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CH 10: Rotation

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**Rotational Kinetic Energy**

Kinetic energy describes the amount of energy an object has when it is moving. We have discussed translational motion, but we must also consider rotational motion. Let us look at the kinetic energy of a system of particles that is rotating about an arbitrary point. We begin by examining the kinetic energy of one particle of that system. Ki – Kinetic energy of ith particle vi – Tangential velocity of ith particle It is more convenient to look at the angular velocity for a system of rotating objects, especially if we assume they all have the same angular velocity. If we look at the entire system, we must sum the kinetic energies for each particle. KR – Rotational kinetic energy [J] I – Rotational Inertia [kg m2] We must consider the distribution of mass as opposed to just the mass. Where

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**Rotational Inertia (Moment of Inertia)**

The rotational inertia is the inertia of a rotating object. This determines how hard it is to change the motion of a rotating object. The mass of the object coupled with the distance from the point of rotation defines the rotational inertia. The more mass and the greater the distance from the point of rotation, the harder it is to change the rotation of the object. Discrete particles Continuous object I – Rotational Inertia r – the distance from the axis of rotation

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**3. Both reach the bottom at the same time.**

Two cylinders of the same size and mass roll down an incline. Cylinder A has most of its weight concentrated at the rim, while cylinder B has most of its weight concentrated at the center. Which reaches the bottom of the incline first? 1. A 2. B 3. Both reach the bottom at the same time. Answer: 2.When the cylinders roll down the incline, gravitational potential energy gets converted to translational and rotational kinetic energy: mgh = 1⁄2 mv2 + 1⁄2 Iω2. We can substitute v/r for ω and write mgh = 1⁄2 (m + I/r2)v2. The values for m and r are the same for both cylinders, and so the cylinder having the smaller rotational inertia has the larger speed. The smaller rotational inertia is obtained when more of the mass is concentrated near the center.

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**1. mring= mdisk, where m is the inertial mass. **

A solid disk and a ring roll down an incline. The ring is slower than the disk if 1. mring= mdisk, where m is the inertial mass. 2. rring = rdisk, where r is the radius. 3. mring = mdisk and rring = rdisk. 4. The ring is always slower regardless of the relative values of m and r. Answer: 4. The ring has more rotational inertia per unit mass than the disk. Therefore as it starts rolling, it has a relatively larger fraction of its total kinetic energy in rotational form and so its translational kinetic energy is lower than that of the disk. As the disk and ring roll down the incline, the potential energy of each is reduced by an amount mgh, where h is the difference in height between the bottom and top of the incline.This energy is converted to kinetic energy (translational and rotational): mgh = 1⁄2 mv2 + 1⁄2 Iω2.We can write the rotational inertia as I= cmR2,where c is a constant equal to 1 for the ring and 1⁄2 for the disk.Because ω= v/R,we have:mgh= 1⁄2mv2+ 1⁄2 cmR2(v/R)2= 1⁄2 (1 + c)mv2. The larger c, therefore, the smaller v.Thus the ring, which has the larger rotational inertia, takes longer to go down the incline, regardless of inertia and radius.

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**Example: Determine the rotational inertia of a cylinder about its central axis.**

z Total volume of cylinder Total mass of cylinder Rotational Inertia of a solid cylinder rotating about its longitudinal axis.

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**Parallel-Axis Theorem**

The rotational inertial of different objects through an axis of symmetry was shown for several objects. If the rotation axis is shifted away from an axis of symmetry the calculation becomes more difficult. A simple method, called the parallel axis theorem, was devised for situations where the rotation axis was shifted some distance from the symmetry axis. The symmetry axis is any axis that passes through the center of mass. I – rotational inertia ICM – Rotational inertia for a rotation axis that passes through the center of mass M – Total mass of the object D – Distance the axis has been shifted by The new rotation axis must be parallel to the symmetry axis that is being used to define ICM.

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Physics 101: Lecture 13, Pg 1 Physics 101: Lecture 13 Rotational Kinetic Energy and Inertia Exam II.

Physics 101: Lecture 13, Pg 1 Physics 101: Lecture 13 Rotational Kinetic Energy and Inertia Exam II.

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