Question

A roller in a printing press turns through an angle \(\theta(t)\) given by \(\theta(t) = \gamma t^2 - \beta t^3\), where \(\gamma =\) 3.20 rad/s\(^2\) and \(\beta =\) 0.500 rad/s\(^3\). (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of t does it occur?

Short answer

Max angular velocity occurs at t = 2.13 s, calculate \( \omega(2.13) \) for exact value.

Step-by-step solution

Differentiate for Angular Velocity

To find the angular velocity \( \omega(t) \), differentiate \( \theta(t) = \gamma t^2 - \beta t^3 \) with respect to time. The derivative is:\[ \omega(t) = \frac{d}{dt}(\gamma t^2 - \beta t^3) = 2\gamma t - 3\beta t^2 \]This gives us the function for angular velocity.

Differentiate for Angular Acceleration

To find the angular acceleration \( \alpha(t) \), differentiate \( \omega(t) = 2\gamma t - 3\beta t^2 \) with respect to time. The derivative is:\[ \alpha(t) = \frac{d}{dt}(2\gamma t - 3\beta t^2) = 2\gamma - 6\beta t \]This gives us the function for angular acceleration.

Find Maximum Angular Velocity

To find the maximum angular velocity, determine when the angular acceleration is zero by setting \( \alpha(t) = 0 \):\[ 2\gamma - 6\beta t = 0 \]Solve for \( t \):\[ t = \frac{2\gamma}{6\beta} = \frac{\gamma}{3\beta} \]Substitute \( \gamma = 3.20 \) rad/s\(^2\) and \( \beta = 0.500 \) rad/s\(^3\) into the equation:\[ t = \frac{3.20}{3 \times 0.500} \]Calculate \( t \) to get \( t = 2.13 \) s.

Calculate Maximum Angular Velocity Value

Substitute \( t = 2.13 \) s into the angular velocity function \( \omega(t) = 2\gamma t - 3\beta t^2 \):\[ \omega(2.13) = 2 \times 3.20 \times 2.13 - 3 \times 0.500 \times (2.13)^2 \]Calculate to find the maximum angular velocity.

Key concepts

Angular Velocity

Angular velocity, denoted as \( \omega(t) \), is a measure of how fast an object rotates or revolves relative to another point. In essence, it tells us how quickly an angle changes in a rotational setting.
For the given problem, the expression for angle \( \theta(t) \) as a function of time is \( \theta(t) = \gamma t^2 - \beta t^3 \). By differentiating this equation with respect to time \( t \), we arrive at the equation for angular velocity:
\[ \omega(t) = \frac{d}{dt}(\gamma t^2 - \beta t^3) = 2\gamma t - 3\beta t^2 \]
This equation reveals how \( \omega(t) \) changes over time.

• It has terms related to \( t \) raised to the first and second powers, indicating that velocity increases and decreases in a non-linear fashion.
• By substituting \( \gamma = 3.20 \) rad/s\(^2\) and \( \beta = 0.500 \) rad/s\(^3\), the angular velocity function becomes specific to our scenario.

Angular Acceleration

Understanding angular acceleration, represented by \( \alpha(t) \), is crucial as it describes how the angular velocity changes over time.
The expression for angular velocity, \( \omega(t) = 2\gamma t - 3\beta t^2 \), is differentiated with respect to \( t \) to provide the formula for angular acceleration:
\[ \alpha(t) = \frac{d}{dt}(2\gamma t - 3\beta t^2) = 2\gamma - 6\beta t \]

• This equation depicts a linear relationship between angular acceleration and time, suggesting that as \( t \) increases, \( \alpha(t) \) decreases.
• Specifically, it starts at a value of \( 2\gamma \) and reduces depending on the coefficient \( \beta \).

When \( \alpha(t) \) becomes zero, \( \omega(t) \) either reaches its maximum or minimum, a key point explored in maximum velocity calculation.

Differentiation in Physics

Differentiation is a fundamental concept in physics used to determine the rate at which a function changes. It's particularly useful in describing motion, such as angular motion in rotational systems.
In this exercise:

• We use differentiation to move from the angular position function, \( \theta(t) \), to obtain angular velocity, \( \omega(t) \).
• Further differentiation of \( \omega(t) \) provides the angular acceleration, \( \alpha(t) \).

This step-by-step progression highlights the interconnectedness of motion variables:

• Angular position, velocity, and acceleration are linked through their respective derivatives.
• This helps us predict and analyze rotational dynamics effectively.

Maximum Velocity Calculation

Calculating the maximum angular velocity entails finding the point in time where this velocity peaks. In our equation for angular acceleration \( \alpha(t) = 2\gamma - 6\beta t \), we set \( \alpha(t) = 0 \) to pinpoint when velocity stops increasing and begins to decrease.
Solve for \( t \):
\[ 2\gamma - 6\beta t = 0 \]
\[ t = \frac{\gamma}{3\beta} \]
Plug in \( \gamma = 3.20 \) rad/s\(^2\) and \( \beta = 0.500 \) rad/s\(^3\):
\[ t = \frac{3.20}{3 \times 0.500} = 2.13 \text{ seconds} \]
Substitute \( t = 2.13 \) s back into \( \omega(t) = 2\gamma t - 3\beta t^2 \) to find the maximum angular velocity. This procedure gives insight into the motion's dynamics, ensuring the roller's peak performance is calculated accurately.