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An electric turntable 0.750 \(\mathrm{m}\) in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 \(\mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of 0.900 \(\mathrm{rev} / \mathrm{s}^{2}\) . (a) Compute the angular velocity of the turntable after 0.200 \(\mathrm{s}\) . (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200\) s? (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{s} ?\)

Short Answer

Expert verified
(a) 0.430 rev/s. (b) 0.085 revolutions. (c) 1.01 m/s. (d) 7.57 m/s².

Step by step solution

01

Understanding the Problem

We are asked to solve four parts of a problem related to an electric turntable's motion. We need to determine the angular velocity, total revolutions, tangential speed, and resultant acceleration of a point on the rim after a certain period of time.
02

Calculating Angular Velocity

The turntable has an initial angular velocity of \( \omega_0 = 0.250 \, \mathrm{rev/s} \) and an angular acceleration of \( \alpha = 0.900 \, \mathrm{rev/s^2} \). We need to find the angular velocity \( \omega \) at \( t = 0.200 \, \mathrm{s} \) using the formula: \[ \omega = \omega_0 + \alpha t \] Substituting the values, we get: \[ \omega = 0.250 + 0.900 \times 0.200 = 0.430 \, \mathrm{rev/s} \]
03

Calculating Total Revolutions

We use the formula for angular displacement \( \theta \) given by: \[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \] Substitute \( \omega_0 = 0.250 \, \mathrm{rev/s} \), \( \alpha = 0.900 \, \mathrm{rev/s^2} \), and \( t = 0.200 \, \mathrm{s} \): \[ \theta = 0.250 \times 0.200 + \frac{1}{2} \times 0.900 \times (0.200)^2 = 0.085 \, \mathrm{rev} \]
04

Calculating Tangential Speed

The tangential speed \( v \) is given by: \[ v = r \times \omega \] where \( r = 0.375 \, \mathrm{m} \) (radius is half the diameter) and \( \omega = 0.430 \, \mathrm{rev/s} \). Convert \( \omega \) from \( \mathrm{rev/s} \) to \( \mathrm{rad/s} \): \( \omega = 0.430 \times 2\pi \). Thus, \[ v = 0.375 \times 0.430 \times 2\pi \approx 1.01 \, \mathrm{m/s} \]
05

Calculating Resultant Acceleration

Resultant acceleration includes radial \( a_r \) and tangential \( a_t \) components. The tangential acceleration \( a_t = r \times \alpha \), where \( \alpha \) must be in \( \mathrm{rad/s^2} \): \( \alpha = 0.900 \times 2\pi \). Thus, \[ a_t = 0.375 \times 0.900 \times 2\pi \approx 2.12 \, \mathrm{m/s^2} \] Radial acceleration \( a_r \) is \( \omega^2 r \) with \( \omega = 0.430 \times 2\pi \): \[ a_r = (0.430 \times 2\pi)^2 \times 0.375 \approx 7.26 \, \mathrm{m/s^2} \] Resultant acceleration \( a = \sqrt{a_t^2 + a_r^2} \): \[ a = \sqrt{2.12^2 + 7.26^2} \approx 7.57 \, \mathrm{m/s^2} \]

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Key Concepts

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

Angular Velocity
Angular velocity \( \omega \) is a measure of how fast an object rotates about an axis. It's similar to the linear velocity, but for rotational motion. An important characteristic of angular velocity is that it's usually measured in rotations per second (rev/s) or radians per second (rad/s). Angular velocity can be affected by angular acceleration \( \alpha \), which is the rate of change of angular velocity over time.

In the example of the electric turntable, the initial angular velocity \( \omega_0 \) was 0.250 rev/s. With a constant angular acceleration of 0.900 rev/s^2, the angular velocity after 0.200 seconds can be found using the equation:

\[ \omega = \omega_0 + \alpha t \]

where
\( \omega_0 = 0.250 \, \text{rev/s} \),
\( \alpha = 0.900 \, \text{rev/s}^2 \), and
\( t = 0.200 \, \text{s} \).

Plug in these values to find:
\[ \omega = 0.250 + 0.900 \times 0.200 = 0.430 \, \text{rev/s} \]

This new value represents how quickly the turntable spins after 0.200 seconds.
Angular Acceleration
Angular acceleration \( \alpha \) determines how quickly the angular velocity changes with time. It's like how acceleration works in linear motion but in the world of rotations.

However, we often need to convert between different units, such as revolutions per second squared (rev/s²) and radians per second squared (rad/s²), because many formulas require angular measurements in radians. This conversion uses \( 2\pi \) since one full revolution equals \( 2\pi \) radians.

To illustrate this, with our electric turntable example, we have:
\( \alpha = 0.900 \, \text{rev/s}^2 \) .

Converting to rad/s² involves:
\[ \alpha = 0.900 \times 2\pi = 5.654 \, \text{rad/s}^2 \]

Knowing angular acceleration helps us understand how rapidly the turntable speeds up or slows down.
Tangential Speed
Tangential speed \( v \) refers to the linear speed of a point located on the edge of a rotating object. It's a direct function of both the object's angular velocity and its radius. Essentially, it tells how fast a point on the edge is moving in a straight line, usually measured in meters per second (m/s).

If we consider the turntable, the radius \( r \) is half its diameter. Here,
\( r = 0.375 \, \text{m} \).

To find the tangential speed, the angular velocity \( \omega \) must be in rad/s. This conversion is:

\[ \omega = 0.430 \times 2\pi = 2.701 \, \text{rad/s} \]

The formula for tangential speed is:
\[ v = r \times \omega = 0.375 \times 2.701 \approx 1.01 \, \text{m/s} \]

This speed shows how fast a point on the turntable's rim is moving linearly.
Resultant Acceleration
In rotational motion, the resultant acceleration of a point on a rotating object includes both tangential and radial components.

Tangential acceleration \( a_t \) tells how fast the tangential speed changes. It's calculated by multiplying angular acceleration by radius:
\[ a_t = r \times \alpha = 0.375 \times 5.654 \approx 2.12 \, \text{m/s}^2 \]

Radial or centripetal acceleration \( a_r \) results from change in direction of the velocity vector and is calculated as follows:
\[ a_r = \omega^2 \times r = (2.701)^2 \times 0.375 \approx 7.26 \, \text{m/s}^2 \]

To find the magnitude of the resultant acceleration \( a \), combine these components using the Pythagorean theorem:
\[ a = \sqrt{a_t^2 + a_r^2} = \sqrt{2.12^2 + 7.26^2} \approx 7.57 \, \text{m/s}^2 \]

This value indicates how the point on the rim of the turntable experiences acceleration in total.

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Most popular questions from this chapter

A metal sign for a car dealership is a thin, uniform right triangle with base length \(b\) and height \(h .\) The sign has mass \(M .\) (a) What is the moment of inertia of the sign for rotation about the side of length \(h ?\) (b) If \(M=5.40 \mathrm{kg}, b=1.60 \mathrm{m},\) and \(h=1.20 \mathrm{m},\) what is the kinetic energy of the sign when it is rotating about an axis along the \(1.20-\mathrm{m}\) side at 2.00 \(\mathrm{rev} / \mathrm{s} ?\)

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Neutron Stars and Supernova Remnants. The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (Fig. P9.98). It is the remnant of a star that underwent a supernova explosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a released by the Crab Nebula at a rate of about \(5 \times 10^{31} \mathrm{W},\) about \(10^{5}\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. This object rotates once every \(0.0331 \mathrm{s},\) and this period is increasing by \(4.22 \times 10^{-13} \mathrm{s}\) for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock \(\left(3000 \mathrm{kg} / \mathrm{m}^{3}\right)\) and to the density of an atomic nucleus (about \(10^{17} \mathrm{kg} / \mathrm{m}^{3} ) .\) Justify the statement that a neutron star is essentially a large atomic nucleus.

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