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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’s 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

A large uniform horizontal turntable rotates freely about a vertical axle at its center. You measure the radius of the turntable to be \(3.00 \mathrm{~m} .\) To determine the moment of inertia \(I\) of the turntable about the axle, you start the turntable rotating with angular speed \(\omega\), which you measure. You then drop a small object of mass \(m\) onto the rim of the turntable. After the object has come to rest relative to the turntable, you measure the angular speed \(\omega_{\mathrm{f}}\) of the rotating turntable. You plot the quantity \(\left(\omega-\omega_{\mathrm{f}}\right) / \omega_{\mathrm{f}}\) (with both \(\omega\) and \(\omega_{\mathrm{f}}\) in rad \(\left./ \mathrm{s}\right)\) as a function of \(m\) (in kg). You find that your data lie close to a straight line that has slope \(0.250 \mathrm{~kg}^{-1}\). What is the moment of inertia \(I\) of the turntable?

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00 -m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The \(0.800 \mathrm{~kg}\) rod has a light basket at the other end to catch the \(0.163 \mathrm{~kg}\) puck. The puck slides across the ice with velocity \(\overrightarrow{\boldsymbol{v}}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

A uniform solid cylinder with mass \(M\) and radius \(2 R\) rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass \(M\) and radius \(R\) that is mounted on a frictionless axle through its center. A block of mass \(M\) is suspended from the free end of the string (Fig. \(\mathbf{P 1 0 . 7 9}\) ). The string doesn't slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after the system is released from rest.

A \(2.20 \mathrm{~kg}\) hoop \(1.20 \mathrm{~m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady \(2.60 \mathrm{rad} / \mathrm{s}\). (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A large wooden turntable in the shape of a flat uniform disk has a radius of \(2.00 \mathrm{~m}\) and a total mass of \(120 \mathrm{~kg}\). The turntable is initially rotating at \(3.00 \mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a \(70.0 \mathrm{~kg}\) parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

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