/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Astronaut in Centrifuge An astro... [FREE SOLUTION] | 91Ó°ÊÓ

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Astronaut in Centrifuge An astronaut is being tested in a centrifuge. The centrifuge has a radius of \(10 \mathrm{~m}\) and, in starting, rotates according to \(\theta=\left(0.30 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}\), where \(t\) is in seconds and \(\theta\) is in radians. When \(t=5.0 \mathrm{~s}\), what are the magnitudes of the astronaut's (a) rotational velocity, (b) translational velocity, (c) tangential acceleration, and (d) radial acceleration?

Short Answer

Expert verified
(a) 3.0 rad/s, (b) 30 m/s, (c) 0.60 rad/s², (d) 90 m/s².

Step by step solution

01

Determine Rotational Velocity

The rotational velocity \(\omega\) can be found by differentiating the given angular position \(\theta = 0.30 \frac{\mathrm{rad}}{\mathrm{s}^2} t^2\) with respect to time. So, \(\omega(t) = \frac{d\theta}{dt} = 0.60 \frac{\mathrm{rad}}{\mathrm{s}^2} t\). Substituting \(t=5.0 \mathrm{s}\), \(\omega = 0.60 \times 5.0 = 3.0 \frac{\mathrm{rad}}{\mathrm{s}}\).
02

Calculate Translational Velocity

Translational velocity \(v\) can be found using the relationship \(v = r \omega\). Given \(r = 10 \mathrm{~m}\) and \(\omega = 3.0 \frac{\mathrm{rad}}{\mathrm{s}}\), \(v = 10 \mathrm{~m} \times 3.0 \frac{\mathrm{rad}}{\mathrm{s}} = 30 \frac{\mathrm{m}}{\mathrm{s}}\).
03

Find Tangential Acceleration

Tangential acceleration \(a_t\) can be found by differentiating rotational velocity with respect to time: \(a_t = \frac{d\omega}{dt}\). Given \(\omega = 0.60 \frac{\mathrm{rad}}{\mathrm{s}^2} t\), \(a_t = 0.60 \frac{\mathrm{rad}}{\mathrm{s}^2}\).
04

Determine Radial Acceleration

Radial acceleration \(a_r\) can be found using \(a_r = \omega^2 r\). Given \(\omega = 3.0 \frac{\mathrm{rad}}{\mathrm{s}}\) and \(r = 10 \mathrm{~m}\), \(a_r = (3.0 \frac{\mathrm{rad}}{\mathrm{s}})^2 \times 10 \mathrm{~m} = 90 \frac{\mathrm{m}}{\mathrm{s}^2}\).

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Key Concepts

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

Rotational Velocity
Rotational velocity refers to how fast an object rotates around a fixed axis. In this exercise, the astronaut's rotational velocity in the centrifuge is given by the equation: \ \( \omega(t) = 0.60 \frac{\mathrm{rad}}{\mathrm{s}^2} t \). \ This equation shows that the rotational velocity increases linearly with time. At \(t = 5.0 \mathrm{s}\), substituting in the equation gives: \ \( \omega = 0.60 \times 5.0 = 3.0 \frac{\mathrm{rad}}{\mathrm{s}} \). \ This means the centrifuge spins at \(3.0 \frac{\mathrm{rad}}{\mathrm{s}} \) after 5 seconds. \( \mathrm{Radians}/ \mathrm{second} \) is the unit of measurement here, showing the angle covered per second in radians. Rotational velocity is key in understanding the overall rotational movement of an object around a point.
Translational Velocity
Translational velocity is the speed at which an object moves through space in a straight line. It is closely linked to rotational velocity by the equation \(v = r \omega\), where \(v\) represents translational velocity, \(r\) is the radius, and \(\omega\) is rotational velocity.
Given that the radius \(r = 10 \mathrm{~m}\) and \(\omega = 3 \frac{\mathrm{rad}}{\mathrm{s}} \), the astronaut's translational velocity can be calculated as: \ \( v = 10 \mathrm{~m} \times 3 \frac{\mathrm{rad}}{\mathrm{s}} = 30 \frac{\mathrm{m}}{\mathrm{s}} \).
This indicates that the astronaut is moving at \(30 \mathrm{m}/ \mathrm{s} \) along the circular path at \(t = 5 \mathrm{~s}\). Translational velocity helps to understand how an orbiting object can translate its circular motion into linear speed if it were to move in a straight line.
Tangential Acceleration
Tangential acceleration is the rate of change of translational (linear) velocity along a path tangent to the curve at the point of interest. This can be found by differentiating the rotational velocity with respect to time: \ \( a_t = \frac{d \omega}{d t} \). \ Using the given rotational velocity \( \omega = 0.60 \frac{\mathrm{rad}}{\mathrm{s}^2} t \), the differentiation gives: \( a_t = 0.60 \frac{\mathrm{rad}}{\mathrm{s}^2} \). \
Hence, the tangential acceleration remains constant at \( 0.60 \frac{\mathrm{rad}}{\mathrm{s}^2} \), which implies that the linear speed of the astronaut is increasing at that constant rate. Understanding tangential acceleration is essential for analyzing how the speed of an object changes over time while moving along a curved path.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, is the rate at which an object changes direction as it moves along a circular path. It always points towards the center of the circular path. The formula for radial acceleration is: \ \( a_r = \omega^2 r \). \ Here, \(\omega = 3.0 \frac{\mathrm{rad}}{\mathrm{s}} \) and \( r = 10 \mathrm{~m} \). Substituting these values: \ \( a_r = (3.0 \frac{\mathrm{rad}}{\mathrm{s}})^2 \times 10 \mathrm{~m} = 90 \frac{\mathrm{m}}{\mathrm{s}^2} \). \ This means the astronaut experiences a radial acceleration of \(90 \frac{\mathrm{m}}{\mathrm{s}^2} \) directed towards the centrifuge's center. Radial acceleration is crucial in analyzing forces that keep an object in circular motion, making it central to understanding rotation dynamics.

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Most popular questions from this chapter

Vinyl Record A vinyl record on a turntable rotates at \(33 \frac{1}{3}\) rev/min. (a) What is its rotational speed in radians per second? What is the translational speed of a point on the record at the needle when the needle is (b) \(15 \mathrm{~cm}\) and (c) \(7.4 \mathrm{~cm}\) from the turntable axis?

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?

A Flywheel Has a Rotational Velocity At \(t_{1}=0\), a flywheel has a rotational velocity of \(4.7 \mathrm{rad} / \mathrm{s}\), a rotational acceleration of \(-0.25\) \(\mathrm{rad} / \mathrm{s}^{2}\), and a reference line at \(\theta_{1}=0 .\) (a) Through what maximum angle \(\theta^{\max }\) will the reference line turn in the positive direction? For what length of time will the reference line turn in the positive direction? At what times will the reference line be at (b) \(\theta=\frac{1}{2} \theta^{\max }\) and (c) \(\theta=-10.5\) rad (consider both positive and negative values of \(t\) )? (d) Graph \(\theta\) versus \(t\), and indicate the answers to \((\mathrm{a}),(\mathrm{b})\), and \((\mathrm{c})\) on the graph.

Torque on a Certain Wheel A torque of \(32.0 \mathrm{~N} \cdot \mathrm{m}\) on a certain wheel causes a rotational acceleration of \(25.0 \mathrm{rad} / \mathrm{s}^{2} .\) What is the wheel's rotational inertia?

Four Identical Particles Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

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