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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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**Vector Product (Cross-product)**

There are two different methods for determining the vector product between any two vectors: The Determinant method and the Cyclic method Determinant Method Cyclic Method Rewrite the j term as to get an identical expression

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The direction of the resultant vector, for a right-handed coordinate system, for a vector product can be determined using the right-hand rule. Fingers point in the direction of the second vector The vector product looks at the product of two vectors which are perpendicular to each other, and who are also perpendicular to the resultant vector. Thumb points in the direction of first vector. Palm points in the direction of resultant vector. Example: Determine the magnitude and direction of the area of a parallelogram described by the vectors r1 and r2 which are used to describe the length of the two sides. y x r1 r2 Out of the page

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Torque When a force is applied to an object it will cause it to accelerate. This acceleration would correspond to a change in the velocity of the object. The velocity could be a translational velocity or a rotational velocity. A torque is a force that causes a change in the angular velocity of the object. r1 r2 q1 q2 The force F1 will cause a counterclockwise rotation about an axis that passes through O. The force F2 will cause a clockwise rotation about an axis that passes through O. r is the distance from the origin O to the point of application of F. d is the lever arm for F which is the distance from O to F such that d is perpendicular to F. t – Torque [Nm] r – distance from axis of rotation [m] F – Force [N] q – angle measured from r to F d = rsinq this is the lever arm, the component of r perpendicular to F The direction of the torque vector is perpendicular to both r and F. The directions clockwise and counterclockwise are used to describe the direction the torque causes the object to rotate.

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Angular Momentum The vector angular momentum of the point mass m about the point P is given by: The position vector of the mass m relative to the point.

Angular Momentum The vector angular momentum of the point mass m about the point P is given by: The position vector of the mass m relative to the point.

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