Problem 66 - University Physics with Modern Physics

Question

A roller in a printing press turns through an angle \(\theta(t)\) given by \(\theta(t)=\gamma t^{2}-\beta t^{3},\) where \(\gamma=3.20 \operatorname{rad} / \mathrm{s}^{2}\) and \(\beta=\) 0.500 \(\mathrm{rad} / \mathrm{s}^{3}\) , (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?

Step-by-Step Solution

Answer

(a) \( \omega(t) = 6.40t - 1.5t^2 \); (b) \( \alpha(t) = 6.40 - 3.0t \); (c) Max \( \omega = 6.826 \) rad/s at \( t = 2.13 \) sec.

Step 1: Differentiate to Find Angular Velocity

The angular velocity \( \omega(t) \) is the first derivative of the angle \( \theta(t) \) with respect to time \( t \). We have \( \theta(t) = \gamma t^{2} - \beta t^{3} \). Differentiate this with respect to \( t \) to get: \[ \omega(t) = \frac{d\theta}{dt} = 2\gamma t - 3\beta t^{2} \]. Substituting \( \gamma = 3.20 \) rad/s² and \( \beta = 0.500 \) rad/s³, \[ \omega(t) = 6.40t - 1.5t^{2} \].

Step 2: Differentiate to Find Angular Acceleration

The angular acceleration \( \alpha(t) \) is the first derivative of angular velocity \( \omega(t) \) with respect to time \( t \). Using \( \omega(t) = 6.40t - 1.5t^{2} \), differentiate it to obtain: \[ \alpha(t) = \frac{d\omega}{dt} = 6.40 - 3.0t \].

Step 3: Find Maximum Positive Angular Velocity

To find the maximum positive angular velocity, set the derivative of \( \omega(t) \) (which is \( \alpha(t) \)) equal to zero and solve for \( t \). Thus, \[ 6.40 - 3.0t = 0 \]. Solving gives \( t = \frac{6.40}{3.0} = 2.13 \) seconds. Substitute \( t = 2.13 \) back into \( \omega(t) \) to find the maximum positive angular velocity: \[ \omega(2.13) = 6.40(2.13) - 1.5(2.13)^{2} = 6.826 \text{ rad/s} \].

Key Concepts

Understanding Angular Velocity

Angular velocity refers to how swiftly an object rotates around an axis. It is a critical concept in rotational motion. Imagine a spinning wheel; the speed at which it spins is its angular velocity.
This speed can change over time, which is what we calculate in exercises like the one given for the roller in a printing press.

• The formula for angular velocity, denoted as \( \omega(t) \), involves taking the first derivative of the angle function \( \theta(t) \) with respect to time \( t \).
• In our example, the angle is \( \theta(t) = \gamma t^2 - \beta t^3 \). By differentiating this, we can determine \( \omega(t) \).
• Once you have \( \omega(t) = 6.40t - 1.5t^2 \), it means at any given instant \( t \), you can plug in the time to find out how fast the roller is spinning.

Calculating the exact value helps understand the variations in speed and how this affects the operation of the printing press.

Exploring Angular Acceleration

Angular acceleration describes the rate of change of angular velocity over time. It's similar to how acceleration in straight-line motion indicates a change in speed. Here, angular acceleration signifies how quickly a rotational object's rate of rotation is changing.

• To determine angular acceleration, we take the derivative of the angular velocity \( \omega(t) \).
• In our case, differentiating \( \omega(t) = 6.40t - 1.5t^2 \) gives us \( \alpha(t) = 6.40 - 3.0t \).
• This function, \( \alpha(t) \), tells us how the roller's spin rate is slowing down or speeding up at different times \( t \).

This is vital for understanding how the roller's rotational speed can change, affecting the print quality or efficiency of the press.

The Role of Calculus in Physics

Calculus is a branch of mathematics centered on change. In physics, calculus helps us precisely calculate changes in motion, like the angular velocity and acceleration we covered.

• Taking derivatives is a fundamental part of calculus used to find rates of change, like how we've derived \( \omega(t) \) and \( \alpha(t) \) from \( \theta(t) \).
• In physics, these derivatives let us understand an object's behavior more precisely than static equations. Dynamic systems, such as rotating objects, are modeled effectively using calculus.
• It allows for a deeper insight into motion, enabling predictions and optimizations in systems like machinery and vehicles.

By solving these equations, we can foresee and manipulate physical phenomena, leading to practical engineering solutions.