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Chapter 10: Problem 20

An object rotates about a fixed axis, and the angular position of a reference line on the object is given by \(\theta=0.40 e^{2 t},\) where \(\theta\) is in radians and \(t\) is in seconds. 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: 0.064 m/s虏; Radial acceleration: 0.0256 m/s虏.

Step by step solution

01

Derive Angular Velocity

Angular velocity \( \omega \) is the time derivative of angular position \( \theta(t) \). Given \( \theta(t) = 0.40 e^{2t} \), we differentiate with respect to time \( t \): \( \omega(t) = \frac{d}{dt}(0.40 e^{2t}) = 0.80 e^{2t} \) rad/s.
02

Derive Angular Acceleration

Angular acceleration \( \alpha \) is the time derivative of angular velocity \( \omega(t) \). Differentiate \( \omega(t) = 0.80 e^{2t} \) with respect to \( t \):\( \alpha(t) = \frac{d}{dt}(0.80 e^{2t}) = 1.60 e^{2t} \) rad/s虏.
03

Find Tangential Acceleration

The tangential acceleration \(a_t\) is the product of angular acceleration \( \alpha \) and the radius \( r \). Here, \( r = 0.04 \) m (converted from cm). Using \( \alpha(0) = 1.6 e^0 = 1.6 \) rad/s虏,\( a_t = \alpha \cdot r = 1.6 \cdot 0.04 = 0.064 \) m/s虏.
04

Determine Radial Acceleration

Radial acceleration \( a_r \) is given by \( \omega^2 \cdot r \). At \( t=0 \), \( \omega(0) = 0.8 e^0 = 0.8 \) rad/s. Thus, \( a_r = (0.8)^2 \cdot 0.04 = 0.0256 \) m/s虏.

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Key Concepts

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

Angular Velocity
Angular velocity, often denoted by \( \omega \), describes the rate at which an object rotates around a fixed axis. Imagine a spinning record; angular velocity tells us how quickly it is turning. Like linear velocity measures how fast an object moves along a straight path, angular velocity measures the rotational speed.
It is the first derivative of the angular position \( \theta \) with respect to time \( t \). In our exercise, the angular position is given by \( \theta(t) = 0.40e^{2t} \). To find the angular velocity, differentiate this function:
  • \( \omega(t) = \frac{d}{dt}(0.40e^{2t}) = 0.80e^{2t} \) rad/s
At any given moment, you can plug in the time \( t \) to find the angular velocity at that instant.
Angular Acceleration
Angular acceleration, represented by \( \alpha \), measures how the angular velocity changes with time. If the spinning speed of our imaginary record increases or decreases, that change is described by the angular acceleration. It is the concept in rotational motion equivalent to linear acceleration, which measures the change of velocity in straight-line motion.
To calculate angular acceleration, differentiate angular velocity with respect to time. From our function, \( \omega(t) = 0.80e^{2t} \), we get:
  • \( \alpha(t) = \frac{d}{dt}(0.80e^{2t}) = 1.60e^{2t} \) rad/s虏
This tells us how fast the spinning speed itself is changing over time.
Tangential Acceleration
Tangential acceleration \( a_t \) is the linear acceleration of a point on a rotating object, directed along the path of motion. In simple terms, it's how quickly the speed of a point on the edge of a spinning object is changing. This is connected to the radius of the rotation and the angular acceleration.
In our scenario, the radius \( r \) is \( 0.04 \) m (or 4 cm). The tangential acceleration is calculated by multiplying the angular acceleration \( \alpha \) by the radius \( r \):
  • \( a_t = \alpha \cdot r = 1.6 \cdot 0.04 = 0.064 \) m/s虏 at \( t = 0 \)
So, at the initial moment, the point on the object is accelerating along its path at \( 0.064 \) m/s虏.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, \( a_r \), is the component of acceleration that points towards the center of the rotation. Imagine swinging a ball on a string; radial acceleration is what keeps the ball moving in a circle.
This acceleration depends on the angular velocity and the radius of the circle. From our exercise, it is calculated by the formula:
  • \( a_r = \omega^2 \cdot r \)
At \( t = 0 \), the angular velocity \( \omega(0) = 0.8 \) rad/s, leading to:
  • \( a_r = (0.8)^2 \cdot 0.04 = 0.0256 \) m/s虏
This is the acceleration needed to keep the point moving in a circle at that instant.

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Most popular questions from this chapter

A gyroscope flywheel of radius \(2.83 \mathrm{~cm}\) is accelerated from rest at \(14.2 \mathrm{rad} / \mathrm{s}^{2}\) until its angular speed is \(2760 \mathrm{rev} / \mathrm{min} .\) (a) What is the tangential acceleration of a point on the rim of the flywheel during this spin-up process? (b) What is the radial acceleration of this point when the flywheel is spinning at full speed? (c) Through what distance does a point on the rim move during the spin-up?

Our Sun is \(2.3 \times 10^{4}\) ly (light-years) from the center of our Milky Way galaxy and is moving in a circle around that center at a speed of \(250 \mathrm{~km} / \mathrm{s}\). (a) How long does it take the Sun to make one revolution about the galactic center? (b) How many revolutions has the Sun completed since it was formed about \(4.5 \times 10^{9}\) years ago?

A small ball of mass \(0.75 \mathrm{~kg}\) is attached to one end of a 1.25 -m-long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is \(30^{\circ}\) from the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

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

The flywheel of an engine is rotating at \(25.0 \mathrm{rad} / \mathrm{s}\). When the engine is turned off, the flywheel slows at a constant rate and stops in \(20.0 \mathrm{~s}\). Calculate (a) the angular acceleration of the flywheel, (b) the angle through which the flywheel rotates in stopping, and (c) the number of revolutions made by the flywheel in stopping.
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