The angular position of a point on the rim of a rotating wheel is given by Theta = 4.0t-3t^2+t^3, where theta is in radians and t is in seconds. At t=0, what are (a.) the point's angular position and (b.) its angular velocity? (c.) What is its angular velocity at t=4.0s? What are the instantaneous angular accelerations at (d.) the beginning and (e.) the end of this time interval?

To find the angular position and angular velocity at a given time, we need to differentiate the equation for angular position with respect to time.

Given: Theta = 4.0t - 3t^2 + t^3

(a.) Angular Position at t=0:
To find the angular position at t=0, substitute t=0 into the equation:
Theta(0) = 4.0(0) - 3(0^2) + (0^3)
Theta(0) = 0 radians

Therefore, the point's angular position at t=0 is 0 radians.

(b.) Angular Velocity:
To find the angular velocity, differentiate the equation for angular position with respect to time (t):

dTheta/dt = d/dt(4.0t - 3t^2 + t^3)
dTheta/dt = 4.0 - 6t + 3t^2

The resulting expression represents the angular velocity, which gives us the rate of change of angular position with respect to time.

(c.) Angular Velocity at t=4.0s:
Substituting t=4.0s into the expression for angular velocity:
Angular velocity at t=4.0s = 4.0 - 6(4.0) + 3(4.0^2)
Angular velocity at t=4.0s = 4.0 - 24.0 + 48.0
Angular velocity at t=4.0s = 28.0 radians per second

Therefore, the angular velocity at t=4.0s is 28.0 radians per second.

(d.) Instantaneous Angular Acceleration at the Beginning:
To find the instantaneous angular acceleration at the beginning of the time interval, we differentiate the angular velocity equation with respect to time:

d^2Theta/dt^2 = d/dt(4.0 - 6t + 3t^2)
d^2Theta/dt^2 = -6 + 6t

Substituting t=0 into the expression for angular acceleration:
Instantaneous angular acceleration at the beginning = -6 + 6(0)
Instantaneous angular acceleration at the beginning = -6 radians per second squared

Therefore, the instantaneous angular acceleration at the beginning of the time interval is -6 radians per second squared.

(e.) Instantaneous Angular Acceleration at the End:
To find the instantaneous angular acceleration at the end of the time interval, we substitute t=4.0s into the angular acceleration equation:

Instantaneous angular acceleration at the end = -6 + 6(4.0)
Instantaneous angular acceleration at the end = -6 + 24.0
Instantaneous angular acceleration at the end = 18 radians per second squared

Therefore, the instantaneous angular acceleration at the end of the time interval is 18 radians per second squared.