in m/s2

wf^2=wi^2 + 2 a d solve for a.

b. wf=wi+at solve for t

c. when, at the end? acceleration=wf^2 *radius

To answer these questions, we need to use the laws of rotational motion.

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

ω^2 = ω0^2 + 2αθ

where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and θ is the angle rotated.

Given:
ω0 = 2π rad/s
ω = 18π rad/s
θ = 17π radians

Plugging in the values, we can rearrange the equation to solve for α:

α = (ω^2 - ω0^2) / (2θ)
= (18π^2 - 4π^2) / (2 * 17π)
= 196π / (34π)
= 98/17 rad/s^2

Therefore, the magnitude of the angular acceleration is 98/17 rad/s^2.

(b) To find the time it takes for the disk to rotate through 17π radians, we can use the formula:

ω = ω0 + αt

First, let's find the angular acceleration, α, using the value we obtained in part (a):

α = 98/17 rad/s^2

Now, we can rearrange the equation to solve for t:

t = (ω - ω0) / α
= (18π - 2π) / (98/17)
= 16π / (98/17)
= 8π / (49/8)
= 128π / 49
≈ 8.22 s

Therefore, it takes approximately 8.22 seconds for the disk to rotate through 17π radians.

(c) To find the tangential acceleration of a point located at a distance of 3 cm from the center of the disk, we can use the formula:

at = rα

where at is the tangential acceleration, r is the radius, and α is the angular acceleration.

Given:
r = 3 cm (which is equal to 0.03 m)
α = 98/17 rad/s^2 (as found in part a)

Plugging in the values, we can calculate the tangential acceleration:

at = (0.03) * (98/17)
≈ 0.176 m/s^2

Therefore, the tangential acceleration of a point located at a distance of 3 cm from the center of the disk is approximately 0.176 m/s^2.