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Moments of Inertia Do NOW : Serway Examples 10.5 – 10.6 – 10.7 – page 304. Homework : Serway Pg : 309 – 311 Examples 10.10 – 10.13 Pg : 313 Table 10.3 Work – Kinetic Energy theorem (10.24) Pg : 314 Examples 10.14 – 10.15 Summary Page 314 Problems Pg : 321 #’s 37, 39.
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Moments of Inertia Do NOW: SerwayExamples 10.5 – 10.6 – 10.7 – page 304 • Homework: Serway • Pg: 309 – 311 Examples 10.10 – 10.13 • Pg: 313 Table 10.3 • Work – Kinetic Energy theorem (10.24) • Pg: 314 Examples 10.14 – 10.15 • Summary Page 314 • Problems • Pg: 321 #’s 37, 39 http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
. Moment of InertiaMoment of inertia is the name given to rotational inertia, the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation. For a point mass the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr2. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses.
Moment of inertia is defined with respect to a specific rotation axis. The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared. The moment of inertia of any extended object is built up from that basic definition. The general form of the moment of inertia involves an integral.
The moment of inertia calculation for a uniform rod involves expressing any mass element in terms of a distance element dralong the rod. To perform the integral, it is necessary to express everything in the integral in terms of one variable, in this case the length variable r. Since the total length L has mass M, then M/L is the proportion of mass to length and the mass element can be expressed as shown. Integrating from -L/2 to +L/2 from the center includes the entire rod. The integral is of Polynomial type: Rod Inertia Calculations
Quick LabKey Whip A string about one meter long has a (relatively heavy) set of keys on one end and a (very light) match box on the other end. The string passes over a pencil with the keys hanging down and the matchbox held horizontal to the pencil with about two/thirds of the string between the pencil and the matchbox. Q: What will happen when the match box is released A: Surprisingly, the keys will not fall to the floor. When the matchbox falls it develops angular momentum. Conservation of angular momentum of the matchbox causes it to rotate very rapidly about the pencil as the string pulls it in. Before the string is used up, the matchbox string actually wraps around the pencil, preventing the keys from falling onto the floor!
Quick LabFooling Gravity PURPOSE: To illustrate gyroscopic precession in a surprising way. EQUIPMENT: Bicycle wheel gyroscope and wheel. DESCRIPTION: Spin the wheel, then hang it from the end of the rope with the rope vertical and the wheel axel in a horizontal orientation. Question(s): What will happen? The wheel will precess in the horizontal with the rope remaining vertical. SUGGESTIONS: Set up the students for this by showing them this same bicycle wheel gyroscope on a rigid pivot. Then ask them if one of the conditions to obtain precession might be a strong, rigid pivot!
Homework • Serway • Pg: 309 – 311 Examples 10.10 – 10.13 • Pg: 313 Table 10.3 • Work – Kinetic Energy theorem (10.24) • Pg: 314 Examples 10.14 – 10.15 • Summary Page 314 • Problems • Pg: 321 #’s 37, 39