problem07_14

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.14: a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the bottom of the circular arc) is ), cos 1 ( θ mgl - where l is the length of the string and θ is the angle the string makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so , ) cos 1 ( 2 2 1 mv θ mgl = - or s m 1 . 2 ) 45 cos 1 ( m) 80 0 ( ) s m 80 9 ( 2 ) cos 1 ( 2 2 = ° - = - = . . θ gl v . b) At ° 45 from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the
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Unformatted text preview: radial component of the weight, or N. 83 . 45 cos ) s m (9.80 kg) 12 . ( cos 2 = ° = θ mg c) At the bottom of the circle, the tension is the sum of the weight and the radial acceleration, N, 1.86 )) 45 cos 1 ( 2 1 ( 2 2 = °-+ = + mg l mv mg or 1.9 N to two figures. Note that this method does not use the intermediate calculation of v ....
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## This document was uploaded on 02/04/2008.

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