ch08-p019 - 19. (a) At Q the block (which is in circular...

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Using the fact that h = 5 R , we find mv 2 = 8 mgR . Thus, the horizontal component of the net force on the block at Q is F = mv 2 / R = 8 mg= 8(0.032 kg)(9.8 m/s 2 )= 2.5 N. and points left (in the same direction as G a ). (b) The downward component of the net force on the block at Q is the downward force of gravity F = mg =(0.032 kg)(9.8 m/s 2 )= 0.31 N. (c) To barely make the top of the loop, the centripetal force there must equal the force of gravity: mv R mg mv mgR t t 2 2 = ¡ = This requires a different value of h than was used above. 2 1 0 2 1 () ( 2 ) 2 PP tt tt KU K U mgh mv mgh mgh mgR mg R += + + =+ Consequently, h = 2.5 R = (2.5)(0.12 m) = 0.30 m. (d) The normal force F N , for speeds v t greater than gR (which are the only possibilities for non-zero F N — see the solution in the previous part), obeys 2 t N mv
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch08-p019 - 19. (a) At Q the block (which is in circular...

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