Problem 4 - University Physics with Modern Physics

Question

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\) \(\gamma-\beta t^{2},\) where \(\gamma=5.00 \mathrm{rad} / \mathrm{s}\) and \(\beta=0.800 \mathrm{rad} / \mathrm{s}^{3} .\) (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_{z}\) at \(t=3.00 \mathrm{s}\) and the average angular acceleration \(\alpha_{\mathrm{axz}}\) for the time interval \(t=0\) to \(t=3.00 \mathrm{s}\) . How do these two quantities compare? If they are different, why are they different?

Step-by-Step Solution

Answer

Angular acceleration: \( \alpha(t) = -1.60t \), Instantaneous at \( t=3s \): \( -4.80 \; \text{rad/s}^{2} \), Average from \( t=0 \) to \( t=3s \): \( -2.40 \; \text{rad/s}^{2} \). They differ because acceleration changes with time.

Step 1: Understanding Angular Acceleration

Angular acceleration, denoted as \( \alpha(t) \), is the derivative of angular velocity \( \omega(t) \) with respect to time \( t \). Thus, to find \( \alpha(t) \), we need to differentiate \( \omega_{z}(t) = \gamma - \beta t^2 \).

Step 2: Calculating Angular Acceleration

Differentiate the given angular velocity function, \( \omega_{z}(t) = \gamma - \beta t^2 \). The derivative is: \[\alpha(t) = \frac{d}{dt}(\gamma - \beta t^2) = -2\beta t\]Substitute the value of \( \beta = 0.800 \) rad/s³:\[\alpha(t) = -2 \times 0.800 \times t = -1.60 t\]Thus, the angular acceleration as a function of time is \( \alpha(t) = -1.60 t \; \text{rad/s}^2 \).

Step 3: Finding Instantaneous Angular Acceleration at t=3s

To find the instantaneous angular acceleration at \( t = 3.00 \) s, substitute \( t = 3.00 \) into the expression for \( \alpha(t) \):\[\alpha(3) = -1.60 \times 3.00 = -4.80 \; \text{rad/s}^2\]The instantaneous angular acceleration at \( t = 3.00 \; \text{s} \) is \( -4.80 \; \text{rad/s}^{2} \).

Step 4: Calculating Average Angular Acceleration

The average angular acceleration over an interval \( [t_0, t_f] \) is given by:\[\alpha_{ ext{avg}} = \frac{\omega(t_f) - \omega(t_0)}{t_f - t_0}\]First, find \( \omega_{z}(0) \) and \( \omega_{z}(3) \):\[\omega_{z}(0) = \gamma = 5.00 \; \text{rad/s} \quad \text{and} \quad \omega_{z}(3) = 5.00 - 0.800 \times (3.00)^2 = 5.00 - 7.20 = -2.20 \; \text{rad/s}\]Now calculate \( \alpha_{ ext{avg}} \):\[\alpha_{ ext{avg}} = \frac{-2.20 - 5.00}{3.00 - 0} = \frac{-7.20}{3.00} = -2.40 \; \text{rad/s}^{2}\]The average angular acceleration over the interval from \( t=0 \) to \( t=3.00 \) s is \( -2.40 \; \text{rad/s}^{2} \).

Step 5: Comparing Instantaneous and Average Angular Acceleration

The instantaneous angular acceleration at \( t = 3.00 \) s is \( -4.80 \; \text{rad/s}^{2} \), while the average angular acceleration over \( 0 \leq t \leq 3 \) s is \( -2.40 \; \text{rad/s}^{2} \). These values differ because the angular acceleration is a linear function of time (\( -1.60t \)), meaning it changes over time, leading to different instantaneous and average values.

Key Concepts

Angular Velocity

Angular velocity is a measure of how fast an object rotates or revolves relative to another point, usually the center of a circle. It describes the rate of change of angular position and is typically measured in radians per second (rad/s). In the given problem, the angular velocity function is presented as \[ \omega_{z}(t) = \gamma - \beta t^{2} \]where \( \gamma \) and \( \beta \) are constants.

This mathematical expression shows how the angle changes with respect to time, emphasizing that angular velocity can depend on time. This dependency reflects how the speed of rotation might slow down or speed up during movement.

• \( \gamma = 5.00 \, \mathrm{rad/s} \) is the initial angular velocity.
• \( \beta = 0.800 \, \mathrm{rad/s^{3}} \) determines how quickly the velocity changes with time squared.

Understanding this helps to predict how the speed or rotation of an object varies over time.

Instantaneous Acceleration

Instantaneous acceleration is the rate of change of velocity at a specific moment in time. For angular motions, instantaneous angular acceleration describes how the rotational velocity changes at an exact point in time and is given by the derivative of the angular velocity function. In this problem, we calculated it using \[\alpha(t) = -2\beta t\] which becomes \(\alpha(3) = -1.60 \times 3 = -4.80 \, \mathrm{rad/s^{2}} \)at \( t = 3.00 \mathrm{s} \).

This instantaneous acceleration tells us how rapidly the fan blade's speed of rotation is changing exactly at this time. It gives a precise snapshot of acceleration, revealing moments where the change is most significant.

Average Acceleration

Average acceleration provides a measurement of the change in velocity over a specific time period, showing the overall trend of how the velocity shifts from start to finish within that interval. It's given by the formula:\[\alpha_{\text{avg}} = \frac{\omega(t_f) - \omega(t_0)}{t_f - t_0}\]For our problem, using \( t_0 = 0 \) and \( t_f = 3 \mathrm{s} \), we calculated:\[\alpha_{\text{avg}} = \frac{-2.20 - 5.00}{3.00} = -2.40 \, \mathrm{rad/s^{2}}\]

Average angular acceleration offers insight into overall trends during rotations or movements over a specified period, contrasting with instantaneous measurements by revealing gradual changes rather than immediate fluctuations.

Derivative of Angular Velocity

The derivative of angular velocity with respect to time is what defines angular acceleration. Differentiation allows us to understand how a function changes at any given point, and in this context, it explains how the rate of rotation evolves over time.

For the exercise, the angular velocity function \( \omega_{z}(t) = \gamma - \beta t^2 \) was differentiated to yield the angular acceleration:\[\alpha(t) = \frac{d}{dt}(\gamma - \beta t^2) = -2\beta t\]This process highlights the importance of calculus in physics, enabling predictions of dynamic behaviors through functional analysis.

Time-Dependent Acceleration

Time-dependent acceleration means that the acceleration of the object changes with time, rather than remaining constant. In the given problem, angular acceleration \( \alpha(t) \) is expressed as \( -1.60 t \, \mathrm{rad/s^{2}} \), which is clearly a time-dependent function. This indicates that as time goes by, the acceleration is not constant but changes linearly.

Understanding this concept is crucial because it tells us that decisions made regarding motion analysis, design, or safety must account for varying accelerations.

• Allows real-time adjustments and precise control for rotating systems.

Knowing the nature of such change helps enhance predictions and improve around mechanical systems or processes.