91Ó°ÊÓ

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 \(\gamma=0.400 \mathrm{rad} / \mathrm{s}\) and \(\beta=\) 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 \(t=5.00\) s and the average angular velocity \(\omega_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=5.00\) s. Show that \(\omega_{\mathrm{av}-z \text { is }}\) not equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{s},\) and explain why it is not.

Step by step solution

01

Differentiate to find Angular Velocity

Angular velocity \(\omega(t)\) is the time derivative of angle \(\theta(t)\). Differentiating \(\theta(t) = \gamma t + \beta t^3\), we get: \[ \omega(t) = \frac{d}{dt}(\gamma t + \beta t^3) = \gamma + 3\beta t^2. \] Substituting the given values \(\gamma = 0.400\, \mathrm{rad/s}\) and \(\beta = 0.0120\, \mathrm{rad/s^3}\), the equation becomes: \[ \omega(t) = 0.400 + 3(0.0120)t^2 = 0.400 + 0.036t^2. \]

02

Calculate Initial Angular Velocity

The initial angular velocity is \(\omega(0)\). Substitute \(t = 0\) into the angular velocity equation: \[ \omega(0) = 0.400 + 0.036 \times 0^2 = 0.400\, \mathrm{rad/s}. \]

03

Find Instantaneous Angular Velocity at t = 5 s

Substitute \(t = 5\, \mathrm{s}\) into the angular velocity equation to find \(\omega(5)\): \[ \omega(5) = 0.400 + 0.036 \times (5)^2 = 0.400 + 0.036 \times 25 = 1.300\, \mathrm{rad/s}. \]

04

Calculate Average Angular Velocity for t = 0 to t = 5 s

Average angular velocity over a time interval is given by: \[ \omega_{\text{av}} = \frac{\theta(t_{\text{final}}) - \theta(t_{\text{initial}})}{t_{\text{final}} - t_{\text{initial}}}. \]\( \theta(5) = \gamma (5) + \beta (5)^3 = 0.400 \times 5 + 0.0120 \times 125 = 2.000 + 1.500 = 3.500 \, \mathrm{rad}\) and \(\theta(0) = 0\). Thus, \[\omega_{\text{av}} = \frac{3.500 - 0}{5 - 0} = 0.700\, \mathrm{rad/s}. \]

05

Compare Average Angular Velocity with Instantaneous Values

The average of instantaneous angular velocities at \(t = 0\) and \(t = 5\) is: \[\frac{\omega(0) + \omega(5)}{2} = \frac{0.400 + 1.300}{2} = 0.850\, \mathrm{rad/s}.\] Since \(0.850\, \mathrm{rad/s} eq 0.700\, \mathrm{rad/s}\), they are not equal. This discrepancy is because average angular velocity considers the entire trajectory, not just start and end points, and \(\omega(t)\) is not linear over time but quadratic.

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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 an essential concept in rotational motion, akin to linear velocity in linear motion. Angular velocity specifies both direction and rate of rotation.

In the exercise, the angular velocity of the merry-go-round is obtained by differentiating the angular displacement function, \(\theta(t)=\gamma t+\beta t^3\), with respect to time. This yields the angular velocity function \(\omega(t) = \gamma + 3\beta t^2\). Here's what it means:

• \(\gamma\) represents the constant part of angular velocity, similar to a starting pace.
• The term \(+ 3\beta t^2\) introduces a time-dependent change, suggesting that the rate of rotation increases as time progresses.

To understand this better, think of the merry-go-round spinning faster over time due to the \(t^2\) term, showing non-linear acceleration in the rotation.

Differentiation in Physics

Differentiation is a powerful mathematical tool used in physics to find rates of change, especially in motion-related problems. In the context of the exercise, differentiation helps derive the angular velocity from the displacement function.

Here's how differentiation works in this scenario:

• The angular displacement \(\theta(t)=\gamma t+\beta t^3\) represents how far the merry-go-round has turned over time.
• When you differentiate \(\theta(t)\) with respect to time \(t\), you obtain the time rate of change of angle, known as angular velocity \(\omega(t)\).

This process highlights how instantaneous changes (e.g., how fast something is spinning at any given moment) are extracted from a time-dependent position or displacement function.

Average Velocity

In rotational motion, average velocity describes how far an object rotates over a certain time period. It provides a simpler, overall picture of motion within an interval, rather than detailing speed at every instant.

For the merry-go-round, the average angular velocity is calculated using the formula:

• \(\omega_{\text{av}} = \frac{\theta(t_{\text{final}}) - \theta(t_{\text{initial}})}{t_{\text{final}} - t_{\text{initial}}}\)

This equation takes the total angular displacement over a given time interval and divides it by the time difference. It’s the rotational counterpart to the linear average velocity, which you'd use on a straight path.

The average velocity is a straightforward way to grasp the overall spin effect during that period without analyzing motion step-by-step.

Instantaneous Velocity

Instantaneous velocity reveals the exact rate of rotation at a specific time. It is akin to checking your speedometer over time during a car ride to see exactly how fast you are going at that moment.

In the merry-go-round problem, the instantaneous angular velocity is found from the angular velocity function \(\omega(t) = 0.400 + 0.036t^2\). To explore this:

• By substituting a specific time into \(\omega(t)\), for instance at \(t=5s\), you determine the merry-go-round's precise spinning speed at that second.

In contrast to average velocity, which covers a whole time span, instantaneous velocity focuses on now, allowing us to capture how the speed varies instantaneously.

Understanding instantaneous velocity is crucial when analyzing motion that changes over time, as it provides insight into the dynamism of rotational motion.