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Unformatted text preview: Taylor 4.8 This is a one-dimensional problem in as defined in the picture above. Were told there are no nonconservative forces so we can say that the total energy E stays constant. Just as the puck is nudged, its kinetic energy T ( = 0) is zero and its potential energy, defined relative to the center of the sphere, is U ( = 0) = mgR . Hence the total energy is E = mgR. (1) As the puck rolls down the sphere, its potential energy changes as a function of U ( ) = mgR cos . (2) Its kinetic energy also changes as a function of the angle, 1 2 mv ( ) 2 = T ( ) = E ( )- U ( ) = mgR (1- cos ) (3) Now lets look at the free body diagram for a minute. Newtons Law for the radial component of the force on the puck gives us N- mg cos =- mv 2 R = F NET r (4) since a particle traveling in a circular path must have such a net force to provide the centripetal acceleration. We can get the normal force, N , as a function of by rearranging this, and by using mv 2 R = 2 T R = 2 mgR (1- cos ) R . (5) So we get N = mg cos - 2 mg (1- cos ) = mg (3 cos - 2) (6) 1 The obvious condition for the angle (...
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