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Homework 6 - Solutions

# Homework 6 - Solutions - homework 06 FRENNEA KYLE Due 4:00...

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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 . 89 cm A 3 . 7 m / s 89 cm B 3 . 7 m / s 40 40 1 . 34 m What is the magnitude of the distance trav- eled | Δ 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 0 = 3 . 7 m / s , θ = 40 , and d = 1 . 34 m . At position A ( x A , y A ) y 0 = d + r sin(40 ) = +1 . 91208 m v y 0 = v 0 cos(40 ) = +2 . 83436 m / s v x 0 = v x = v 0 sin(40 ) = 2 . 37831 m / s . Consider the kinematic equation v 2 y = v 2 y 0 2 g ( y y 0 ) . Set y = 0 and y 0 is given above. Then, v y = radicalBig v 2 y 0 + 2 g y 0 . To find the time of flight, consider the kine- matic equation v y = v y 0 g t . Solving for t yields, t A = v y + v y 0 g = 1 g braceleftBigradicalBig v 2 y 0 + 2 g y 0 + v y 0 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 Δ x ) of the stone when released at A is, x A = v 0 sin( θ ) t A = (3 . 7 m / s) sin(40 ) (0 . 977602 s) = 2 . 32505 m . Question 2 part 2 of 5 10 points Given: y 0 = d + r sin θ . What is the magnitude of the distance trav- eled | Δ 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 Δ x ) of the stone when released at B is found by a similar pro- cess as above, except that v y 0 = v 0 cos θ = (3 . 7 m / s) cos(40 ) = 2 . 83436 m / s .

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homework 06 – FRENNEA, KYLE – Due: Feb 28 2007, 4:00 am 2 Then t B = 1 g parenleftBigradicalBig v 2 y 0 + 2 g y 0 + v y 0 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 Δ x ) of the stone when released at B is x B = v 0 (sin θ ) t B = (3 . 7 m / s) (sin 40 ) (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 )? Correct answer: 11 . 7833 m / s 2 (tolerance ± 1 %). Explanation: The centripetal acceleration in the hori- zontal direction of the stone just before it is released at A is a c = v 2 0 r = (3 . 7 m / s) 2 0 . 89 m = 15 . 382 m / s 2 and a x = a c cos θ = (15 . 382 m / s 2 ) cos 40 = 11 . 7833 m / s 2 | a x | = 11 . 7833 m / s 2 . Question 4 part 4 of 5 10 points What is the magnitude of the total acceler- ation of the stone just before it is released at A , ( x A , y A )? Correct answer: 16 . 6219 m / s 2 (tolerance ± 1 %). Explanation: From the free-body diagram and Eq. 1, the acceleration in the tangential ( θ ) direction is a θ = g sin θ (2) = (9 . 8 m / s 2 ) sin(40 ) = 6 . 29932 m / s 2 | a θ | = 6 . 29932 m / s 2 , The magnitude of the total acceleration of the stone just before it is released at A is the magnitude of the vector sum of the centripetal acceleration (Eq. 1) and the gravitational ac-
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