# E-sol - Chapter E Problems Blinn College Physics 2425 Terry...

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Chapter E - Problems Blinn College - Physics 2425 - Terry Honan Problem E.1 (a) What is the centripetal (radial) acceleration of a point on the earth's equator? (b) Give an expression for the centripetal acceleration as a function of the latitude angle, q L . What is this at the latitude of Bryan, Texas, at q L = 30.7 °? (c) What are the speeds of a point at the equator and at Bryan, Texas due to the eaarth's rotation? (d) The earth's rotation is slowing at a rate of 2.2 s every 100,000 years. . T t = D T D t = 2.3 s 100,000 years What is the (very small) tangential component of the acceleration of a point on the earth's equator? Hint: use the chain rule. t 2 p r T = - 2 p r T 2 T t Solution to E.1 a c = v 2 r and v = 2 p r T ï a c = J 2 p T N 2 r For the earth: T = 1 day = 24 ÿ 3600 s (a) At the equator: r = R E ï a c = K 2 p 24 ÿ 3600 O 2 6.37 μ 10 6 = 0.0337 m s 2 Comment: Compared to the gravitational acceleration of g, this is small but not negligible. (b) r is the radius of the circle but something on the rotating earth moves along a path that traces a latitude line. At some point other that the equator this is: r = R E cos q L ï a c = 0.0337 m s 2 μ cos q L q L = 30.7 ° ï a c = 0.0290 m s 2 (c) The speed is found form the period: v = 2 p r T . At equator: v = 2 p 24 ÿ 3600 6.37 μ 10 6 = 463 m s At Bryan: v = 2 p 24 ÿ 3600 6.37 μ 10 6 cos 30.7 ° = 398 m s (d) To find the tangential acceleration use a t = „ v ê t and the hints.

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a t = v t = t 2 p r T = - 2 p r T 2 T t = - 2 p r T 2 D T D t = - 2 p R E H 1 day L 2 2.3 s 100,000 years = - 2 p R E H 1 day L 2 2.3 s 100,000 year = - 2 p 6.37 μ 10 6 H 24 μ 3600 L 2 2.3 100,000 μ H 365.24 μ 24 μ 3600 L = 3.91 μ 10 - 15 m s 2 Problem E.2 A car drives around a 200 m radius circle with a speed that decreases uniformly from 30 m ê s to 20 m ê s in 8 s. At the instant the speed is 25 m ê s then: (a) what is the centripetal acceleration, (b) what is the tangential acceleration and (c) what is the magnitude of the acceleration? Solution to E.2 (a) a c = v 2 r = 25 2 200 = 3.125 m s 2 (b) a t = v t = D v D t = 20 - 30 8 = - 1.25 m s 2 (c) a = a c 2 + a t 2 = 3.37 m s 2 Problem E.3 While moving in a circle with a 12 m radius the speed of a particle varies with time by v H t L = 2 + 10 t - 4 t 2 in SI units. At t = 2 s
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E-sol - Chapter E Problems Blinn College Physics 2425 Terry...

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