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During a certain period of time, the angular position of a swinging door is described by \(\theta=5.00+10.0 t+2.00 t^{2}\) , where \(\theta\) is in radians and \(t\) is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at \(t=0\) (b) at \(t=3.00 \mathrm{s}\) .

Short Answer

Expert verified
At t=0, \(\theta = 5.00\) radians, \(\omega = 10.0\) rad/s, \(\alpha = 4.00\) rad/s^2. At t=3.00 s, \(\theta = 53.00\) radians, \(\omega = 22.0\) rad/s, \(\alpha = 4.00\) rad/s^2.

Step by step solution

01

Identify the given equation

Note the given angular position equation of the swinging door: \(\theta = 5.00 + 10.0t + 2.00t^2\). This equation shows that \(\theta\) is a function of time, represented as a quadratic equation with time-dependent coefficients.
02

Find the angular position at t=0

Substitute \(t = 0\) into the given equation to find the initial angular position: \(\theta(0) = 5.00 + 10.0(0) + 2.00(0)^2 = 5.00\) radians.
03

Compute the angular speed (velocity)

Differentiate the given equation with respect to time to obtain the angular speed: \(\omega(t) = \frac{d\theta}{dt} = 0 + 10.0 + 2(2.00t) = 10.0 + 4.00t\).
04

Find the angular speed at t=0

Substitute \(t = 0\) into the angular speed equation: \(\omega(0) = 10.0 + 4.00(0) = 10.0\) radians per second.
05

Determine the angular acceleration

Differentiate the angular speed equation with respect to time to get the angular acceleration: \(\alpha(t) = \frac{d\omega}{dt} = 0 + 4.00 = 4.00\) radians per second squared, which is constant.
06

Find the angular position at t=3.00s

Substitute \(t = 3.00\) seconds into the given equation to find the angular position at \(t=3.00\) seconds: \(\theta(3.00) = 5.00 + 10.0(3.00) + 2.00(3.00)^2 = 5.00 + 30.00 + 18.00 = 53.00\) radians.
07

Find the angular speed at t=3.00s

Substitute \(t = 3.00\) seconds into the angular speed equation: \(\omega(3.00) = 10.0 + 4.00(3.00) = 10.0 + 12.00 = 22.0\) radians per second.

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Key Concepts

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

Angular Position
The concept of angular position is central to understanding rotary motion. It is the angle in radians or degrees that an object, like a swinging door, has rotated from a reference orientation. In physics, angular position is denoted by the symbol \( \theta \) and is particularly useful for objects moving in a circular path.

For example, if a door opens from a closed position (our reference orientation) to form a 30-degree angle with the frame, its angular position is 30 degrees. It's important to note that angular position is different from linear position; it doesn't tell you how far away something is, but rather how far it's rotated around a point or axis.
Angular Speed
Angular speed is the rate at which an object changes its angular position, indicating how fast it is spinning or rotating. This is similar to linear speed but for rotational motion. It's measured in radians per second (rad/s) or degrees per second.

To calculate angular speed, one can derive the equation for angular position with respect to time. The equation \( \omega(t) = 10.0 + 4.00t \), obtained from the derivative of the angular position equation, provides us with the angular speed as a function of time.
Angular Acceleration
Angular acceleration, denoted by \( \alpha \), measures the rate of change of angular speed. In other words, it describes how quickly the rotation of an object is speeding up or slowing down. It's expressed in radians per second squared (rad/s^2).

The angular acceleration can be constant or time-dependent, as is the case for many types of angular motion. In the given swinging door example, the angular acceleration is a constant \( 4.00 \) rad/s^2, which means that the angular speed of the door is increasing at a steady rate.
Quadratic Equation in Physics
Quadratic equations are commonly used in physics to describe motions where the acceleration is constant. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents an unknown variable.

In our exercise, the angular position equation \( \theta(t) = 5.00 + 10.0t + 2.00t^2 \) resembles a quadratic equation, with \( t \) being the variable that represents time. This indicates that the door's rotation includes an acceleration component, which can be seen in the term with \( t^2 \). Solving quadratic equations helps us understand the motion of objects under uniform acceleration.
Time-Dependence in Motion
Time-dependence in motion refers to how an object’s motion properties such as position, speed, and acceleration change over time. When these properties are expressed as functions of time, we can predict the future state of the moving object.

In rotational motion, time-dependent expressions allow us to understand how the angular position and speed of an object such as our swinging door evolve. Initial conditions, like being at rest with \( \theta(0) = 5.00 \) radians, determine how subsequent calculations for specific times are made. This shows how, with an understanding of time-dependent motion, one can accurately describe the complete state of a moving object at any given moment.

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