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Chapter 9: Problem 4

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?

Short Answer

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

Step by step solution

01

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.
02

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 \[\alpha_z(t) = -2(0.800)t = -1.60t.\]
03

Evaluating Instantaneous Angular Acceleration at t = 3.00s

Substitute \(t = 3.00\) seconds into the angular acceleration function \[\alpha_z(3) = -1.60 \times 3.00 = -4.80\text{ rad/s}^2.\]
04

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 \(\alpha_{av-z}\):\[\alpha_{av-z} = \frac{\omega_z(3) - \omega_z(0)}{3 - 0} = \frac{2.80 - 5.00}{3} = -0.733\overline{3} \text{ rad/s}^2.\]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

An advertisement claims that a centrifuge takes up only \(0.127 \mathrm{~m}\) of bench space but can produce a radial acceleration of \(3000 \mathrm{~g}\) at 5000 rev \(/ \mathrm{min}\). Calculate the required radius of the centrifuge. Is the claim realistic?

Two metal disks, one with radius \(R_1 =\) 2.50 cm and mass \(M_1 =\) 0.80 kg and the other with radius \(R_2 =\) 5.00 cm and mass \(M_2 =\) 1.60 kg, are welded together and mounted on a frictionless axis through their common center (\(\textbf{Fig. P9.77}\)). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block is suspended from the free end of the string. If the block is released from rest at a distance of 2.00 m above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.

You must design a device for shooting a small marble vertically upward. The marble is in a small cup that is attached to the rim of a wheel of radius 0.260 m; the cup is covered by a lid. The wheel starts from rest and rotates about a horizontal axis that is perpendicular to the wheel at its center. After the wheel has turned through 20.0 rev, the cup is the same height as the center of the wheel. At this point in the motion, the lid opens and the marble travels vertically upward to a maximum height \(h\) above the center of the wheel. If the wheel rotates with a constant angular acceleration \(\alpha\), what value of a is required for the marble to reach a height of \(h =\) 12.0 m?

The angle \(\theta\) through which a disk drive turns is given by \(\theta(t) = a + bt - ct^3\), where \(a\), \(b\), and \(c\) are constants, \(t\) is in seconds, and \(\theta\) is in radians. When \(t =\) 0, \(\theta =\) \(\pi\)/4 rad and the angular velocity is 2.00 rad/s. When \(t =\) 1.50 s, the angular acceleration is 1.25 rad/s\(^2\). (a) Find \(a\), \(b\), and \(c\), including their units. (b) What is the angular acceleration when \(\theta =\) \(\pi\)/4 rad? (c) What are u and the angular velocity when the angular acceleration is 3.50 rad/s\(^2\)?

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta{(t) = \gamma t + \beta t^3}\), where \(\gamma =\) 0.400 rad/s and \(\beta =\) 0.0120 rad/s\(^3\). (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega$$_z\) at \(t =\) 5.00 s and the average angular velocity \(\omega_{av-z}\) for the time interval \(t =\) 0 to \(t =\) 5.00 s. Show that \(\omega_{av-z}\) is not equal to the average of the instantaneous angular velocities at \(t =\) 0 and \(t =\) 5.00 s, and explain.
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