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Practice Homework 5 Solutions

Practice Homework 5 Solutions - practice 05 ALIBHAI ZAHID...

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practice 05 – ALIBHAI, ZAHID – Due: Feb 18 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A copper block rests 46 . 9 cm from the cen- ter of a steel turntable. The coefficient of static friction between block and surface is 0 . 49. The turntable starts from rest and ro- tates with a constant angular acceleration of 0 . 616 rad / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . After what interval will the block start to slip on the turntable? Correct answer: 5 . 1945 s (tolerance ± 1 %). Explanation: Before the copper block slips, the force of static friction supplies the centripetal force that keeps the block in place. Therefore, when the block is about to slip f = ( f s ) max = μ m g = m r ω 2 , from which ω = radicalbigg μ g r = radicalbigg (0 . 49)(9 . 8 m / s 2 ) 0 . 469 m = 3 . 19981 rad / s . The block will begin to slip when ω reaches the value of 3 . 19981 rad / s. Using ω = ω 0 + α t = α t, we find that when ω = 3 . 19981 rad / s, t = ω α = 3 . 19981 rad / s 0 . 616 rad / s 2 = 5 . 1945 s . Question 2 part 1 of 2 10 points The coefficient of static friction between the person and the wall is 0 . 77 . The radius of the cylinder is 5 . 22 m . The acceleration of gravity is 9 . 8 m / s 2 . An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away. ω 5 . 22 m What is the minimum angular velocity ω min needed to keep the person from slip- ping downward? Correct answer: 1 . 56147 rad / s (tolerance ± 1 %). Explanation: Let : R = 5 . 22 m and μ = 0 . 77 . Basic Concepts: Centripetal force F = m v 2 r . Frictional force f μ N = f max . Solution: The maximum force due to static friction is f max = μ N , where N is the inward directed normal force exerted by the wall of the cylinder on the person. To support the person vertically, the maximal friction force must be larger than the force of gravity m g , so that the actual force, which is equal to or less than the maximum μ N , is allowed to take on the value m g in the pos- itive vertical direction. In other words, the “ceiling” μ N on the frictional force has to be raised high enough to allow for the value m g . The normal force supplies the centripetal ac- celeration v 2 R on the person, so from Newton’s second law, N = mv 2 R . Since f max = μ N = μ m v 2 R m g ,
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practice 05 – ALIBHAI, ZAHID – Due: Feb 18 2007, 4:00 am 2 the minimum speed required to keep the per- son supported is at the limit of this inequality, which is μ m v 2 min R = m g , or v min = radicalBigg g R μ . From this we immediately find the angular speed ω min v min R = radicalbigg g μ R = radicalBigg 9 . 8 m / s 2 (0 . 77) (5 . 22 m) = 1 . 56147 rad / s . Question 3 part 2 of 2 10 points Suppose the person, whose mass is m , is being held up against the wall with an angular velocity of ω = 2 ω min . The magnitude of the frictional force be- tween the person and the wall is 1. F = 1 2 m g .
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