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The angle through which a rotating wheel has turned in time \(t\) is given by \(\theta=8.5 t-15.0 t^{2}+1.6 t^{4},\) where \(\theta\) is in radians and \(t\) in seconds. Determine an expression \((a)\) for the instantaneous angular velocity \(\omega\) and \((b)\) for the instantaneous angular acceleration \(\alpha .(c)\) Evaluate \(\omega\) and \(\alpha\) at \(t=3.0 \mathrm{~s} . \quad(d)\) What is the average angular velocity, and (e) the average angular acceleration between \(t=2.0 \mathrm{~s}\) and \(t=3.0 \mathrm{~s} ?\)

Short Answer

Expert verified
Expressions are \(\omega(t) = 8.5 - 30.0t + 6.4t^3\) and \(\alpha(t) = -30.0 + 19.2t^2\); \(\omega(3.0) = 91.3\,\text{rad/s}\), \(\alpha(3.0) = 142.8\,\text{rad/s}^2 \); average \(\omega = 37.5\,\text{rad/s}\), average \(\alpha = 91.6\,\text{rad/s}^2\).

Step by step solution

01

Find Expression for Instantaneous Angular Velocity

To find the instantaneous angular velocity \( \omega \), take the derivative of the given position function \( \theta(t) = 8.5t - 15.0t^2 + 1.6t^4 \) with respect to time \( t \). The derivative is: \[ \omega(t) = \frac{d\theta}{dt} = 8.5 - 30.0t + 4(1.6)t^3 = 8.5 - 30.0t + 6.4t^3 \].
02

Find Expression for Instantaneous Angular Acceleration

To find the instantaneous angular acceleration \( \alpha \), take the derivative of the angular velocity \( \omega(t) = 8.5 - 30.0t + 6.4t^3 \) with respect to time \( t \). The derivative is: \[ \alpha(t) = \frac{d\omega}{dt} = -30.0 + 3(6.4)t^2 = -30.0 + 19.2t^2 \].
03

Evaluate Instantaneous Angular Velocity at t=3.0 seconds

Substitute \( t = 3.0 \) seconds into the expression for \( \omega(t) \): \[ \omega(3.0) = 8.5 - 30.0(3.0) + 6.4(3.0)^3 \] Calculate the expression: \[ \omega(3.0) = 8.5 - 90.0 + 6.4 \times 27 \] \[ \omega(3.0) = 8.5 - 90.0 + 172.8 = 91.3 \text{ radians per second} \].
04

Evaluate Instantaneous Angular Acceleration at t=3.0 seconds

Substitute \( t = 3.0 \) seconds into the expression for \( \alpha(t) \): \[ \alpha(3.0) = -30.0 + 19.2(3.0)^2 \] Calculate the expression: \[ \alpha(3.0) = -30.0 + 19.2 \times 9 \] \[ \alpha(3.0) = -30.0 + 172.8 = 142.8 \text{ radians per second squared} \].
05

Calculate Average Angular Velocity between t=2.0 and t=3.0 seconds

The average angular velocity \( \bar{\omega} \) is the change in \( \theta \) over the change in \( t \). Evaluate \( \theta \) at \( t = 3.0 \) and \( t = 2.0 \), then find the difference. \[ \theta(3.0) = 8.5(3.0) - 15.0(3.0)^2 + 1.6(3.0)^4 \] \[ = 25.5 - 135.0 + 129.6 = 20.1 \] \[ \theta(2.0) = 8.5(2.0) - 15.0(2.0)^2 + 1.6(2.0)^4 \] \[ = 17.0 - 60.0 + 25.6 = -17.4 \] Change in \( \theta = 20.1 - (-17.4) = 37.5 \text{ radians} \), \[ t = 3.0 - 2.0 = 1.0 \text{ second} \]. \[ \bar{\omega} = \frac{37.5}{1.0} = 37.5 \text{ radians per second} \].
06

Calculate Average Angular Acceleration between t=2.0 and t=3.0 seconds

The average angular acceleration \( \bar{\alpha} \) is the change in \( \omega \) over the change in \( t \). Use the values from Step 3 and compute \( \omega(2.0) \). \[ \omega(2.0) = 8.5 - 30.0(2.0) + 6.4(2.0)^3 \] \[ = 8.5 - 60.0 + 51.2 = -0.3 \] Change in \( \omega = 91.3 - (-0.3) = 91.6 \text{ radians/s} \), \[ t = 3.0 - 2.0 = 1.0 \text{ second} \]. \[ \bar{\alpha} = \frac{91.6}{1.0} = 91.6 \text{ radians/s}^2 \].

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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 spins around a particular axis. It describes the rate of change of angular displacement and is typically expressed in radians per second. Angular velocity can provide insight into the dynamics of rotational systems, such as wheels or gears in motion.
To find angular velocity, you need to take the derivative of the angular position with respect to time. This tells us how fast the angle is changing at any specific moment. For the given problem, the formula for instantaneous angular velocity is obtained by differentiating the angle function \( \theta(t) = 8.5t - 15.0t^2 + 1.6t^4 \) resulting in \( \omega(t) = \frac{d\theta}{dt} = 8.5 - 30.0t + 6.4t^3 \).
Calculating angular velocity at a specific time, like \( t = 3.0 \) seconds, gives \( \omega(3.0) = 91.3 \text{ radians per second} \). This tells us that at \( t = 3 \) seconds, the wheel's rotation speed is 91.3 radians every second.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. It indicates how quickly an object is speeding up or slowing down its rotation and is measured in radians per second squared. Understanding angular acceleration is key to comprehending the dynamics of rotating systems as it can affect stability and efficiency.
In the exercise, angular acceleration is found by taking the derivative of the angular velocity function. Given \( \omega(t) = 8.5 - 30.0t + 6.4t^3 \), the angular acceleration \( \alpha(t) \) is \( \frac{d\omega}{dt} = -30.0 + 19.2t^2 \). This formula tells us how the speed of rotation is changing at any given instant.
At \( t = 3.0 \) seconds, the angular acceleration is calculated as \( \alpha(3.0) = 142.8 \text{ radians per second squared} \). This value reveals that at this moment, the wheel is accelerating quite rapidly.
Average Angular Velocity
The average angular velocity is a useful concept in understanding the overall rotational motion between two points in time. It is calculated by dividing the total angular displacement by the time interval over which the motion occurs.
For the time period from \( t = 2.0 \) seconds to \( t = 3.0 \) seconds, the change in angular displacement \( \Delta \theta \) is calculated as follows:
  • \( \theta(3.0) = 20.1 \text{ radians} \)
  • \( \theta(2.0) = -17.4 \text{ radians} \)
The resulting change \( \Delta \theta = 20.1 - (-17.4) = 37.5 \text{ radians} \).
Thus, the average angular velocity is \( \bar{\omega} = \frac{37.5}{1.0} = 37.5 \text{ radians per second} \). This tells us that, on average, the wheel rotates 37.5 radians every second between the two time intervals.
Average Angular Acceleration
Average angular acceleration gives an understanding of how a rotational system's speed changes over a time interval. It is defined as the change in angular velocity divided by the time duration.
Using the calculated angular velocities:
  • \( \omega(3.0) = 91.3 \text{ radians per second} \)
  • \( \omega(2.0) = -0.3 \text{ radians per second} \)
We find the change in angular velocity \( \Delta \omega = 91.3 - (-0.3) = 91.6 \text{ radians per second} \).
The average angular acceleration is then determined as \( \bar{\alpha} = \frac{91.6}{1.0} = 91.6 \text{ radians per second squared} \). This shows that, on average, the rotational speed increases by 91.6 radians every second squared over the given time period.

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