Phys105A_W09_HW4sols

# Phys105A_W09_HW4sols - Taylor 4.8 This is a one-dimensional...

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Unformatted text preview: Taylor 4.8 This is a one-dimensional problem in θ as defined in the picture above. We’re told there are no nonconservative forces so we can say that the total energy E stays constant. Just as the puck is nudged, it’s kinetic energy T ( θ = 0) is zero and it’s potential energy, defined relative to the center of the sphere, is U ( θ = 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 θ U ( θ ) = mgR cos θ. (2) Its kinetic energy also changes as a function of the angle, 1 2 mv ( θ ) 2 = T ( θ ) = E ( θ )- U ( θ ) = mgR (1- cos θ ) (3) Now let’s look at the free body diagram for a minute. Newton’s Law for the radial component of the force on the puck gives us N- mg cos θ =- 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 θ by rearranging this, and by using mv 2 R = 2 T R = 2 mgR (1- cos θ ) R . (5) So we get N = mg cos θ- 2 mg (1- cos θ ) = mg (3 cos θ- 2) (6) 1 The obvious condition for the angle (...
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Phys105A_W09_HW4sols - Taylor 4.8 This is a one-dimensional...

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