HW03-solutions

# HW03-solutions - Le (tl8426) HW03 gentle (56245) 1 This...

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Unformatted text preview: Le (tl8426) HW03 gentle (56245) 1 This print-out should have 33 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Consider the following situations. A) An object travels as a projectile in a gravitational field with negligible air re- sistance. B) An object moves in a straight line at con- stant speed. C) An object moves with uniform circular motion. In which of the situations would the object be accelerated? 1. B only 2. A and C only correct 3. B and C only 4. A and B only 5. C only 6. All exhibit acceleration. 7. A only 8. None exhibits acceleration. Explanation: A) The projectile undergoes gratitational acceleration. B) The velocity of the object (its direction and magnitude) is unchanged, so it is not accelerated. C) The direction of the velocity constantly changes; the centripetal acceleration is di- rected toward the center of the motion. 002 10.0 points A space station in the form of a large wheel, 309 m in diameter, rotates to provide an artificial gravity of 5 . 6 m / s 2 for people located at the outer rim. What is the frequency of the rotational mo- tion for the wheel to produce this effect? Correct answer: 1 . 81803 rev / min. Explanation: d = v t And the frequency (where T is the period) is f = 1 T = v d . Since bardbl vectora r bardbl = v 2 r = 2 v 2 d or v = radicalbigg a r d 2 Therefore, we have f = radicalbigg a r 2 2 d = radicalBigg (5 . 6 m / s 2 ) 2 2 (309 m) 60 sec 1 min = 1 . 81803 rev / min . 003 10.0 points A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 44 m. The plane of the circle is 1 . 39 m above the ground. The string breaks and the ball lands 2 . 51 m away from the point on the ground directly beneath the balls location when the string breaks. The acceleration of gravity is 9 . 8 m / s 2 . Find the centripetal acceleration of the ball during its circular motion. Correct answer: 50 . 475 m / s 2 . Explanation: In order to find the centripetal acceleration of the ball, we need to find the initial velocity of the ball. Let y be the distance above the ground. After the string breaks, the ball has no initial velocity in the vertical direction, so the time spent in the air may be deduced from the kinematic equation, y = 1 2 g t 2 . Le (tl8426) HW03 gentle (56245) 2 Solving for t , t = radicalbigg 2 y g . Let d be the distance traveled by the ball. Then v x = d t = d radicalbigg 2 y g . Hence, the centripetal acceleration of the ball during its circular motion is a c = v 2 x r = d 2 g 2 y r = 50 . 475 m / s 2 . 004 (part 1 of 2) 10.0 points Young David, who slew Goliath, experi- mented with slings before tackling the gi- ant. He found that with a sling of length . 391 m, he could revolve the sling at the rate of 9 . 94 rev / s. If he increased the length to 0 . 742 m, he could revolve the sling only 7 . 04 rev / s....
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## This note was uploaded on 12/05/2011 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas at Austin.

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HW03-solutions - Le (tl8426) HW03 gentle (56245) 1 This...

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