Circular Motion -solutions

Circular Motion -solutions - le (txl65) – Circular Motion...

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Unformatted text preview: le (txl65) – Circular Motion – mohanty – (81440) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A metal block of mass 99 g rests at a point 2 . 4 m from the center of a horizontal rotat- ing wooden platform. The coefficient of static friction between the block and the platform is 0 . 346. The platform initially rotates very slowly but the rotation rate is gradually in- creasing. The acceleration of gravity is 9 . 8 m / s 2 . At what minimum angular velocity of the platform would the block slide away? Correct answer: 1 . 18863 rad / s. Explanation: The platform is horizontal and all the mo- tion is horizontal. Consequently, the normal force between the platform and the block is N = m g and the maximal friction force is F max f = μ s N = μ s m g . Since the friction is the only force capable of producing the centripetal acceleration, the latter is limited a c = F f m ≤ ( μ s m g ) m = μ s g . Note: This limit does not depend on the block’s mass. As long as the block does not slide, its centripetal acceleration must be a c = ω 2 r . Since this is possible only when a c ≤ μ s g , we must have ω < ω max = radicalbigg μ s g r = radicalbigg . 346(9 . 8 m / s 2 ) 2 . 4 m = 1 . 18863 rad / s . When the platform angular velocity exceeds this limit, the block will slide. 002 10.0 points A space station in the form of a large wheel, 287 m in diameter, rotates to provide an “ artificial gravity ” of 7 . 5 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 . 18311 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 (7 . 5 m / s 2 ) 2 π 2 (287 m) · 60 sec 1 min = 2 . 18311 rev / min . 003 10.0 points A fighter plane flying at constant speed 460 m / s and constant altitude 4800 m makes a turn of curvature radius 3600 m. On the ground, the plane’s pilot weighs (56 kg) (9 . 8 m / s 2 ) = 548 . 8 N. What is his/her apparent weight during the plane’s turn? Correct answer: 3336 . 99 N. Explanation: The plane’s altitude is not important per se , but the fact that it is constant tells us that the plane moves in a horizontal plane and its normal acceleration a N = v 2 R = 58 . 7778 m / s 2 le (txl65) – Circular Motion – mohanty – (81440) 2 has a horizontal direction. Furthermore, con- stant speed implies zero tangential accelera- tion, hence vectora = vectora N ....
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This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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Circular Motion -solutions - le (txl65) – Circular Motion...

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