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Chapter 9: Problem 24

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 \(t=0.200 \mathrm{~s} ?\) (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) 0.068 revolutions; (c) 1.01 m/s; (d) 3.45 m/s虏

Step by step solution

01

Understanding the Problem

We have a turntable of diameter 0.750 m rotating with initial angular velocity \( \omega_0 = 0.250 \, \mathrm{rev/s} \) and a constant angular acceleration \( \alpha = 0.900 \, \mathrm{rev/s^2} \). We need to solve for the angular velocity after 0.200 s, the number of revolutions in this time, tangential speed of a point on the rim, and the magnitude of resultant acceleration.
02

Calculate the final angular velocity

Using the formula for angular velocity \( \omega = \omega_0 + \alpha \cdot t \), where \( t = 0.200 \, \mathrm{s} \), \( \omega_0 = 0.250 \, \mathrm{rev/s} \), and \( \alpha = 0.900 \, \mathrm{rev/s^2} \), we can compute the final angular velocity:\[ \omega = 0.250 + 0.900 \times 0.200 = 0.250 + 0.180 = 0.430 \, \mathrm{rev/s} \]
03

Calculate the number of revolutions

The number of revolutions can be found using the formula for angular displacement \( \theta = \omega_0 \cdot t + 0.5 \cdot \alpha \cdot t^2 \):\[ \theta = 0.250 \times 0.200 + 0.5 \times 0.900 \times (0.200)^2 = 0.050 + 0.018 = 0.068 \, \mathrm{rev} \]
04

Calculate the tangential speed

Tangential speed \( v_t \) is given by \( v_t = r \times \omega \), where the radius \( r = 0.750/2 = 0.375 \, \mathrm{m} \) and \( \omega = 0.430 \, \mathrm{rev/s} = 0.430 \times 2\pi \, \mathrm{rad/s} \):\[ \omega = 0.430 \times 2\pi = 2.701 \approx 2.70\, \mathrm{rad/s} \]\[ v_t = 0.375 \times 2.70 \approx 1.01 \, \mathrm{m/s} \]
05

Calculate the resultant acceleration

The resultant acceleration consists of both tangential \( a_t \) and radial (centripetal) \( a_c \) components.For tangential acceleration, use \( a_t = r \cdot \alpha \) with \( \alpha = 0.900 \times 2\pi \):\[ a_t = 0.375 \times 0.900 \times 2\pi = 2.12 \approx 2.12 \, \mathrm{m/s^2} \]Centripetal acceleration \( a_c \) is \( a_c = r \cdot \omega^2 \):\[ a_c = 0.375 \times (2.70)^2 \approx 2.73 \, \mathrm{m/s^2} \]Resultant acceleration \( a = \sqrt{a_t^2 + a_c^2} \):\[ a = \sqrt{2.12^2 + 2.73^2} \approx 3.45 \approx 3.45 \, \mathrm{m/s^2} \]

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Key Concepts

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

Tangential Speed
Understanding tangential speed is essential when dealing with rotational motion. When an object is moving in a circular path, tangential speed refers to how fast the object is traveling along that path at any given point. It's the linear speed of a point on the circumference of a rotating object.

Calculating Tangential Speed:
  • The formula for tangential speed (\( v_t \)) is: \( v_t = r \cdot \omega \). Here, \( r \) is the radius of the path, and \( \omega \) is the angular velocity of the rotating object.
  • In the given exercise, with the diameter of 0.750 m, the radius is half of that, 0.375 m. The angular velocity at \( t = 0.200 \) seconds is computed as 0.430 rev/s, which needs conversion to rad/s to use in calculations because the equation for tangential speed uses SI units.
  • Converting the angular velocity: \( 0.430 \times 2\pi \approx 2.70 \, \mathrm{rad/s} \).
  • Now, the tangential speed: \( 0.375 \times 2.70 \approx 1.01 \, \mathrm{m/s} \).
In essence, tangential speed tells us how quickly a point on the rim of a spinning disk would move if it were suddenly set free to travel in a straight line.
Angular Velocity
Angular velocity is a measure of how much an object spins around a specific axis over a certain amount of time. It answers the question of how fast the angle is changing in a rotational motion.

Comprehending Angular Velocity:
  • Represented by \( \omega \), angular velocity is the rate of change of the angle with which an object revolves around a circle or an axis.
  • Its unit is typically radians per second (rad/s), though it may also be expressed in revolutions per second (rev/s), as seen in the exercise where \( \omega \) initially is 0.250 rev/s.
  • To calculate the final angular velocity given constant angular acceleration, use: \( \omega = \omega_0 + \alpha \cdot t \).
  • In the exercise, applying this formula: \( \omega = 0.250 + 0.900 \times 0.200 = 0.430 \, \text{rev/s} \).
A key point to remember about angular velocity is that it's directly related to both the speed at which an object rotates and the radius of its path.
Centripetal Acceleration
Centripetal acceleration is crucial in rotational dynamics as it describes the acceleration required to keep an object moving in a circular path, constantly pulling it towards the center of the circle.

Explaining Centripetal Acceleration:
  • It maintains an object's circular motion and is defined by \( a_c = r \cdot \omega^2 \), where \( \omega \) is the angular velocity and \( r \) is the radius of the circle.
  • In our problem, \( \omega = 2.70 \, \mathrm{rad/s} \) and \( r = 0.375 \, \mathrm{m} \), allowing us to calculate \( a_c \) as: \( a_c = 0.375 \times (2.70)^2 \approx 2.73 \, \mathrm{m/s^2} \).
  • Centripetal acceleration is always directed towards the center of the circular path, hence "centripetal," meaning "center-seeking."
This type of acceleration is essential for understanding how forces act on objects moving in circular paths. It shows that without centripetal acceleration, objects would move off in a tangent straight line due to inertia.

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Most popular questions from this chapter

You are to design a rotating cylindrical axle to lift 800-N buckets of cement from the ground to a rooftop 78.0 m above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s when it is turning at 7.5 rpm? (b) If instead the axle must give the buckets an upward acceleration of 0.400 m/s\(^2\), what should the angular acceleration of the axle be?

A fan blade rotates with angular velocity given by \(\omega_z\)(\(t\)) \(= \gamma - \beta t^2\), where \(\gamma =\) 5.00 rad/s and \(\beta =\) 0.800 rad/s\(^3\). (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_z\) at \(t =\) 3.00 s and the average angular acceleration \(\alpha_{av-z}\) for the time interval \(t =\) 0 to \(t =\) 3.00 s. How do these two quantities compare? If they are different, why?

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

A circular saw blade with radius 0.120 m starts from rest and turns in a vertical plane with a constant angular acceleration of 2.00 rev/s\(^2\). After the blade has turned through 155 rev, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

Energy is to be stored in a 70.0-kg flywheel in the shape of a uniform solid disk with radius \(R =\) 1.20 m. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 m/s\(^2\). What is the maximum kinetic energy that can be stored in the flywheel?
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