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Chapter 11: Problem 30

Fixed Axis An object rotates about a fixed axis, and the rotational position of a reference line on the object is given by \(\theta=\) \((0.40 \mathrm{rad}) e^{\left(2.0 \mathrm{~s}^{-1}\right) t}\). Consider a point on the object that is \(4.0 \mathrm{~cm}\) from the axis of rotation. At \(t=0\), what are the magnitudes of the point's (a) tangential component of acceleration and (b) radial component of acceleration?

Short Answer

Expert verified
Tangential acceleration is 0.064 m/s虏, and radial acceleration is 0.0256 m/s虏

Step by step solution

01

Understand the given equation

The rotational position of a reference line on the object is given by \[\theta(t) = (0.40 \text{ rad}) e^{(2.0 \text{ s}^{-1}) t}\]This equation describes how the rotational position changes with time.
02

Find angular velocity

Angular velocity \(\omega\) is the time derivative of \(\theta\). Compute \(\omega\) by differentiating \(\theta(t)\) with respect to \(t\):\[\omega(t) = \frac{d\theta(t)}{dt} = (0.40 \text{ rad}) \frac{d}{dt} e^{(2.0 \text{ s}^{-1}) t} = (0.40 \text{ rad}) (2.0 \text{ s}^{-1}) e^{(2.0 \text{ s}^{-1}) t} = 0.80 \text{ rad} \cdot \text{s}^{-1} e^{(2.0 \text{ s}^{-1}) t}\]
03

Evaluate angular velocity at t = 0

Substitute \(t = 0\) into the expression for \(\omega(t)\):\[\omega(0) = 0.80 \text{ rad} \cdot \text{s}^{-1} e^{(2.0 \text{ s}^{-1}) \cdot 0} = 0.80 \text{ rad} \cdot \text{s}^{-1}\]
04

Find angular acceleration

Angular acceleration \(\alpha\) is the time derivative of angular velocity \(\omega(t)\). Compute \(\alpha\) by differentiating \(\omega(t)\) with respect to \(t\):\[\alpha(t) = \frac{d\omega(t)}{dt} = 0.80 \text{ rad} \cdot \text{s}^{-1} \cdot 2.0 \text{ s}^{-1} e^{(2.0 \text{ s}^{-1}) t} = 1.60 \text{ rad} \cdot \text{s}^{-2} e^{(2.0 \text{ s}^{-1}) t}\]
05

Evaluate angular acceleration at t = 0

Substitute \(t = 0\) into the expression for \(\alpha(t)\):\[\alpha(0) = 1.60 \text{ rad} \cdot \text{s}^{-2} e^{(2.0 \text{ s}^{-1}) \cdot 0} = 1.60 \text{ rad} \cdot \text{s}^{-2}\]
06

Calculate the tangential component of acceleration

The tangential component of acceleration \(a_t\) for a point a distance \(r\) from the axis is given by \(a_t = r\alpha\). Given \(r = 0.04 \text{ m}\) and using \(\alpha(0)\), we get:\[a_t = 0.04 \text{ m} \cdot 1.60 \text{ rad} \cdot \text{s}^{-2} = 0.064 \text{ m} \cdot \text{s}^{-2}\]
07

Calculate the radial component of acceleration

The radial component of acceleration \(a_r\) for a point a distance \(r\) from the axis is given by \(a_r = r\omega^2\). Given \(r = 0.04 \text{ m}\) and using \(\omega(0)\), we get:\[a_r = 0.04 \text{ m} \cdot (0.80 \text{ rad} \cdot \text{s}^{-1})^2 = 0.04 \text{ m} \cdot 0.64 \text{ rad}^2 \cdot \text{s}^{-2} = 0.0256 \text{ m} \cdot \text{s}^{-2}\]

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Key Concepts

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

Angular Velocity
Angular velocity describes how fast an object rotates or spins around a particular axis. It is often symbolized as \(\omega\) and measured in radians per second (rad/s). To find angular velocity from the given rotational position equation \(\theta(t) = (0.40 \text{ rad}) e^{(2.0 \text{ s}^{-1}) t}\), we take the derivative of \(\theta\) with respect to time. This yields \(\omega(t) = 0.80 \text{ rad} \cdot \text{s}^{-1} e^{(2.0 \text{ s}^{-1}) t}\) as the expression for angular velocity. Evaluating this at \(t = 0\), the angular velocity is \(0.80 \text{ rad} \cdot \text{s}^{-1}\). This tells us the object is rotating at a constant rate initially.
Angular Acceleration
Angular acceleration refers to the rate at which the angular velocity of an object changes with time. This is represented as \(\alpha\) and measured in radians per second squared (rad/s虏). To find angular acceleration from the angular velocity \(\omega(t)\), we differentiate \(\omega\) with respect to time. This yields \(\alpha(t) = 1.60 \text{ rad} \cdot \text{s}^{-2} e^{(2.0 \text{ s}^{-1}) t}\). At \(t = 0\), \(\alpha(0) = 1.60 \text{ rad} \cdot \text{s}^{-2}\), indicating that the object's rotation is initially speeding up at this rate.
Tangential Acceleration
Tangential acceleration is the component of linear acceleration that is tangent to the circular path of a rotating object. It arises due to the change in angular velocity. The formula to calculate tangential acceleration \(a_t\) is \(a_t = r\alpha\), where \(r\) is the radius (distance from the axis of rotation). Given \(\alpha(0) = 1.60 \text{ rad} \cdot \text{s}^{-2}\) and \(r = 0.04 \text{ m}\), the tangential acceleration at \(t = 0\) is \(0.064 \text{ m} \cdot \text{s}^{-2}\). This value indicates how fast the linear speed of the point is changing along the circular path.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, is directed towards the center of the circular path. It arises due to the change in direction of the velocity vector as the object moves in a circle. The formula to calculate radial acceleration \(a_r\) is \(a_r = r\omega^2\), using angular velocity. Considering \(\omega(0) = 0.80 \text{ rad} \cdot \text{s}^{-1}\) and \(r = 0.04 \text{ m}\), the radial acceleration at \(t = 0\) is \(0.0256 \text{ m} \cdot \text{s}^{-2}\). This value describes how much the point on the rotating object is accelerating towards the axis of rotation.

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Most popular questions from this chapter

Communications \(\quad\) Satellite \(A\) communications satellite is a solid cylinder with mass \(1210 \mathrm{~kg}\), diameter \(1.21 \mathrm{~m}\), and length \(1.75 \mathrm{~m}\). Prior to launching from the shuttle cargo bay, it is set spinning at \(1.52 \mathrm{rev} / \mathrm{s}\) about the cylinder axis (Fig. \(11-27)\). Calculate the satellite's (a) rotational inertia about the rotation axis and (b) rotational kinetic energy.

Spaceship What are the magnitudes of (a) the rotational velocity, (b) the radial acceleration, and (c) the tangential acceleration of a spaceship taking a circular turn of radius \(3220 \mathrm{~km}\) at a speed of \(29000 \mathrm{~km} / \mathrm{h} ?\)

Fly on an LP An old-fashioned record player spins a disk at approximately a constant angular velocity, \(\omega\). A fly of mass \(m\) is sitting on the disk as it turns, at a point a distance \(R\) from the center. (a) What force keeps the fly from sliding off the rotating disk? What direction does the force point? How big is it? For the last question, express your answer in terms of the symbols given in the description above. (b) If the fly has a mass of \(0.5\) grams, is sitting \(10 \mathrm{~cm}\) from the center of the disk, and the disk is turning at a rate of \(33 \mathrm{rev} / \mathrm{min}\), what is the coefficient of friction?

Coupled Wheels In Fig. \(11-25\). wheel \(A\) of radius \(r_{A}=10 \mathrm{~cm}\) is coupled by belt \(B\) to wheel \(C\) of radius \(r_{C}=25 \mathrm{~cm}\). The rotational speed of wheel \(A\) is increased from rest at a constant rate of \(1.6 \mathrm{rad} / \mathrm{s}^{2}\) Find the time for wheel \(C\) to reach a rotational speed of \(100 \mathrm{rev} / \mathrm{min}\),assuming the belt does not slip. (Hint: If the belt does not slip, the translational speeds at the rims of the two wheels must be equal.)

Oxygen Molecule The oxygen molecule \(\mathrm{O}_{2}\) has a mass of \(5.30 \times\) \(10^{-26} \mathrm{~kg}\) and a rotational inertia of \(1.94 \times 10^{-46} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about an axis through the center of the line joining the atoms and perpendicular to that line. Suppose the center of mass of an \(\mathrm{O}_{2}\) molecule in a gas has a translational speed of \(500 \mathrm{~m} / \mathrm{s}\) and the molecule has a rotational kinetic energy that is \(\frac{2}{3}\) of the translational kinetic energy of its center of mass. What then is the molecule's rotational speed about the center of mass?
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