/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A uniform rod of mass \(1.0 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform rod of mass \(1.0 \mathrm{kg}\) and length \(2.0 \mathrm{m}\) is free to rotate about one end (see the following figure). If the rod is released from rest at an angle of \(60^{\circ}\) with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?

Short Answer

Expert verified
The speed of the tip of the rod as it passes the horizontal position is \(3.58\, m/s\).

Step by step solution

01

Determine the initial potential energy of the rod

The initial potential energy is the gravitational potential energy when the rod is at a 60-degree angle to the horizontal. This can be calculated as: \[ U = mgh \] where: \(U\) = potential energy, \(m\) = mass of the rod (1.0 kg), \(g\) = acceleration due to gravity (9.81 m/s²), \(h\) = height of the rod's center of mass above the horizontal position. To find the height, we can use the length of the rod (2 m) and the angle (60 degrees) with respect to the horizontal. The distance from the pivot to the center of mass is half the length of the rod, so 1 m. The height can be found as: \[ h = L \sin(\theta) \] where: \(L\) = half-length of the rod (1 m), \(\theta\) = angle from the horizontal (60 degrees). Calculating \(h\): \(h = 1 \sin(60^\circ) = 1\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\,m \) Now, we can calculate the potential energy: \(U = (1\, kg)(9.81\, m/s^2)(\frac{\sqrt{3}}{2}\,m) = 8.49\,J \)
02

Conservation of Mechanical Energy

As the rod rotates to the horizontal position, the potential energy converts into kinetic energy. Since mechanical energy is conserved, the initial potential energy equals the final kinetic energy: \[ U = \frac{1}{2}I\omega^2 \] where: \(I\) = moment of inertia of the rod about the pivot, \(\omega\) = angular speed of the rod when it reaches the horizontal position. For a uniform rod, the moment of inertia about one end is given by: \[ I = \frac{1}{3}mL^2 \] Substituting the given values to find the moment of inertia: \(I = \frac{1}{3}(1\, kg)(2\, m)^2 = \frac{4}{3}\, kg\cdot m^2 \) Now, we can find the angular speed \(\omega\): \[ \omega^2 = \frac{2U}{I} \] \[ \omega^2 = \frac{2 \cdot 8.49\, J}{\frac{4}{3}\, kg\cdot m^2} = 3.19\, rad^2/s^2 \] \[ \omega = \sqrt{3.19\, rad^2/s^2} = 1.79\, rad/s \]
03

Calculate the linear speed of the rod's tip

Now that we have the angular speed of the rod, we can find the linear speed of the tip as it passes the horizontal position. This can be found using the following equation: \[ v = \omega L \] where: \(v\) = linear speed of the rod's tip, \(\omega\) = angular speed (1.79 rad/s), \(L\) = length of the rod (2 m). Calculating the linear speed: \(v = (1.79\, rad/s)(2\, m) = 3.58\, m/s \) So, the speed of the tip of the rod as it passes the horizontal position is \(3.58\, m/s\).

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Key Concepts

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

Gravitational Potential Energy
Imagine holding a book above the ground. The moment you release it, the book falls. This is because it had energy due to its position high above the ground—an energy known as gravitational potential energy (GPE). GPE is the energy an object possesses because of its position in a gravitational field. Objects at a height have the potential to fall, converting this stored energy into kinetic energy.

The formula to calculate GPE is \( U = mgh \), where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height above the reference point. In our textbook exercise, we calculated the GPE of a rod at a certain height by considering its center of mass and the angle at which it's held. Recognizing the center of mass and its height helps us determine the initial energy that later transforms into motion. Through these concepts, we can predict movement without observing it in real life, which is highly beneficial in understanding real-world applications like the energy in roller coasters or pendulums.
Conservation of Mechanical Energy
The concept of the conservation of mechanical energy is a fundamental principle in physics that essentially tells us that the total mechanical energy (potential plus kinetic energy) of an isolated system remains constant if only conservative forces are acting on it. A conservative force, like gravity, doesn't dissipate energy; it just converts it from one form to another.

In the context of our rod being dropped, as the rod falls, the mechanical energy transitions from potential to kinetic form, as demonstrated by the equation \( U = \frac{1}{2}I\omega^2 \). Here, potential energy is being converted to rotational kinetic energy, and thanks to energy conservation, we know the total amount of energy remains constant, giving us a way to calculate the movement of objects, just like finding the angular velocity of falling objects. By thoroughly understanding this concept, students can see the invisible interplay of energy in everyday phenomena—an essential topic in physics that is not only captivating but also incredibly useful.
Moment of Inertia
When you try to spin a wheel or twirl a baton, you're experiencing a property known as moment of inertia (MOI). It's a measure of an object's resistance to changes in its rotation, akin to mass in linear motion. But instead of just mass, MOI depends on both the mass of the object and how that mass is distributed relative to the axis of rotation.

The MOI for different objects can be calculated using specific formulas. For example, for a rod rotating about one end, the formula is \( I = \frac{1}{3}mL^2 \). In our exercise, harnessing this formula allows us to find the rod's MOI and, subsequently, its angular speed as it drops. Understanding MOI isn't only important for solving textbook problems but also crucial in real-world applications like engineering, where ensuring the proper rotation of machinery components could be the difference between a smooth operation and a mechanical failure.

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