HW03-solutions

# HW03-solutions - lee(cyl327 – HW03 – gentle –(56245 1...

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Unformatted text preview: lee (cyl327) – 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 moves in a straight line at con- stant speed. B) An object travels as a projectile in a gravitational field with negligible air re- sistance. C) An object moves with uniform circular motion. In which of the situations would the object be accelerated? 1. A and B only 2. A only 3. A and C only 4. None exhibits acceleration. 5. All exhibit acceleration. 6. B and C only correct 7. C only 8. B only Explanation: A) The velocity of the object (its direction and magnitude) is unchanged, so it is not accelerated. B) The projectile undergoes gratitational acceleration. 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, 358 m in diameter, rotates to provide an “ artificial gravity ” of 8 . 8 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: 2 . 11732 rev / min. Explanation: d = v t And the frequency (where T is the period) is f = 1 T = v π 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 π 2 d = radicalBigg (8 . 8 m / s 2 ) 2 π 2 (358 m) · 60 sec 1 min = 2 . 11732 rev / min . 003 10.0 points A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 416 m. The plane of the circle is 1 . 35 m above the ground. The string breaks and the ball lands 2 . 07 m away from the point on the ground directly beneath the ball’s 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: 37 . 3861 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 . lee (cyl327) – 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 = 37 . 3861 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 . 274 m, he could revolve the sling at the rate of 9 . 13 rev / s. If he increased the length to 0 . 564 m, he could revolve the sling only 5 . 33 rev / s....
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HW03-solutions - lee(cyl327 – HW03 – gentle –(56245 1...

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