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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.

Short Answer

Expert verified
(a) \( \omega(t) = 0.400 + 3 \times 0.0120 \times t^2 \). (b) 0.400 rad/s. (c) Instantaneous: 1.300 rad/s, Average: 0.7 rad/s. These differ due to non-linear acceleration.

Step by step solution

01

Expression for Angular Velocity

The angular velocity \( \omega(t) \) is the derivative of the angle \( \theta(t) \) with respect to time. Thus, we have: \[ \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt} (\gamma t + \beta t^3) \]Differentiating gives:\[ \omega(t) = \gamma + 3 \beta t^2 \]
02

Initial Angular Velocity

To find the initial angular velocity, substitute \( t = 0 \) into the expression for \( \omega(t) \):\[ \omega(0) = \gamma + 3 \beta (0)^2 = \gamma = 0.400 \, \text{rad/s} \]
03

Instantaneous Angular Velocity at t = 5 s

Substitute \( t = 5.00 \,s \) into the expression for \( \omega(t) \) to find the instantaneous angular velocity:\[ \omega(5) = 0.400 + 3(0.0120)(5)^2 \]Calculate:\[ \omega(5) = 0.400 + 3 \times 0.0120 \times 25 = 0.400 + 0.900 = 1.300 \, \text{rad/s} \]
04

Average Angular Velocity Over Interval

The average angular velocity \( \omega_{av-z} \) over the interval from \( t = 0 \) to \( t = 5 \) is given by:\[ \omega_{av-z} = \frac{\theta(5) - \theta(0)}{5 - 0} \]Calculate \( \theta(5) \):\[ \theta(5) = 0.400 \times 5 + 0.0120 \times (5)^3 = 2.0 + 1.5 = 3.5 \, \text{radians} \] Since \( \theta(0) = 0 \),\[ \omega_{av-z} = \frac{3.5 - 0}{5} = 0.7 \, \text{rad/s} \]
05

Compare Averages

The average of the instantaneous angular velocities at \( t = 0 \) and \( t = 5 \) is:\[ \frac{\omega(0) + \omega(5)}{2} = \frac{0.400 + 1.300}{2} = 0.850 \, \text{rad/s} \]This differs from \( \omega_{av-z} = 0.7 \, \text{rad/s} \) because the average angular velocity is affected by the acceleration term \( \beta t^3 \), which changes the rate of rotation non-linearly over time.

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Key Concepts

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

Understanding Rotational Kinematics
Rotational kinematics is the branch of physics that deals with the motion of objects that rotate or spin. Objects like merry-go-rounds rotate around an axis, and rotational kinematics helps describe this motion.
  • Key terms include angular position, angular velocity, and angular acceleration.
  • The equations and concepts are similar to those used in linear kinematics but are adapted for rotation.
By knowing the angular position of an object over time, we can calculate its angular velocity and acceleration, giving us a complete picture of its rotational motion.
Exploring Angular Displacement
Angular displacement refers to the change in the angle through which an object has rotated, measured in radians. It helps us understand how much rotation has occurred.For a merry-go-round, if the initial angle is zero, the angular displacement can be calculated using the function \[ \theta(t) = \gamma t + \beta t^3 \]where \( \gamma \) and \( \beta \) are coefficients that represent the constant and timedependent components of the motion, respectively.
  • This equation integrates time to determine how the angle changes as the object turns.
  • Understanding angular displacement helps in predicting future positions and in calculating other dynamic factors like velocity and acceleration.
Instantaneous Velocity Clarified
Instantaneous velocity is the speed of an object at a specific moment in time. For rotational motion, this is called angular velocity and is denoted as \( \omega(t) \).The angular velocity indicates how quickly the object is rotating at any given point, and can be derived by differentiating the expression for angular displacement:\[ \omega(t) = \frac{d\theta}{dt} = \gamma + 3\beta t^2 \]
  • This formula reveals that angular velocity changes with time due to the term \( 3\beta t^2 \).
  • At \( t = 5\) seconds, for instance, this function gives us the precise rotational speed at that moment, helping gauge performance or synchronization in mechanical systems.
Calculating Average Velocity
Average angular velocity over a period gives a broad picture of how an object has been rotating, smoothing out its speed variations into a single value.This is calculated by the change in angular displacement over time:\[ \omega_{av-z} = \frac{\theta(5) - \theta(0)}{5} \]
  • This formula computes the average rate of rotation from start to finish of a given time span.
  • Crucially, this value can differ from the mean of instantaneous velocities at the start and end times due to the nature of non-uniform motion.
Understanding both instantaneous and average angular velocities allows us to analyze rotational motion in different contexts, from simple machinery to complex engineering systems.

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Most popular questions from this chapter

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?

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm. The power is off for 30.0 s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

A compound disk of outside diameter 140.0 cm is made up of a uniform solid disk of radius 50.0 cm and area density 3.00 g/cm\(^2\) surrounded by a concentric ring of inner radius 50.0 cm, outer radius 70.0 cm, and area density 2.00 g/cm\(^2\). Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is 60.0 cm long and has mass 0.400 kg. (a) What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod? (b) One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a 60.0\(^\circ\) angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the V at its vertex?

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

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