91Ó°ÊÓ

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).

Step by step solution

01

Calculate Tangential Speed for Each Object

The tangential speed \( v \) of an object in circular motion is given by \( v = \omega r \), where \( \omega \) is the angular speed and \( r \) is the radius. Here \( \omega = 6.00 \mathrm{rad/s} \).For object 1:\[ v_1 = 6.00 \times 2.00 = 12.00 \mathrm{m/s} \]For object 2:\[ v_2 = 6.00 \times 1.50 = 9.00 \mathrm{m/s} \]For object 3:\[ v_3 = 6.00 \times 3.00 = 18.00 \mathrm{m/s} \]

02

Calculate Total Kinetic Energy

The kinetic energy of each object is given by \( \mathrm{KE} = \frac{1}{2} mv^2 \). Sum the kinetic energy of all three objects:For object 1:\[ \mathrm{KE}_1 = \frac{1}{2} \times 6.00 \times (12.00)^2 = 432.00 \mathrm{J} \]For object 2:\[ \mathrm{KE}_2 = \frac{1}{2} \times 4.00 \times (9.00)^2 = 162.00 \mathrm{J} \]For object 3:\[ \mathrm{KE}_3 = \frac{1}{2} \times 3.00 \times (18.00)^2 = 486.00 \mathrm{J} \]Total kinetic energy:\[ \mathrm{KE} = 432.00 + 162.00 + 486.00 = 1080.00 \mathrm{J} \]

03

Calculate Moment of Inertia

The moment of inertia \( I \) for point masses is given by \( I = \sum m_i r_i^2 \).For object 1:\[ I_1 = 6.00 \times (2.00)^2 = 24.00 \mathrm{kg \cdot m^2} \]For object 2:\[ I_2 = 4.00 \times (1.50)^2 = 9.00 \mathrm{kg \cdot m^2} \]For object 3:\[ I_3 = 3.00 \times (3.00)^2 = 27.00 \mathrm{kg \cdot m^2} \]Total moment of inertia:\[ I = 24.00 + 9.00 + 27.00 = 60.00 \mathrm{kg \cdot m^2} \]

04

Calculate Rotational Kinetic Energy

The rotational kinetic energy \( \mathrm{KE}_{\mathrm{R}} \) is given by \( \mathrm{KE}_{\mathrm{R}} = \frac{1}{2} I \omega^2 \).Substitute for \( I = 60.00 \mathrm{kg \cdot m^2} \) and \( \omega = 6.00 \mathrm{rad/s} \):\[ \mathrm{KE}_{\mathrm{R}} = \frac{1}{2} \times 60.00 \times (6.00)^2 = 1080.00 \mathrm{J} \]The rotational kinetic energy matches the total kinetic energy calculated in Step 2.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Tangential Speed

In rotational dynamics, the concept of tangential speed is crucial for understanding the motion of objects in circular paths. Tangential speed, often denoted as \( v \), is the linear speed of a point located at a distance from the axis of rotation. It provides the measure of how fast the point is moving along its circular path. To find the tangential speed, we use the formula \( v = \omega r \), where \( \omega \) is the angular speed, measured in radians per second, and \( r \) is the radius, the distance from the rotation axis to the point.

• For object 1: \( v_1 = 6.00 \times 2.00 = 12.00 \, \mathrm{m/s} \)
• For object 2: \( v_2 = 6.00 \times 1.50 = 9.00 \, \mathrm{m/s} \)
• For object 3: \( v_3 = 6.00 \times 3.00 = 18.00 \, \mathrm{m/s} \)

Thus, the speed depends directly on both the angular speed and the radius.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. In the context of rotating systems, each object in the rotation has its own kinetic energy. For objects undergoing circular motion at a tangential speed, their kinetic energy is calculated with the formula \( \mathrm{KE} = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the tangential speed.Let's calculate for each object:

• Object 1: \( \mathrm{KE}_1 = \frac{1}{2} \times 6.00 \times (12.00)^2 = 432.00 \, \mathrm{J} \)
• Object 2: \( \mathrm{KE}_2 = \frac{1}{2} \times 4.00 \times (9.00)^2 = 162.00 \, \mathrm{J} \)
• Object 3: \( \mathrm{KE}_3 = \frac{1}{2} \times 3.00 \times (18.00)^2 = 486.00 \, \mathrm{J} \)

By summing these values, the total kinetic energy is found to be 1080.00 J. This energy quantifies the system's ability to perform work due to its motion.

Moment of Inertia

The moment of inertia, often symbolized by \( I \), is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation; the farther the mass is from the axis, the greater the moment of inertia.For point masses, the moment of inertia is calculated using \( I = \sum m_i r_i^2 \), where each mass \( m_i \) is multiplied by the square of its distance \( r_i \) from the rotational axis.

• Object 1: \( I_1 = 6.00 \times (2.00)^2 = 24.00 \, \mathrm{kg \cdot m^2} \)
• Object 2: \( I_2 = 4.00 \times (1.50)^2 = 9.00 \, \mathrm{kg \cdot m^2} \)
• Object 3: \( I_3 = 3.00 \times (3.00)^2 = 27.00 \, \mathrm{kg \cdot m^2} \)

Adding these gives a total moment of inertia \( I = 60.00 \, \mathrm{kg \cdot m^2} \). This quantity is crucial for understanding how different parts of the system contribute to overall rotational resistance.

Rotational Kinetic Energy

Rotational kinetic energy is the energy due to the rotation of an object and is part of its total kinetic energy. It is calculated using the formula \( \mathrm{KE}_{\mathrm{R}} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular speed.In this exercise:- With \( I = 60.00 \, \mathrm{kg \cdot m^2} \) and \( \omega = 6.00 \, \mathrm{rad/s} \), replacing these into the formula yields:\[ \mathrm{KE}_{\mathrm{R}} = \frac{1}{2} \times 60.00 \times (6.00)^2 = 1080.00 \, \mathrm{J} \]The value matches the total kinetic energy calculated earlier, demonstrating that the system's energy due to its rotational motion correlates directly with its inertia and angular speed. This highlights the interconnected nature of rotational and translational energies in dynamic systems.