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Unformatted text preview: dang (ttd73) – HW04 – Tsoi – (58020) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A distant star has a single planet circling it in a circular orbit of radius 3 . 6 × 10 11 m. The period of the planet’s motion about the star is 924 days. What is the mass of the star? The value of the universal gravitational constant is 6 . 67259 × 10 − 11 N · m 2 / kg 2 . Correct answer: 4 . 33113 × 10 30 kg. Explanation: Let : G = 6 . 67259 × 10 − 11 N · m 2 / kg 2 , R B = 3 . 6 × 10 11 m , and T B = 924 day . T B = (924 day) parenleftbigg 24 h 1 day parenrightbigg 3600 s 1 h = 7 . 98336 × 10 7 s . According to Newton’s explanation of Ke pler’s third law R 3 B T 2 B = GM s 4 π 2 = const. The mass of the star is thus M s = 4 π 2 G R 3 B T 2 B = 4 π 2 6 . 67259 × 10 − 11 N · m 2 / kg 2 × (3 . 6 × 10 11 m) 3 (7 . 98336 × 10 7 s) 2 = 4 . 33113 × 10 30 kg . 002 (part 1 of 5) 10.0 points A car travels at a speed of 18 m / s around a curve of radius 36 m. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 2 M g μ 26 ◦ What is the net centripetal force needed to keep the car from skidding sideways? Correct answer: 19800 N. Explanation: Let : m = 2200 kg , v = 18 m / s , r = 36 m , θ = 26 ◦ , and μ = 0 . 297416 Part5 . The centripetal acceleration of the car rounding a curve is a c = v 2 R and the net centripetal force needed to provide such ac celeration is F c = ma c = mv 2 R = 19800 N . 003 (part 2 of 5) 10.0 points Were there no friction between the car’s tires and the road, what centripetal force could be provided just by the banking of the road? Correct answer: 10515 . 5 N. Explanation: In the absence of friction, there are only two forces acting on the car, namely its weight mvectorg and the normal force vector N provided by the road. The weight is directed vertically down while the normal force is directed perpendicular to the banked road surface and thus at the angle θ from the vertical. The centripetal force due to banking comes from the horizontal component of the normal force, F banking c = N sin θ . The free body diagram in the vertical di rection gives N cos θ = mg or N = mg cos θ and horizontally gives F banking c = N sin θ dang (ttd73) – HW04 – Tsoi – (58020) 2 = mg cos θ sin θ = mg tan θ = 10515 . 5 N . 004 (part 3 of 5) 10.0 points Now suppose the friction force is sufficient to keep the car from skidding. Calculate the magnitude of the normal force exerted on the car by the road’s surface. Hint: Check the correctness of your answer to the first question before proceeding with this and the following questions....
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 Spring '08
 Kaplunovsky
 Force, Friction, Mass, Correct Answer, Tperiod

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