Homework 6 - Solutions - homework 06 FRENNEA, KYLE Due: Feb...

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Unformatted text preview: homework 06 FRENNEA, KYLE Due: Feb 28 2007, 4:00 am 1 Question 1 part 1 of 5 10 points A stone at the end of a sling is whirled in a vertical circle of radius 89 cm. At points A and B the string is inclined from the hor- izontal axis at an angle 40 and the stone has speed 3 . 7 m / s, as shown in the figure. The center of the circle is 1 . 34 m above level ground. The acceleration of gravity is 9 . 8 m / s 2 . 8 9 c m A 3 . 7 m / s 8 9 c m B 3 . 7 m / s 4 4 1 . 34 m What is the magnitude of the distance trav- eled | x | in the horizontal direction before it hits the ground if the stone is released at A , ( x A , y A )? Correct answer: 2 . 32505 m (tolerance 1 %). Explanation: Given : r = 0 . 89 m , v = 3 . 7 m / s , = 40 , and d = 1 . 34 m . At position A ( x A , y A ) y = d + r sin(40 ) = +1 . 91208 m v y = v cos(40 ) = +2 . 83436 m / s v x = v x = v sin(40 ) = 2 . 37831 m / s . Consider the kinematic equation v 2 y = v 2 y 2 g ( y y ) . Set y = 0 and y is given above. Then, v y = radicalBig v 2 y + 2 g y . To find the time of flight, consider the kine- matic equation v y = v y g t. Solving for t yields, t A = v y + v y g = 1 g braceleftBigradicalBig v 2 y + 2 g y + v y bracerightBig = 1 9 . 8 m / s 2 braceleftBigbracketleftBig (2 . 83436 m / s) 2 + 2 (9 . 8 m / s 2 ) (1 . 91208 m) bracketrightBig 1 / 2 + 2 . 83436 m / s bracerightBig = 0 . 977602 s . Hence the range (magnitude of x ) of the stone when released at A is, x A = v sin( ) t A = (3 . 7 m / s) sin(40 ) (0 . 977602 s) = 2 . 32505 m . Question 2 part 2 of 5 10 points Given: y = d + r sin . What is the magnitude of the distance trav- eled | x | in the horizontal direction before it hits the ground if the stone is released at B , ( x A , x B )? Correct answer: 0 . 94933 m (tolerance 1 %). Explanation: The range (magnitude of x ) of the stone when released at B is found by a similar pro- cess as above, except that v y = v cos = (3 . 7 m / s) cos(40 ) = 2 . 83436 m / s . homework 06 FRENNEA, KYLE Due: Feb 28 2007, 4:00 am 2 Then t B = 1 g parenleftBigradicalBig v 2 y + 2 g y + v y parenrightBig = 1 (9 . 8 m / s 2 ) parenleftBig bracketleftbig (2 . 83436 m / s) 2 +2 (9 . 8 m / s 2 ) (1 . 91208 m) bracketrightbig 1 / 2 2 . 83436 m / s parenrightBig = 0 . 399161 s . The range (magnitude of x ) of the stone when released at B is x B = v (sin ) t B = (3 . 7 m / s) (sin40 ) (0 . 399161 s) = 0 . 94933 m . Question 3 part 3 of 5 10 points What is the magnitude of the centripetal acceleration in the horizontal direction of the stone just before it is released at A , ( x A , y A )?...
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This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Homework 6 - Solutions - homework 06 FRENNEA, KYLE Due: Feb...

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