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Unformatted text preview: pasha (sep635) – HW 52 – Antoniewicz – (56445) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A circular pendulum of length 1 . 3 m goes around at an angle of 30 degrees to the verti cal. Predict the speed of the mass at the end of the string. Use g = 9 . 8 m / s 2 . Correct answer: 1 . 91774 m / s. Explanation: Application of the momentum principle in the vertical direction gives F T cos θ = m g, while application of the momentum principle in the horizontal direction gives F T sin θ = m  ~ v  2 R . Combining these two equations, we obtain m g tan θ = m  ~ v  2 R ⇒ g tan θ =  ~ v  2 R ⇒  ~ v  = p R g tan θ = p ( L sin θ ) g tan θ = q (0 . 65 m)(9 . 8 m / s 2 ) tan30 ◦ = 1 . 91774 m / s . 002 (part 2 of 2) 10.0 points Also predict the period, the time it takes to go around once. Correct answer: 2 . 12963 s. Explanation: We find the period by writing  ~ v  = 2 π R T ⇒ T = 2 π R  ~ v  = 2 . 12963 s . 003 (part 1 of 2) 10.0 points By “weight” we usually mean the gravita tional force exerted on an object by the Earth. However, when you sit in a chair your own perception of you own “weight” is based on the contact force the chair exerts upward on your rear end rather than on the gravitational force. The smaller this contact force is, the less “weight” you perceive, and if the contact force is zero, you feel peculiar and “weight less” (an odd word to describe a situation when the only force acting on you is the gravi tational force exerted by the Earth!). Also, in this condition you internal organs no longer press on each other, which presumably con tributes to the odd sensation in your stomach. How fast must a roller coaster car go over the top of a circular arc for you to feel “weight less”? The center of the car moves along a circular arc of radius R as in the following figure). ~ v R 1.  ~ v  = p 2 g R 2.  ~ v  = g R 3.  ~ v  = 2 g R 4.  ~ v  = p g R correct 5.  ~ v  = r g R 2 Explanation: The force by the seat on the rider can be up ward (+ y direction) or downward ( y direc tion) depending on how fast the roller coaster goes over the hill. The gravitational force by pasha (sep635) – HW 52 – Antoniewicz – (56445) 2 the Earth on the rider is downward ( y direc tion). At the top of the hill, the net force on the rider is in the downward direction, toward the center of the circle. To feel weightless, the force by the seat on the rider must be zero at the top of the hill: ~ F net = ~ F seat + ~ F grav h , m  ~ v  2 R , i = h , m g, i m  ~ v  2 R = m g ⇒  ~ v  = p g R ....
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This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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