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Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

Step by step solution

01

Determine the Moment of Inertia for Rod A

For Rod A, the moment of inertia is calculated using the formula for a point mass at a distance from the axis: \[ I_A = m imes L^2 \]where \(m = 0.66 \, \text{kg}\) and \(L = 0.75 \, \text{m}\).Substitute the values to get:\[ I_A = 0.66 imes (0.75)^2 = 0.37125 \, \text{kg} \, \text{m}^2 \]

02

Determine the Moment of Inertia for Rod B

For Rod B, the moment of inertia is calculated using the formula for a uniform rod rotating about an end:\[ I_B = \frac{1}{3} m L^2 \]where \(m = 0.66 \, \text{kg}\) and \(L = 0.75 \, \text{m}\).Substitute the values to get:\[ I_B = \frac{1}{3} imes 0.66 imes (0.75)^2 = 0.12375 \, \text{kg} \, \text{m}^2 \]

03

Compare Moments of Inertia

Rod A has \( I_A = 0.37125 \, \text{kg} \, \text{m}^2 \) and Rod B has \( I_B = 0.12375 \, \text{kg} \, \text{m}^2 \). Since \( 0.37125 > 0.12375 \), Rod A has the greater moment of inertia.

04

Calculate Rotational Kinetic Energy for Rod A

Rotational kinetic energy is given by the formula:\[ KE_A = \frac{1}{2} I_A \omega^2 \]Substitute \(I_A = 0.37125 \, \text{kg} \, \text{m}^2\) and \(\omega = 4.2 \, \text{rad/s}\):\[ KE_A = \frac{1}{2} \times 0.37125 \times 4.2^2 = 3.267585 \, \text{J} \]

05

Calculate Rotational Kinetic Energy for Rod B

Using the same formula:\[ KE_B = \frac{1}{2} I_B \omega^2 \]Substitute \(I_B = 0.12375 \, \text{kg} \, \text{m}^2\) and \(\omega = 4.2 \, \text{rad/s}\):\[ KE_B = \frac{1}{2} \times 0.12375 \times 4.2^2 = 1.089195 \, \text{J} \]

06

Compare Rotational Kinetic Energies

Rod A has \( KE_A = 3.267585 \, \text{J} \) and Rod B has \( KE_B = 1.089195 \, \text{J} \). Since \( 3.267585 > 1.089195 \), Rod A has the greater rotational kinetic energy.

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Key Concepts

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

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. For rotating bodies, it is given by the formula: \( KE = \frac{1}{2} I \omega^2 \) where \( I \) is the moment of inertia and \( \omega \) is the angular speed. The larger the moment of inertia, the more energy is needed to achieve the same angular speed. This is why in the given exercise, Rod A, which has a greater moment of inertia, also has a greater rotational kinetic energy compared to Rod B. Even though both rods are rotating at the same speed, Rod A's energy is higher because the mass is concentrated at the end, increasing the moment of inertia.

Angular Speed

Angular speed \( \omega \) refers to how fast something rotates, expressed in radians per second (rad/s). It tells us how quickly an object completes its rotation around a specific axis. In this problem, it's given as 4.2 rad/s for both rods. Angular speed is a crucial part of calculating rotational kinetic energy because even if two objects rotate at the same speed, their energy might differ based on how their mass is distributed. This demonstrates the relationship between angular speed and energy, as well as the moment of inertia, which affects how parts of an object move through space.

Uniform Rod

A uniform rod is a rod whose mass is evenly distributed along its length. For physical problems, such as the one described, it is important to note that the moment of inertia for a uniform rod rotating about one of its ends is given by the formula: \( I = \frac{1}{3} m L^2 \) where \( m \) is the mass of the rod and \( L \) is its length. Because of this distribution, Rod B has a smaller moment of inertia compared to when mass is concentrated at one end. Thus, its rotational kinetic energy is less compared to Rod A when both are spinning at the same angular speed.

Point Mass

A point mass is an idealized concept where an entire body's mass is concentrated at a single point in space. This is used to simplify calculations when the distribution of mass doesn't significantly affect the results. For Rod A, the mass is assumed to be concentrated at the end farthest from the axis. In calculations, the moment of inertia for a point mass at distance \( L \) from the pivot is \( I = mL^2 \). This concept illustrates how distributing mass affects rotational motion, as concentrating the mass at one end of Rod A leads to a higher moment of inertia and subsequently, a higher rotational kinetic energy compared to Rod B.