The angular position of a point on the rim of a rotating wheel is described by ∅ = (4.0 rad/s)t – (3.0 rad/s²)t² + (1.0 rad/s³)t³, (a) What is the angular velocity at t = 2.0 s and at t = 4.0 s? (b) What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? (c) What is the instantaneous angular acceleration at the beginning and end of this time interval?

The angular position of a point on a rotating wheel is given by a polynomial function of time. To find the angular velocity and acceleration, we can use calculus to derive the necessary values from this position function. Let's break down the problem step by step.

Finding Angular Velocity

The angular position is given by the equation:

∅(t) = (4.0 rad/s)t – (3.0 rad/s²)t² + (1.0 rad/s³)t³

To find the angular velocity, we need to take the first derivative of the angular position with respect to time:

Angular Velocity (ω): ω(t) = d(∅)/dt

Calculating the derivative:

• The derivative of (4.0 rad/s)t is 4.0 rad/s.
• The derivative of (−3.0 rad/s²)t² is −6.0 rad/s² * t.
• The derivative of (1.0 rad/s³)t³ is 3.0 rad/s³ * t².

Thus, the angular velocity function is:

ω(t) = 4.0 - 6.0t + 3.0t²

Now we can evaluate this function at t = 2.0 s and t = 4.0 s:

• At t = 2.0 s: ω(2) = 4.0 - 6.0(2) + 3.0(2)² = 4.0 - 12.0 + 12.0 = 4.0 rad/s.
• At t = 4.0 s: ω(4) = 4.0 - 6.0(4) + 3.0(4)² = 4.0 - 24.0 + 48.0 = 28.0 rad/s.

Calculating Average Angular Acceleration

The average angular acceleration over a time interval is defined as the change in angular velocity divided by the change in time:

Average Angular Acceleration (α_avg): α_avg = (ω(t2) - ω(t1)) / (t2 - t1)

Using our values:

• ω(2) = 4.0 rad/s
• ω(4) = 28.0 rad/s

Substituting these into the formula gives us:

α_avg = (28.0 - 4.0) / (4.0 - 2.0) = 24.0 / 2.0 = 12.0 rad/s².

Instantaneous Angular Acceleration

To find the instantaneous angular acceleration, we need to take the derivative of the angular velocity function:

Angular Acceleration (α): α(t) = d(ω)/dt

Taking the derivative of ω(t):

• For ω(t) = 4.0 - 6.0t + 3.0t²:
• The derivative of 4.0 is 0.
• The derivative of -6.0t is -6.0.
• The derivative of 3.0t² is 6.0t.

So, the angular acceleration function is:

α(t) = -6.0 + 6.0t.

Now we can evaluate this function at the beginning and end of the interval:

• At t = 2.0 s: α(2) = -6.0 + 6.0(2) = -6.0 + 12.0 = 6.0 rad/s².
• At t = 4.0 s: α(4) = -6.0 + 6.0(4) = -6.0 + 24.0 = 18.0 rad/s².

In summary:

• At t = 2.0 s, angular velocity is 4.0 rad/s, and instantaneous angular acceleration is 6.0 rad/s².
• At t = 4.0 s, angular velocity is 28.0 rad/s, and instantaneous angular acceleration is 18.0 rad/s².
• The average angular acceleration between t = 2.0 s and t = 4.0 s is 12.0 rad/s².