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Chapter 9: Problem 32

A clay vase on a potter's wheel experiences an angular acceleration of \(8.00 \mathrm{rad} / \mathrm{s}^{2}\) due to the application of a \(10.0-\mathrm{N} \cdot \mathrm{m}\) net torque. Find the total moment of inertia of the vase and potter's wheel.

Short Answer

Expert verified
The total moment of inertia is 1.25 kg鈰卪虏.

Step by step solution

01

Understand Relationship Between Torque and Angular Acceleration

The relationship between torque \(\tau\), moment of inertia \(I\), and angular acceleration \(\alpha\) is given by the equation \(\tau = I \alpha\). This equation will allow us to solve for the moment of inertia.
02

Substitute Known Values into the Torque Formula

Substitute \(\tau = 10.0 \mathrm{\ N}\cdot \mathrm{m}\) and \(\alpha = 8.00 \mathrm{\ rad/s}^2\) into the equation \(\tau = I \alpha\). This gives us: \(10.0 = I \times 8.00\).
03

Solve for the Moment of Inertia

Rearrange the equation \(10.0 = I \times 8.00\) to solve for \(I\). This results in: \[ I = \frac{10.0}{8.00} \].
04

Calculate the Moment of Inertia

Compute the division to find \(I\). \(I = \frac{10.0}{8.00} = 1.25 \mathrm{\ kg \cdot m^2}\).

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Key Concepts

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

Angular Acceleration
Angular acceleration describes how quickly an object speeds up or slows down its rotation. It is denoted by the Greek letter \( \alpha \) and is measured in radians per second squared (\( \mathrm{rad/s}^2 \)). In the given problem, the clay vase on the potter's wheel undergoes an angular acceleration of \( 8.00 \mathrm{rad/s}^2 \).
This means its rotational speed is increasing rapidly every second. Angular acceleration is crucial in understanding how rotational speed changes over time.
  • Suppose a potter's wheel starts from rest and has a constant angular acceleration. The longer it accelerates, the faster the wheel spins.
  • The rate of change of angular velocity due to angular acceleration is consistent, similar to linear acceleration in a straight line.
Understanding angular acceleration helps in predicting how rotational systems evolve under the influence of forces, such as torque.
Torque
Torque is the rotational equivalent of force and causes objects to rotate. It is measured in newton-meters (\( \mathrm{N\cdot m} \)). In our problem, the clay vase and potter's wheel experience a net torque of \( 10.0 \mathrm{N \cdot m} \).
Torque can be thought of as a twisting force. It plays a significant role in determining how rotational motion changes.
  • The equation \( \tau = I \alpha \) shows how torque relates to angular acceleration. Here, \(\tau\) is torque, \(I\) is the moment of inertia, and \(\alpha\) is angular acceleration.
  • Increasing torque usually increases the speed at which the object spins, assuming the moment of inertia is constant.
  • The greater the torque applied to an object, the greater the change in its angular acceleration.
Torque allows us to calculate how much rotational effect we can expect from a given force.
Rotational Motion
Rotational motion is how an object turns about an axis. This type of motion is everywhere, from wheels to celestial bodies spinning in space. In our scenario, the clay vase on a potter's wheel is an excellent example.
Here, the wheel rotates thanks to a net torque applied, resulting in angular acceleration.
  • Rotational motion involves quantities like angular velocity, angular displacement, and angular acceleration.
  • Understanding rotational motion enables us to apply concepts like torque and moment of inertia to predict and manipulate rotation.
Just like linear motion, rotational motion obeys its own set of rules and equations. Knowing these helps you understand how objects behave when they spin or rotate.

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Most popular questions from this chapter

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+\) \(M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, \(M\) is the total mass of the object, and \(h\) is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius \(R\) relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of \(6.6 \mathrm{m} / \mathrm{s} .\) Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

A bowling ball encounters a \(0.760-\mathrm{m}\) vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is \(3.50 \mathrm{m} / \mathrm{s}\) at the bottom of the rise. Find the translational speed at the top.

A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod can rotate freely on the tabletop. Two forces, both parallel to the tabletop, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of \(38.0 \mathrm{N}\). The second force has a magnitude of \(55.0 \mathrm{N}\) and is directed at an angle \(\theta\) with respect to the rod. If the sum of the torques due to the two forces is zero, what must be the angle \(\theta ?\)

Three objects lie in the \(x, y\) plane. Each rotates about the \(z\) axis with an angular speed of \(6.00 \mathrm{rad} / \mathrm{s}\). The mass \(m\) of each object and its perpendicular distance \(r\) from the \(z\) axis are as follows: (1) \(m_{1}=6.00 \mathrm{kg}\) and \(r_{1}=\) \(2.00 \mathrm{m},(2) m_{2}=4.00 \mathrm{kg}\) and \(r_{2}=1.50 \mathrm{m},(3) m_{3}=3.00 \mathrm{kg}\) and \(r_{3}=3.00 \mathrm{m}\) (a) Find the tangential speed of each object. (b) Determine the total kinetic energy of this system using the expression \(\mathrm{KE}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+\frac{1}{2} m_{3} v_{3}^{2}\) (c) Obtain the moment of inertia of the system. (d) Find the rotational kinetic energy of the system using the relation \(\mathrm{KE}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}\) to verify that the answer is the same as the answer to (b).
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