4 - homework 04 ARMSTRONG, DOMINIC Due: Oct 2 2007, 3:00 am...

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homework 04 – ARMSTRONG, DOMINIC – Due: Oct 2 2007, 3:00 am 1 Question 1, chap 6, sect 3. part 1 of 5 10 points A car travels at a speed of 29 m / s around a curve of radius 47 m. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 8 Mg μ 29 What is the net centripetal force needed to keep the car from skidding sideways? Correct answer: 50102 . 1 N (tolerance ± 1 %). Explanation: Let : m = 2800 kg , v = 29 m / s , r = 47 m , θ = 29 , and μ = 0 . 631961 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 = 50102 . 1 N . Question 2, chap 6, sect 3. part 2 of 5 10 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: 15210 . 2 N (tolerance ± 1 %). Explanation: In the absence of friction, there are only two forces acting on the car, namely its weight mVg and the normal force V 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-
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This note was uploaded on 04/07/2008 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas at Austin.

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4 - homework 04 ARMSTRONG, DOMINIC Due: Oct 2 2007, 3:00 am...

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