181_PartUniversity Physics Solution

181_PartUniversity Physics Solution - Applying Newtons Laws...

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Applying Newton&s Laws 5-53 5.119. IDENTIFY: The analysis is the same as for Problem 5.96. SET UP: The speed is related to the period by 22 ( t a n ) / vR Th T ππ β == , or 2 (tan )/ v πβ = . EXECUTE: The maximum and minimum speeds are the same as those found in Problem 5.96, s max s cos sin tan sin cos vg h βμ β βμ + = and s min s cos sin tan sin cos h = + . The minimum and maximum values of the period T are then s min s tan sin cos 2 cos sin h T g ββ μ π = + and s max s tan sin cos 2 cos sin h T g + = . EVALUATE: For s 0 = the results for the speeds reduce to min max vv g h . tan R h = . The result for v then agrees with the result in Example 5.23, if we take into account that in this problem is measured from the vertical whereas in Example 5.23 it is measured relative to the horizontal. 5.120. IDENTIFY: Apply m = F a ! ! to the block and to the wedge. SET UP: For both parts, take the x -direction to be horizontal and positive to the right, and the y -direction to be vertical and positive upward. The normal force between the block and the wedge is n ; the normal force between the wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The horizontal acceleration of the wedge is A , and the components of acceleration of the block are x a and y a . EXECUTE: (a) The equations of motion are then sin MA n α = − , sin x ma n = and cos y ma n mg =− . Note that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are three equations in four unknowns, A , , x y aa and n . Solution is possible with the imposition of the relation between A , x a and y a . An observer on the wedge is not in an inertial frame, and should not apply Newton&s laws, but the kinematic relation between the components of acceleration are not so restricted. To such an observer, the vertical acceleration of the block is , y a but the horizontal acceleration of the block is . x aA To this observer, the block descends at an angle , so the relation needed is tan . y x a α At this point, algebra is unavoidable. A possible approach is to eliminate x a by noting that x M m , using this in the kinematic constraint to eliminate y a and then eliminating n . The results are : () t a n ( t a n ) gm A Mm M = ++ t a n ( t a n ) x gM a M = (m ) t a n t a n ( t a n ) y gM a M −+ = (b) When , 0, A >> as expected (the large block won&t move). Also, 2 tan sin cos (1 tan ) tan 1 x g ag g αα →= = which is the acceleration of the block ( sin g in this case), with the factor of cos giving the horizontal component. Similarly, 2 sin y →− . (c) The trajectory is a spiral. EVALUATE: If mM >> , our general results give 0 x a = and y = − . The massive block accelerates straight downward, as if it were in free-fall. 5.121. IDENTIFY: Apply m = F a ! ! to the block and to the wedge.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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181_PartUniversity Physics Solution - Applying Newtons Laws...

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