physics prelab 4

physics prelab 4 - forces of the static masses. The slope...

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The acceleration of a rotating object is not zero for uniform circular motion because the velocity of that object is tangential to the circle. To continue to move around the circle with constant speed, a force must be acting on the object. That force is the centripetal force. Newton's second law (F=ma) describes how there must be an acceleration if a force is constantly being placed on the object. The mass of the rotating mass is used in the equation F = mr(omega)^2. The static mass allows us to calculate the force applied by the spring. This, in turn, allows us to calculate the centripetal force placed on the rotating mass. Using a known static mass gives the force pulling opposite the centripetal force. By rotating the apparatus until the rotating mass is centered between the spring and the static mass, we are able to determine when the centripetal force is equal in magnitude to the known
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Unformatted text preview: forces of the static masses. The slope of the graph of omega^2 vs. F (angular speed squared vs. force) will represent the radius of the rotating mass. Centripetal force = ma where a = v^2/r, so Force = mv^2/r. Since m is constant, it can be simplified to radius = velocity squared/force. The apparatus not being level will introduce systematic error into the experiment. The forces will work at different angles, especially the normal force, rather than perpendicular to the ground, and some of the vertical component of gravity will be translated into horizontal components. In order to reduce systematic error, we are going to check to make sure that the apparatus is in fact level. In addition, the weight of the static masses and rotating masses will be measured, rather than just taken from the label on the mass....
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This note was uploaded on 11/27/2011 for the course PHYS 105 taught by Professor Walker during the Fall '08 term at UNC.

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