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Chapter_11_09 - .8 In the Figure a solid brass ball of mass...

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Chapter 11 11.8 In the Figure, a solid brass ball of mass 0.280 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 14.0 cm and the ball has radius r R . (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop if the ball is release at height h = 6.00 R , what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball at point Q. mg Q h i R 2R When the ball is just about to come off the track, the centripetal force is entirely due to gravity. This allows us to solve for the velocity mv 2 R = mg v 2 = Rg Now that we know the velocity, we can use energy conservation. The initial state is where we release the ball, the final state is at the top of the loop. E i = mgh E f = 1 2 mv 2 + 1 2 I ω 2 + mg (2 R )
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E i = E f mgh = mg (2 R ) + 1 2 mv 2 + 1 2 I ω 2 mgh = mg (2 R ) + 1 2 mv 2 + 1 2 ( 2 5 mr 2 ) v 2 r 2 mgh = mg (2 R ) + 1 2 mv 2 + 1 2 ( 2 5 m ) v 2 h = g (2 R ) + 1 2 v 2 + 1 2 ( 2 5 ) v 2 g = g (2 R ) + 1 2 Rg + 1 2 ( 2 5 ) Rg g = (2 R ) + 1 2 R + 1 5 R = 27 10 R In the second part of this problem, we need the velocity so that we can compute the centripetal force again, since it is the net inward force. In this case, we know the initial height. The force at Q will be to the left. E i = E f mg (6 R ) = mg ( R ) + 1 2 mv 2 + 1 2 I ω 2 mg (6 R ) = mg ( R ) + 1 2 mv 2 + 1 2 ( 2 5 mr 2 ) v 2 r 2 mg (5 R ) = 1 2 mv 2 + 1 2 ( 2 5 m ) v 2 v 2 = g (5 R ) 1 2 + 1 2 ( 2 5 ) = 10 7 g (5 R ) = 50 7 gR F = mv 2 R = m 50 7 gR R = 50 7 m g
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11.9 A solid cylinder of radius 10cm and mass 12kg starts from rest and rolls without slipping a distance of 6m down a house roof that is inclined at 30 degrees (a)
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