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Unformatted text preview: EXPERIMENT 9B Rotational Motion 2 Moment of inertia Objectives to familiarize yourself with the concept of moment of inertia, I, which plays the same role in the description of the rotation of a rigid body as mass plays in the description of linear motion to investigate how changing the moment of inertia of a body affects its rotational motion APPARATUS See Figure 3a. THEORY If we apply a single unbalanced force, F, to an object, the object will undergo a linear acceleration, a, which is determined by the unbalanced force acting on the object and the mass of the object. The mass is a measure of an object's inertia, or its resistance to being accelerated. Newtons Second Law expresses this relationship: F = ma If we consider rotational motion, we find that a single unbalanced torque = (Force)(lever arm) # produces an angular acceleration, , which depends not only on the mass of the object but on how that mass is distributed . The equation which is analogous to F = ma for an object that is rotationally accelerating is = I . (1) where the Greek letter tau ( ) represents the torque in Newtonmeters, is the angular acceleration in radians/sec 2 and I is the moment of inertia in kg*m 2 . The moment of inertia is a measure of the way the mass is distributed on the object and determines its resistance to angular acceleration. # In this lab the lever arm will be the radius at which the force is applied (the radius of the axle). This is due to the fact that the forces will be applied tangentially, i.e., perpendicular to the radius. The general form of this relationship is sin * * arm lever force = , where is the angle between the force and the lever arm. However, in this experiment is 90 and sin(90) = 1. Every rigid object has a definite moment of inertia about any particular axis of rotation. Here are a couple of examples of the expression for I for two special objects: One point mass m on a weightless rod of radius r ( I = mr 2 ): x y z O Figure 1 Two point masses on a weightless rod ( I = m 1 r 1 2 + m 2 r 2 2 ): y x z Figure 2 To illustrate we will calculate the moment of inertia for a mass of 2 kg at the end of a massless rod that is 2 m in length (object #1 above): I = mr 2 = (2 kg) (2 m) 2 = 8 kg*m...
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This note was uploaded on 07/25/2008 for the course PHY 251 taught by Professor J.t. during the Summer '08 term at Michigan State University.
 Summer '08
 J.T.
 Physics, Inertia, Mass

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