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
(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
b. wf=wi+at solve for t
c. when, at the end? acceleration=wf^2 *radius
(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.