/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 A fan blade rotates with angular... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\gamma-\beta t^{2}, \quad\) where \(\quad \gamma=5.00 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.800 \mathrm{rad} / \mathrm{s}^{3}\) (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 \(\alpha_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=3.00 \mathrm{~s}\). How do these two quantities compare? If they are different, why?

Short Answer

Expert verified
The angular acceleration function is \(\alpha(t) = -2\beta t\). The instantaneous angular acceleration at \(t=3.00s\) and the average angular acceleration from \(t=0\) to \(t=3.00s\) are computed based on this function. A comparison between these two quantities reveals their relationship and variations during a specific time interval.

Step by step solution

01

Derive the angular acceleration function

First, we should recall that angular acceleration \( \alpha \) is the rate of change of angular velocity \( \omega \) with respect to time. In terms of calculus, it's the derivative of angular velocity with respect to time. Therefore, we can derive the angular acceleration function by taking the derivative of the given angular velocity function \( \omega = \gamma - \beta t^{2} \). After taking the derivative with respect to time, we get: \( \alpha(t) = -2\beta t \) .
02

Calculate the instantaneous angular acceleration at t=3s

Substitute \(t=3.00s\) into the angular acceleration function \(\alpha(t) = -2\beta t\), we get the instantaneous angular acceleration at t=3s: \(\alpha(3) = -2\beta * 3 \)
03

Calculate the average angular acceleration between t=0 and t=3s

The average angular acceleration in a time interval is defined as the change in angular velocity divided by the change in time. So, since the angular velocity function is given by \(\omega(t) = \gamma - \beta t^{2}\), we calculate the change in angular velocity between t=0 and t=3.00s as \( \Delta \omega = \omega(3) - \omega(0) \), then divide by the time interval to find the average angular acceleration between t=0 and t=3s: \(\alpha_{av} = \frac{\Delta \omega}{3} \)
04

Comparison of the instantaneous and average angular accelerations

After calculating the instantaneous and average angular acceleration, we need to compare these two quantities. The differences, if any, are likely due to the fact that angular velocity varies with time, suggesting a changing angular acceleration over the time interval of interest.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates or revolves around an axis. It is usually denoted by the Greek letter omega (\(\omega\)) and is typically expressed in radians per second (rad/s). For any rotating object, angular velocity represents the rate of change of the angular position with respect to time.
In the context of the original exercise, the angular velocity of the fan blade is given by:
  • \(\omega_{z}(t) = \gamma - \beta t^{2}\)
Here, \(\gamma\) is the initial angular velocity, and \(\beta\) is a constant that affects how the angular velocity changes over time.
The minus sign before \(\beta t^2\) indicates that as time increases, the angular velocity decreases quadratically due to the \(t^2\) term. Understanding angular velocity is crucial for analyzing the rotational motion of objects and predicting their future states.
Angular Acceleration
Angular acceleration is the rate at which the angular velocity of an object changes with time. It is represented by \(\alpha\) and is measured in radians per second squared (rad/s²). Angular acceleration can be calculated by taking the derivative of the angular velocity with respect to time.
In the provided problem, the angular velocity equation is:
  • \(\omega_{z}(t) = \gamma - \beta t^{2}\)
To find the angular acceleration, we derive this equation with respect to time, resulting in:
  • \(\alpha(t) = -2\beta t\)
This function tells us that the angular acceleration is directly proportional to time \(t\), and the negative sign indicates that the acceleration is decreasing. This means as time progresses, the angular velocity slows down at a steady rate determined by the constants \(\beta\) and \(t\). Recognizing how angular acceleration works helps to predict how quickly or slowly the angular velocity can change over a specific timeframe.
Calculus in Physics
Calculus plays a vital role in understanding and solving physics problems, especially those involving motion. It allows us to handle changing rates, which are common in real-world scenarios. In the context of rotational motion, calculus helps us transition from angular velocity (\(\omega\)) to angular acceleration (\(\alpha\)).
Laid simply, calculus enables us to take derivatives and integrals:
  • Derivative: Provides the rate of change of a function. For this exercise, the derivative of angular velocity with respect to time gives us angular acceleration.
  • Integral: Provides the accumulated change. Although not directly used in this problem, integrals can calculate changes over time intervals, such as finding displacement.
Understanding differential calculus in physics is essential because it connects various motion descriptors, helping us move from one property (like velocity) to another (like acceleration) efficiently. This provides a more complete picture of an object's state of motion over time.
Instantaneous vs Average Values
Understanding the difference between instantaneous and average values is crucial when analyzing changes over time.
Instantaneous values refer to those that exist at a specific moment, giving precise immediate behavior, while average values represent the overall behavior over a time interval.
  • Instantaneous Value: For angular acceleration, it describes the rate of change of angular velocity at a particular time, say \(t = 3s\). Here, you substitute \(t = 3s\) into the acceleration function \(\alpha(t) = -2\beta t\) to get \(\alpha(3s)\).
  • Average Value: This is the total change in angular velocity divided by the total time period (from \(t = 0\) to \(t = 3s\)), given in the formula \(\alpha_{av} = \frac{\Delta \omega}{3}\).
Differences between instantaneous and average angular acceleration stem from the fact that angular velocity changes non-linearly over time. Thus, the instantaneous acceleration at a specific time may differ from the average over a period due to cumulative effects and variability in the rate of change. Comprehending these differences helps one analyze rotational motion more accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(v=1.25 \mathrm{~m} / \mathrm{s} .\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the \(\mathrm{CD}\) is played. (See Exercise \(9.20 .\) ) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta)=r_{0}+\beta \theta,\) where \(r_{0}\) is the radius of the spiral at \(\theta=0\) and \(\beta\) is a constant. On a \(\mathrm{CD}, r_{0}\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d \theta,\) the distance scanned along the track is \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) since the track is scanned at a constant linear speed \(v,\) the distance \(s\) found in part (a) is equal to vi. Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta ;\) choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_{z}\) and the angular acceleration \(\alpha_{z}\) as functions of time. Is \(\alpha_{z}\) constant? (d) On a CD, the inner radius of the track is \(25.0 \mathrm{~mm}\), the track radius increases by \(1.55 \mu \mathrm{m}\) per revolution, and the playing time is \(74.0 \mathrm{~min} .\) Find \(r_{0}, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_{z}\) (in rad/s) versus \(t\) and \(\alpha_{z}\) (in rad/s \(^{2}\) ) versus \(t\) between \(t=0\) and \(t=74.0 \mathrm{~min}\)

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius \(12.0 \mathrm{~cm} .\) If the angular speed of the front sprocket is 0.600 rev \(/ \mathrm{s},\) what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be \(5.00 \mathrm{~m} / \mathrm{s} ?\) The rear wheel has radius \(0.330 \mathrm{~m}\).

A thin, light wire is wrapped around the rim of a wheel as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius \(0.180 \mathrm{~m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{~kg} \cdot \mathrm{m}^{2}\). A small block with mass \(0.340 \mathrm{~kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-9.00 \mathrm{~J}\) of work as the block descends \(3.00 \mathrm{~m}\). What is the magnitude of the angular velocity of the wheel after the block has descended \(3.00 \mathrm{~m} ?\)

A bucket of mass \(m\) is tied to a massless cable that is wrapped around the outer rim of a uniform pulley of radius \(R,\) on a frictionless axle, similar to the system shown in Fig. E9.47. In terms of the stated variables, what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket?

Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7} \mathrm{~m}\) deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are \(25.0 \mathrm{~mm}\) and \(58.0 \mathrm{~mm}\), respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(1.25 \mathrm{~m} / \mathrm{s}\). (a) What is the angular speed of the \(\mathrm{CD}\) when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration \(\mathrm{CD}\) if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum-duration CD during its 74.0 min playing time? Take the direction of rotation of the disc to be positive.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.