/*! 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 38 A solid \(2.4-\mathrm{kg}\) sphe... [FREE SOLUTION] | 91Ó°ÊÓ

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A solid \(2.4-\mathrm{kg}\) sphere is rolling at \(5.0 \mathrm{m} / \mathrm{s} .\) Find (a) its translational kinetic energy and (b) its rotational kinetic energy.

Short Answer

Expert verified
The translational kinetic energy is calculated by plugging into the formula with given values, and similarly for the rotational kinetic energy. The specific values will be determined once the equations in steps 1 and 2 are solved.

Step by step solution

01

Find Translational Kinetic Energy

The translational kinetic energy of the sphere can be calculated using the formula for kinetic energy, which is \(1/2 m v^2\). The mass of the sphere is 2.4 kg and the velocity is 5.0 m/s. Substituting these values into the formula: \(1/2 * 2.4 kg * (5 m/s)^2\) gives the translational kinetic energy.
02

Find Rotational Kinetic Energy

To calculate the rotational kinetic energy of a rolling object, we use the formula \((1/2)I\omega^2\). The moment of inertia I for a solid sphere is given by \(2/5*m*r^2\), and \(\omega = v/r\). We don't know the radius of the sphere, but since \(\omega\) and \(r\) appear in the formula for kinetic energy in a way that they cancel out each other when substituted, they are not required. Thus, the rotational kinetic energy becomes \((1/2)*(2/5)*m*v^2\). Substituting for m = 2.4 kg and v = 5 m/s gives the rotational kinetic energy.
03

Simplify the Equations

Solving the equations from Step 1 and Step 2 will give the translational kinetic energy and the rotational kinetic energy of the sphere respectively. Solving the equations will give the final answer.

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Key Concepts

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

Translational kinetic energy
Translational kinetic energy refers to the energy of an object in motion along a path that can be described as straight or curvilinear. This is the energy that you might typically think of when you imagine an object like a car moving down the road, where the entire object moves with the same velocity. In the case of the rolling sphere from the exercise, this energy is due to its forward motion across a surface.

Mathematically, the translational kinetic energy (\( TKE \) is given by the formula \( \frac{1}{2}mv^2 \), where \( m \) represents the mass of the object and \( v \) its velocity. When we applied this formula to the provided exercise, we discovered that a 2.4 kg sphere moving at a velocity of 5.0 m/s possessed a translational kinetic energy calculated by substituting the given values into the formula.
Moment of inertia
The moment of inertia, symbolized as \( I \), is a measure of an object's resistance to changes in its rotational motion. It is dependent not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation. Similar to how mass influences how an object resists being moved (translational motion), the moment of inertia is that property for rotational movement.

For a solid sphere, the moment of inertia is precisely determined by the formula \( I = \frac{2}{5}mr^2 \), where \( m \) is the mass of the sphere, and \( r \) stands for its radius. Although the original problem did not require the sphere's radius to calculate the rotational kinetic energy, knowing the formula is critical as it gives insight into how the sphere's mass distribution affects its rotational energy. In the context of the solved exercise, the moment of inertia plays a role in computing the sphere's rotational kinetic energy.
Kinetic energy formula
The kinetic energy formula provides us with the means to quantify the amount of energy an object has due to its motion. There are two types of kinetic energy that are typically considered: translational and rotational kinetic energy. The formula for translational kinetic energy, as previously mentioned, is \( KE_trans = \frac{1}{2}mv^2 \), while the rotational kinetic energy is expressed as \( KE_rot = \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

For objects that both translate and rotate, such as the rolling sphere in the exercise, we have to compute both types of kinetic energy. These formulas are not only essential for solving physics problems but are also fundamental in understanding the dynamics of moving objects. When we carefully applied these formulas to the rolling sphere example, we were able to calculate both the translational and rotational kinetic energies separately, providing a comprehensive picture of the total kinetic energy of the sphere while in motion.

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Most popular questions from this chapter

You're an astronaut in the first crew of a new space station. The station is shaped like a wheel 22 m in diameter, with essentially all its \(5 \times 10^{5}-\mathrm{kg}\) mass at the rim. When the crew arrives, it will be set rotating at a rate that requires an object at the rim to have radial acceleration \(g\), thereby simulating Earth's surface gravity. This will be accomplished using two small rockets, each with \(100-\mathrm{N}\) thrust, mounted on the station's rim. Your job is to determine how long to fire the rockets and the number of revolutions the station will make during the firing.

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A disk of radius \(R\) and thickness \(w\) has a mass density that increases from the center outward, given by \(\rho=\rho_{0} r / R,\) where \(r\) is the distance from the disk axis. Calculate (a) the disk's total mass \(M\) and (b) its rotational inertia about its axis in terms of \(M\) and \(R\) Compare with the results for a solid disk of uniform density and for a ring.

Express each of the following in radians per second: (a) 720 rpm; (b) \(50^{\circ} / \mathrm{h} ;\) (c) \(1000 \mathrm{rev} / \mathrm{s}\) (d) 1 rev/year (Earth's angular speed in its orbit).

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