/*! 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 5 A child is pushing a merry-go-ro... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\theta(t)=\gamma t+\beta t^{3}, \quad\) where \(\quad \gamma=0.400 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) (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 \(t=5.00 \mathrm{~s}\) and the average angular velocity \(\omega_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{~s}\) Show that \(\omega_{\mathrm{av}-z}\) is \(n o t\) equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{~s},\) and explain.

Short Answer

Expert verified
The angular velocity as a function of time is given by the equation \(\omega(t) = \gamma + 3\beta t^2\), the initial angular velocity (at \(t=0\)) is \(\gamma\), the instantaneous angular velocity at \(t=5\) is \(\omega(5) = \gamma + 3\beta(5)^2\), and the average angular velocity between \(t = 0\) and \(t = 5\) is \(\omega_{av} = \frac{\theta(5)-\theta(0)}{5}\). The averaged instantaneous angular velocities at \(t=0\) and \(t=5\) will not be equal to the calculated average angular velocity since the angular velocity \(\omega(t)\) isn't constant and changes over time.

Step by step solution

01

Angular velocity as a function of time

The angular velocity is the rate of change of angle with respect to time. So, differentiate the function \(\theta(t)=\gamma t +\beta t^3\) with respect to time. Therefore, the angular velocity, denoted by \(\omega(t)\), is given as \(\omega(t) = \frac{d\theta(t)}{dt} = \gamma + 3\beta t^2\)
02

Initial value of the angular velocity

Now calculate the initial value of the angular velocity. This means calculating the value of \(\omega(t)\) at \(t=0\). Substituting \(t = 0\) in \(\omega(t)\) will provide the initial angular velocity.
03

Calculating Instantaneous and Average Angular velocity

Calculate the instantaneous angular velocity at time \(t = 5.00 s\) by substituting \(t = 5\) in the \(\omega(t)\) equation. \n To find the average angular velocity, use the formula: \(\omega_{av} = \frac{\theta}{t}\). To calculate for \(t = 0\) to \(t = 5.00 s\), substitute these values into the function \(\theta(t)\) to get \(\theta\), and then divide by the total time.
04

Verifying Omega and explaining

Substitute \(t = 0\) and \(t=5\) into \(\omega(t)\) to get the instantaneous angular velocities at these times. Average these two angular velocities and compare it with the earlier calculated average angular velocity. If they're not same, it shows that the instantaneous angular velocity isn't constant and changes over time.

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Key Concepts

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

Instantaneous Angular Velocity
Instantaneous angular velocity is the angular velocity of a rotating object at a particular moment in time. It's defined as the rate at which the angle (theta) changes with respect to time, denoted as \(\theta'(t)\) or \(\frac{d\theta}{dt}\). In simple terms, it tells you how fast the object is rotating at that specific instant.
For example, if we consider a merry-go-round, its instantaneous angular velocity would be how fast it is spinning at a given second. If a child pushes the merry-go-round harder, the instantaneous angular velocity would increase. Whereas, if the merry-go-round encounters resistance, like friction, the instantaneous velocity could decrease.
In our exercise, the instantaneous angular velocity \(\omega(t)\) is computed by differentiating the provided angular position function \(\theta(t)\) with respect to time. This differentiation yields \(\omega(t) = \gamma + 3\beta t^2\), where \(\gamma\) and \(\beta\) are constants specific to how the merry-go-round was pushed.
Average Angular Velocity
Unlike instantaneous angular velocity which considers a specific moment, average angular velocity is concerned with the overall rate of change of the angular position over a given time interval. It is calculated by taking the total angle through which an object has rotated and dividing it by the total time it took to rotate that angle.
The formula for average angular velocity \(\omega_{\text{av}}\) can be given by \(\omega_{\text{av}} = \frac{\theta_{\text{final}} - \theta_{\text{initial}}}{t_{\text{final}} - t_{\text{initial}}}\), where \(\theta_{\text{final}}\) and \(\theta_{\text{initial}}\) are the angular positions at the final and initial times, respectively.

Understanding Through Exercise

Returning to our merry-go-round scenario, for a given time period from \(t = 0\) to \(t = 5.00 s\), the average angular velocity is not simply the 'mean' of the instantaneous angular velocities at the start and end times. This is because the merry-go-round doesn't have a uniform acceleration due to the \(\beta t^3\) term in the angle equation. Therefore, we must calculate the average angular velocity by considering the angle turned over the entire 5 seconds—not just two snapshots.
Kinematics of Rotational Motion
Kinematics of rotational motion describes the motion of objects that rotate about an axis, such as wheels, gears, or merry-go-rounds. This field of physics concerns itself with quantities like rotational angle, angular velocity, and angular acceleration, without delving into the forces that cause the motion.

Angular Displacement and Velocity

In rotational motion, angular displacement represents how far an object has rotated and is measured in radians. Angular velocity, discussed earlier, corresponds to how swiftly this displacement occurs. Meanwhile, angular acceleration describes how quickly the angular velocity itself changes over time.
When we analyze the kinematics of an accelerating merry-go-round, as in our exercise, we include terms in our equations for both constant angular velocity and for angular acceleration. This lets us predict future positions and velocities of the merry-go-round, which is incredibly important in designing and understanding mechanical systems.
Remember, the equations we employ in kinematics, just like the one used in our example exercise, \(\theta(t) = \gamma t + \beta t^3\), are indispensable tools for describing the rich landscape of rotational dynamics in a clear and precise way.

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Most popular questions from this chapter

A fan blade rotates with angular velocity given by \(\omega_{z}(t)=\gamma-\beta t^{2}, \quad\) 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?

A wheel of diameter \(40.0 \mathrm{~cm}\) starts from rest and rotates with a constant angular acceleration of \(3.00 \mathrm{rad} / \mathrm{s}^{2}\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{\mathrm{rad}}=\omega^{2} r\) and (b) \(a_{\mathrm{rad}}=v^{2} / r\)

The earth is approximately spherical, with a diameter of \(1.27 \times 10^{7} \mathrm{~m} .\) It takes 24.0 hours for the earth to complete one revolution. What are the tangential speed and radial acceleration of a point on the surface of the earth, at the equator?

Three small blocks, each with mass \(m\), are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point onefourth of the length from one end.

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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