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The angular acceleration of a wheel is \(\alpha=6.0 t^{4}-4.0 t^{2}\), with \(\alpha\) in radians per second-squared and \(t\) in seconds. At time \(t=0\), the wheel has an angular velocity of \(+2.0 \mathrm{rad} / \mathrm{s}\) and an angular position of \(+1.0 \mathrm{rad}\). Write expressions for (a) the angular velocity \((\mathrm{rad} / \mathrm{s})\) and (b) the angular position (rad) as functions of time (s).

Short Answer

Expert verified
The angular velocity is \(\omega(t) = \frac{6.0}{5} t^5 - \frac{4.0}{3} t^3 + 2.0\; \text{rad/s}\). The angular position is \(\theta(t) = \frac{1.0}{5} t^6 - \frac{1.0}{3} t^4 + 2.0t + 1.0\; \text{rad}\).

Step by step solution

01

Understand Initial Conditions

At time \(t=0\), the angular velocity \(\omega\) is \(+2.0 \, \text{rad/s}\) and the angular position \(\theta\) is \(+1.0 \, \text{rad}\). These conditions will be used to determine the constants of integration when solving for \(\omega(t)\) and \(\theta(t)\).
02

Determine Angular Velocity Expression

To find \(\omega(t)\), integrate the angular acceleration \(\alpha(t) = 6.0 t^4 - 4.0 t^2\) with respect to time \(t\):\[ \omega(t) = \int (6.0 t^4 - 4.0 t^2) \, dt = \frac{6.0}{5} t^5 - \frac{4.0}{3} t^3 + C \]Using the initial condition \(\omega(0) = 2.0\), solve for \(C\):\[ \omega(0) = \frac{6.0}{5} (0)^5 - \frac{4.0}{3} (0)^3 + C = 2.0 \implies C = 2.0 \]Thus, the expression for angular velocity is:\[ \omega(t) = \frac{6.0}{5} t^5 - \frac{4.0}{3} t^3 + 2.0 \]
03

Determine Angular Position Expression

To find \(\theta(t)\), integrate the angular velocity \(\omega(t) = \frac{6.0}{5} t^5 - \frac{4.0}{3} t^3 + 2.0\) with respect to time \(t\):\[ \theta(t) = \int \left(\frac{6.0}{5} t^5 - \frac{4.0}{3} t^3 + 2.0\right) \, dt = \frac{6.0}{30} t^6 - \frac{4.0}{12} t^4 + 2.0t + D \]Simplifying the coefficients:\[ \theta(t) = \frac{1.0}{5} t^6 - \frac{1.0}{3} t^4 + 2.0t + D \]Using the initial condition \(\theta(0) = 1.0\), solve for \(D\):\[ \theta(0) = \frac{1.0}{5} (0)^6 - \frac{1.0}{3} (0)^4 + 2.0(0) + D = 1.0 \implies D = 1.0 \]Thus, the expression for angular position is:\[ \theta(t) = \frac{1.0}{5} t^6 - \frac{1.0}{3} t^4 + 2.0t + 1.0 \]

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Key Concepts

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

Angular Acceleration
Angular acceleration (\( \alpha \)) describes how the angular velocity of an object changes over time. It is a vector quantity, measured in radians per second squared (rad/s²), and indicates whether the object is speeding up or slowing down its rotation. In the exercise, the angular acceleration is given by the function \( \alpha = 6.0 t^{4} - 4.0 t^{2} \). Here, the acceleration changes depending on time due to the terms involving power of \( t \). If you were to evaluate it at a specific moment, you would substitute that moment's time into the function to find the exact angular acceleration at that moment. Angular acceleration is fundamental in rotational dynamics and it helps in understanding how rotational speed is altered.
Angular Velocity
Angular velocity (\( \omega \)) refers to the rate of change of an angular position with time. It tells you how fast the object is spinning and in which direction. In our exercise, it's calculated by integrating the angular acceleration with respect to time:
  • \( \omega(t) = \int (6.0 t^4 - 4.0 t^2) \, dt = \frac{6.0}{5} t^5 - \frac{4.0}{3} t^3 + C \).
It is important to remember that, like in this case, angular velocity can initially be provided, \( \omega(0) = 2 \) rad/s, and is used to find the constant \( C \) when solving the integral. Understanding angular velocity helps understand how quickly an object rotates through an angle at any given moment.
Angular Position
Angular position (\( \theta \)) indicates the location of a rotating object at any point in time based on its starting orientation. To determine it as a function of time, we integrate the angular velocity:
  • \( \theta(t) = \int \left(\frac{6.0}{5} t^5 - \frac{4.0}{3} t^3 + 2.0\right) \, dt = \frac{1.0}{5} t^6 - \frac{1.0}{3} t^4 + 2.0t + D \).
The initial angular position \( \theta(0) \) given in the problem helps solve for the constant \( D \), setting it to 1 in this case. Calculating the angular position over time allows tracking the object's rotation and plotting its movement pattern.
Integration in Physics
Integration is a fundamental tool used in physics to calculate some quantities from others. It can be used to find quantities like displacement from velocity or angular position from angular velocity. In our exercise, this involved integrating a given angular acceleration to find angular velocity, and further integrating to obtain angular position. The process of integration allows the determination of cumulative effects over time.In each integration:
  • We determine the constants of integration (\( C \) or \( D \)) by applying initial conditions, which provide specific values at \( t = 0 \).
By using integration, we can deeply understand the behavior of rotating systems over time and predict how objects will move based on initial conditions and applied forces.

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Most popular questions from this chapter

In a judo foot-sweep move, you sweep your opponent's left foot out from under him while pulling on his gi (uniform) toward that side. As a result, your opponent rotates around his right foot and onto the mat. Figure \(10-41\) shows a simplified diagram of your opponent as you face him, with his left foot swept out. The rotational axis is through point \(O\). The gravitational force \(\vec{F}_{g}\) on him effectively acts at his center of mass, which is a horizontal distance \(d=28 \mathrm{~cm}\) from point \(O\). His mass is \(70 \mathrm{~kg}\), and his rotational inertia about point \(O\) is 65 \(\mathrm{kg} \cdot \mathrm{m}^{2}\). What is the magnitude of his initial angular acceleration about point \(O\) if your pull \(\vec{F}_{g}\) on his gi is (a) negligible and (b) horizontal with a magnitude of \(300 \mathrm{~N}\) and applied at height \(h=1.4 \mathrm{~m}\) ?

Calculate the rotational inertia of a wheel that has a kinetic energy of \(24400 \mathrm{~J}\) when rotating at \(602 \mathrm{rev} / \mathrm{min}\).

(a) Show that the rotational inertia of a solid cylinder of mass \(M\) and radius \(R\) about its central axis is equal to the rotational inertia of a thin hoop of mass \(M\) and radius \(R / \sqrt{2}\) about its central axis. (b) Show that the rotational inertia \(I\) of any given body of mass \(M\) about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass \(M\) and a radius \(k\) given by $$ k=\sqrt{\frac{I}{M}} . $$ The radius \(k\) of the equivalent hoop is called the radius of gyration of the given body.

Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the \(20 \mathrm{~cm}\) mark. (Treat the stick as a thin rod.)

Two uniform solid spheres have the same mass of \(1.65 \mathrm{~kg}\), but one has a radius of \(0.226 \mathrm{~m}\) and the other has a radius of \(0.854 \mathrm{~m}\). Each can rotate about an axis through its center. (a) What is the magnitude \(\tau\) of the torque required to bring the smaller sphere from rest to an angular speed of \(317 \mathrm{rad} / \mathrm{s}\) in \(15.5 \mathrm{~s} ?(\mathrm{~b})\) What is the magnitude \(F\) of the force that must be applied tangentially at the sphere's equator to give that torque? What are the corresponding values of (c) \(\tau\) and (d) \(F\) for the larger sphere?

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