Problem 7 - University Physics with Modern Physics

Question

The angle \(\theta\) through which a disk drive turns is given by \(\theta(t) = a + bt - ct^3\), where \(a\), \(b\), and \(c\) are constants, \(t\) is in seconds, and \(\theta\) is in radians. When \(t =\) 0, \(\theta =\) \(\pi\)/4 rad and the angular velocity is 2.00 rad/s. When \(t =\) 1.50 s, the angular acceleration is 1.25 rad/s\(^2\). (a) Find \(a\), \(b\), and \(c\), including their units. (b) What is the angular acceleration when \(\theta =\) \(\pi\)/4 rad? (c) What are u and the angular velocity when the angular acceleration is 3.50 rad/s\(^2\)?

Step-by-Step Solution

Answer

(a) \(a = \frac{\pi}{4}, b = 2.00, c = -0.139 \). (b) 0 rad/s². (c) Angular velocity \(\approx -1.42\) rad/s at \( t = 4.20\). u is undefined.

Step 1: Set Up Equations for Initial Conditions

The angle function is given by \( \theta(t) = a + bt - ct^3 \). Using the initial condition \( \theta(0) = \frac{\pi}{4} \), substitute \( t = 0 \) into the equation: \[ a + b(0) - c(0)^3 = \frac{\pi}{4} \]. Therefore, \( a = \frac{\pi}{4} \).

Step 2: Establish Equation for Angular Velocity

The angular velocity \( \omega(t) \) is the derivative of \( \theta(t) \) with respect to \( t \): \[ \omega(t) = \frac{d\theta}{dt} = b - 3ct^2 \]. Use the condition \( \omega(0) = 2.00 \text{ rad/s} \) \[ b - 3c(0)^2 = 2.00 \]. Thus, \( b = 2.00 \).

Step 3: Create Equation for Angular Acceleration

The angular acceleration \( \alpha(t) \) is the derivative of \( \omega(t) \) with respect to \( t \): \[ \alpha(t) = \frac{d\omega}{dt} = -6ct \]. For the condition \( \alpha(1.50) = 1.25 \text{ rad/s}^2 \), substitute \( t = 1.50 \): \[ -6c(1.50) = 1.25 \]. Solving, \( c = -\frac{1.25}{9} \approx -0.139 \) rad/s\(^3\).

Step 4: Determine Angular Acceleration when \( \theta = \frac{\pi}{4} \)

Find \( t \) such that \( \theta(t) = \frac{\pi}{4} \). With \( a = \frac{\pi}{4} \), \( \theta(t) = \frac{\pi}{4} + 2t + 0.139t^3 \) give an equation \( 0 = 2t - 0.139t^3 \). Solve for \( t \), we get \( t = 0 \) since we know \( \theta(0) = \frac{\pi}{4} \). At \( t = 0 \), \( \alpha = -6c(0) = 0 \).

Step 5: Solve for Angular Velocity and \( u \) when \( \alpha = 3.50 \) rad/s$^2$

From \( \alpha(t) = -6ct \), solve \( -6c t = 3.50 \). Using \( c = -0.139 \), \( t = \frac{3.50}{6 \times 0.139} \approx 4.20 \). Now let's determine the angular velocity: \( \omega = 2 - 3(0.139)(4.20)^2 \approx -1.42 \) rad/s. Unfortunately, without knowing \( u \) explicitly defined in the question context, the remaining calculations focus on known variables.

Key Concepts

Angular Velocity

Angular velocity, denoted by \( \omega(t) \), describes how fast an object rotates or spins around a point. It's crucial to understand this concept because it tells us the rate at which the angle (in radians) through which the object rotates is changing with time. In the context of our exercise, the angular velocity can be found by differentiating the angle function \( \theta(t) \) with respect to time \( t \). This involves calculating the first derivative:\[\omega(t) = \frac{d\theta}{dt} = b - 3ct^2\]Here, \( b \) is a constant representing the initial angular velocity when \( t = 0 \), and \( -3ct^2 \) represents the change in angular velocity over time, given the influence of the constant \( c \). The conditions given in the problem, such as \( \omega(0) = 2.00 \text{ rad/s} \), help determine these constants accurately.

Angular Acceleration

Angular acceleration, symbolized by \( \alpha(t) \), is the rate at which the angular velocity changes over time. Think of it as the rotational equivalent of linear acceleration but for spinning objects. We calculate angular acceleration by taking the derivative of the angular velocity: \[\alpha(t) = \frac{d\omega}{dt} = -6ct\]This expression indicates that angular acceleration is directly proportional to time \( t \), with a factor determined by the constant \( c \). For instance, in our exercise, the constant \( c \) was found using specific conditions (\( \alpha(1.50) = 1.25 \text{ rad/s}^2 \)). Such conditions are often applied to adjust or "tune" the formula to reflect what's happening in the scenario at a particular moment.

Rotational Kinematics

Rotational kinematics refers to the study of motion in objects that rotate, much like linear kinematics involves translational motion. This area of study involves quantities like angle \( \theta \), angular velocity \( \omega \), and angular acceleration \( \alpha \), which parallel the concepts of displacement, velocity, and acceleration in linear motion.Rotational kinematics helps us describe the motion of rotating bodies in a structured way:

• Angle (\( \theta \)): Represents the rotation, often measured in radians.
• Angular Velocity (\( \omega \)): Describes how fast the angle changes.
• Angular Acceleration (\( \alpha \)): Shows how quickly the change of angular velocity is occurring.

Our exercise follows the principles of rotational kinematics by setting up equations for angle \( \theta(t) \), derivative for angular velocity \( \omega(t) \), and further derivative for angular acceleration \( \alpha(t) \), employing initial conditions to solve for unknowns. Recognize the symmetry with linear kinematics: each equation corresponds directly with measures such as position, speed, and acceleration in translation.