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Chapter 10: Problem 28

Two uniform solid balls are rolling without slipping at a constant speed. Ball 1 has twice the diameter, half the mass, and one-third the speed of ball 2 . The kinetic energy of ball 2 is \(27.0 \mathrm{~J}\). What is the kinetic energy of ball \(1 ?\)

Short Answer

Expert verified
After performing the calculations, you will find that the kinetic energy of Ball 1 is 5 J.

Step by step solution

01

Analyzing the problem

Here, you are given two balls with varying diameters, masses, speeds, and kinetic energy. The energy is directly proportional to the mass \(m\) and the square of the speed \(v^2\), and since Ball 1 has twice the diameter, half the mass, and one-third the speed of Ball 2, you'll need to adjust the equation for Ball 1's kinetic energy accordingly.
02

Find the speed of Ball 2

The kinetic energy of Ball 2 is given as 27 J. Given that the total kinetic energy \(K_2\) is \(\frac{7}{10}mv^2\), we can solve for the speed \(v_2\).
03

Calculate the kinetic energy of Ball 1

Now, let's substitute the given values into the kinetic energy formula for Ball 1. The speed of Ball 1 is one-third the speed of Ball 2, \(v_1= v_2/3\). The mass of Ball 1 is half the mass of Ball 2, \(m_1=m_2/2\). Hence the kinetic energy of Ball 1 is \(\frac{7}{10}(m_2/2)((v_2/3)^2)\).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes to its rotation. Specifically, it is the resistance to angular acceleration based on the distribution of an object's mass. It鈥檚 similar to mass in linear motion but for rotational motion. For instance, in the exercise, the moment of inertia can be used to understand the rotational kinetic energy of the rolling balls.

The moment of inertia ('I') of a solid ball is given by the formula \( I = \frac{2}{5}mr^2 \) where 'm' is the mass and 'r' is the radius of the ball. Since Ball 1 has twice the diameter of Ball 2, its radius is also doubled which will greatly affect its moment of inertia. An increased moment of inertia means that Ball 1 will have a different energy distribution than Ball 2 when rolling without slipping, playing a crucial role in determining its kinetic energy.
Rotational Motion
Rotational motion involves an object spinning around an internal axis, and it's governed by similar principles to those of linear motion, but with some key differences. One such principle is the conservation of angular momentum, which dictates that in the absence of external torques, the total angular momentum of a system remains constant.

In the case of our rolling balls, this motion is a form of kinetic energy, known as rotational kinetic energy, represented by the equation \( K_{rot} = \frac{1}{2}Iw^2 \), where 'I' is the moment of inertia and 'w' is the angular velocity. Notably, as Ball 1 has a larger diameter and different speed, this affects its angular velocity. Due to its size, Ball 1 will have a smaller angular velocity than Ball 2, assuming they cover the same linear distance in a given time. In general, understanding the concepts of rotational motion is essential in calculating the kinetic energy for rolling objects.
Energy Conservation
Energy conservation is a fundamental principle of physics stating that the total energy in a closed system remains constant over time. In terms of mechanics, this means the sum of potential energy and kinetic energy remains invariant if there's no energy transfer to or from the system. The exercise touches on this concept through the calculation of kinetic energy.

It's important to note that for rolling objects, the total kinetic energy is the sum of translational kinetic energy \( K_{trans} = \frac{1}{2}mv^2 \) and rotational kinetic energy \( K_{rot} = \frac{1}{2}Iw^2 \). Therefore, when computing Ball 1's kinetic energy, we not only consider its slower speed v_1, which decreases its translational kinetic energy, but also its larger moment of inertia, which affects its rotational kinetic energy. In conclusion, energy conservation guides us to correctly account for all forms of energy in the system to solve for the unknown kinetic energy of Ball 1.

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Most popular questions from this chapter

Stabilization of the Hubble Space Telescope. The Hubble Space Telescope is stabilized to within an angle of about 2 -millionths of a degree by means of a series of gyroscopes that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass \(2.0 \mathrm{~kg}\) and diameter \(5.0 \mathrm{~cm},\) spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of \(1.0 \times 10^{-6}\) degree during a 5.0 hour exposure of a galaxy?

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3 ). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\). If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{\mathrm{k}}\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0\) Rolling without slipping sets in when \(v_{x}=r \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius \(0.600 \mathrm{~m}\) and mass \(1.60 \mathrm{~kg}\) and is initially at rest. A bullet with mass \(0.0200 \mathrm{~kg}\) is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of \(0.600 \mathrm{~m}\) from the axle. After being struck by the bullet, the disk rotates at \(4.00 \mathrm{rad} / \mathrm{s}\). What is the horizontal velocity of the bullet just before it strikes the disk?

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to \(80.0 \mathrm{~N}\) is applied to the rim of the wheel. The wheel has radius \(0.120 \mathrm{~m}\) Starting from rest, the wheel has an angular speed of \(12.0 \mathrm{rev} / \mathrm{s}\) after \(2.00 \mathrm{~s}\). What is the moment of inertia of the wheel?

The mechanism shown in Fig. \(\mathbf{P} \mathbf{1 0 . 6 4}\) is used to raise a crate of supplies from a ship's hold. The crate has total mass \(50 \mathrm{~kg} .\) A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius \(0.25 \mathrm{~m}\) and moment of inertia \(I=2.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius \(0.12 \mathrm{~m},\) the cylinder turns, and the crate is raised. What magnitude of the force \(\vec{F}\) applied tangentially to the rotating crank is required to raise the crate with an acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2} ?\) (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)
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