A wheel rotating with an angular velocity of 16rad/sec is slowed to an angular velocity of 4rad/sec by an angular acceleration of 2rad/sec^2.find

a. The angle turned in the process of slowing down
b. The time required

a. V^2 = Vo^2 + 2a*d.

d = (V^2-Vo^2)/2a.
d = (16-256)/-4 = 60 Rad.

b. t = (V-Vo)/a = (4-16)/-2 = 6 s.

To find the answers, we can use the kinematic equation for angular motion:

ω^2 = ω0^2 + 2αθ

Where:
- ω is the final angular velocity
- ω0 is the initial angular velocity
- α is the angular acceleration
- θ is the angle turned

a. The angle turned in the process of slowing down:

We know the initial angular velocity (ω0 = 16 rad/sec), final angular velocity (ω = 4 rad/sec), and angular acceleration (α = -2 rad/sec^2, negative because it is slowing down).

Using the kinematic equation, we can rearrange to solve for θ:

θ = (ω^2 - ω0^2) / (2α)
θ = (4^2 - 16^2) / (2 * -2)
θ = (16 - 256) / (-4)
θ = (-240) / (-4)
θ = 60 radians

Therefore, the angle turned in the process of slowing down is 60 radians.

b. The time required:

We can use another kinematic equation for angular motion:

ω = ω0 + αt

We know the initial angular velocity (ω0 = 16 rad/sec), final angular velocity (ω = 4 rad/sec), and angular acceleration (α = -2 rad/sec^2, negative because it is slowing down).

Rearranging the equation to solve for time (t):

t = (ω - ω0) / α
t = (4 - 16) / (-2)
t = (-12) / (-2)
t = 6 seconds

Therefore, the time required to slow down from an angular velocity of 16 rad/sec to 4 rad/sec is 6 seconds.