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 rad/s and \(\beta =\) 0.800 rad/s\(^3\). (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_z\) at \(t =\) 3.00 s and the average angular acceleration \(\alpha_{av-z}\) for the time interval \(t =\) 0 to \(t =\) 3.00 s. How do these two quantities compare? If they are different, why?

Step-by-Step Solution

Answer

Instantaneous acceleration is -4.80 rad/s²; average is -0.733 rad/s². The difference is because instantaneous measures a specific point, while average is over an interval.

Step 1: Understanding Angular Velocity Function

The angular velocity is given as \(\omega_z(t) = \gamma - \beta t^2\). We know \(\gamma = 5.00\) rad/s and \(\beta = 0.800\) rad/s\(^3\). We will use this equation to find angular acceleration.

Step 2: Finding Angular Acceleration Function

Angular acceleration \(\alpha_z(t)\) is the derivative of angular velocity with respect to time \(t\). Thus, we differentiate \(\omega_z(t)\): \[\alpha_z(t) = \frac{d}{dt}(\gamma - \beta t^2) = 0 - 2\beta t = -2\beta t.\] Plugging in the value of \(\beta\), we get \[\alpha_z(t) = -2(0.800)t = -1.60t.\]

Step 3: Evaluating Instantaneous Angular Acceleration at t = 3.00s

Substitute \(t = 3.00\) seconds into the angular acceleration function \[\alpha_z(3) = -1.60 \times 3.00 = -4.80\text{ rad/s}^2.\]

Step 4: Calculating Average Angular Acceleration for Interval 0 to 3s

The average angular acceleration \(\alpha_{av-z}\) over a time interval is the change in angular velocity divided by the time interval duration. Firstly, evaluate \(\omega_z(0)\) and \(\omega_z(3)\):\[\omega_z(0) = \gamma - \beta (0)^2 = 5.00 \text{ rad/s},\]\[\omega_z(3) = \gamma - \beta (3)^2 = 5.00 - 0.800 \times 9 = 2.80 \text{ rad/s}.\]Then, calculate \(\alpha_{av-z}\):\[\alpha_{av-z} = \frac{\omega_z(3) - \omega_z(0)}{3 - 0} = \frac{2.80 - 5.00}{3} = -0.733\overline{3} \text{ rad/s}^2.\]

Step 5: Comparing Instantaneous and Average Angular Acceleration

The instantaneous angular acceleration at \(t = 3.00\) seconds is \(-4.80 \text{ rad/s}^2\), whereas the average angular acceleration over the interval from \(t = 0\) to \(t = 3.00\) seconds is \(-0.733\overline{3} \text{ rad/s}^2\). These values are different because instantaneous acceleration is calculated at a specific moment, while average acceleration considers the entire interval, smoothing out changes over time.

Key Concepts

angular velocity

Angular velocity refers to how quickly an object rotates or revolves around a point or axis. It's like speed but for rotation. In the context of circular motion, angular velocity, often denoted as \( \omega \), measures how much an angle changes over a certain period, typically in radians per second (rad/s). In our exercise, the angular velocity of the fan blade is expressed mathematically as \( \omega_z(t) = \gamma - \beta t^2 \). This shows that the angular velocity changes with time due to the subtraction of \( \beta t^2 \) from \( \gamma \), the initial constant velocity.

• \( \gamma \) acts as a starting point or initial angular velocity.
• \( \beta t^2 \) represents the deceleration factor depending on time \( t \).

A negative \( \beta t^2 \) indicates that over time, the fan blade's rotation slows down because \( \beta \) increases as \( t \) increases, reducing \( \omega \). This results in a changing rate of spin, which is further analyzed when exploring angular acceleration.

instantaneous acceleration

Instantaneous acceleration is a snapshot of how fast the rate of angular velocity is changing at a specific point in time. Unlike average acceleration, which considers changes over a period, instantaneous acceleration focuses on a particular moment.To find this rate of change for the fan blade, we take the derivative of the angular velocity function. For our exercise, the angular velocity function is given by \( \omega_z(t) = \gamma - \beta t^2 \). When we differentiate this, we are essentially finding the instantaneous angular acceleration \( \alpha_z(t) \) such that:\[\alpha_z(t) = \frac{d}{dt} (\gamma - \beta t^2) = -2\beta t.\]At \( t = 3.00 \) seconds, this gives us a specific numerical value for \( \alpha_z \), calculated as \( \alpha_z(3) = -4.80 \text{ rad/s}^2 \). This reveals precisely how fast the angular velocity of the fan blade is changing at that very second. Remember, this is an instantaneous measure, meaning it's for that one exact point in time.

average acceleration

Average acceleration offers a broader view of how angular velocity changes over a given time span. It's the overall change smoothened out over the time interval. In the exercise, the average angular acceleration from \( t = 0 \) to \( t = 3 \) seconds is computed using the angular velocities at these two times.The formula used here is:\[\alpha_{av-z} = \frac{\omega_z(3) - \omega_z(0)}{3 - 0}.\]By substituting in the values previously calculated for \( \omega_z(3) = 2.80 \text{ rad/s} \) and \( \omega_z(0) = 5.00 \text{ rad/s} \), we find the average angular acceleration to be \( -0.733\overline{3} \text{ rad/s}^2 \). This result helps understand the fan's performance across that interval. Unlike instantaneous acceleration, average acceleration does not tell us about changes at any specific instant, but rather how changes are spread out across time. This means smooth fluctuations in the angular velocity are interpreted as one constant rate of change, making it easier to understand long-term behavior.