HW5 solutions - HomeWork 5 solutions Physics 203 1. Chapter...

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HomeWork 5 solutions Physics 203 1. Chapter 5 problem 12. “Centripetal acceleration” is another name for “radial component of acceleration”. It’s better to use the latter term since it’s much more descriptive of what it means. REASONING AND SOLUTION a. At the equator a person travels in a circle whose radius equals the radius of the earth, r = R e = 6.38 × 10 6 m, and whose period of rotation is T = 1 day = 86 400 s. We have v = 2 π R e / T The centripetal acceleration is ( ) 2 2 –2 2 c 6 464 m/s 3.37 10 m/s 6.38 10 m v a r = = = × × b. At 30.0° latitude a person travels in a circle of radius, r = R e cos 30.0° = 5.53 × 10 6 m Thus, v = 2 π r / T and a c = v 2 / r = 2.92 × 10 –2 m/s 2 2. Problem 20 REASONING When the penny is rotating with the disk (and not sliding relative to it), it is the static frictional force that provides the centripetal force required to keep the penny moving on a circular path. The magnitude MAX s f of the maximum static frictional force is given by MAX s s N fF µ = (Equation 4.7), where F N is the magnitude of the normal force and s is the coefficient of static friction. Solving this relation for s gives MAX s s N f F = (1) Since the maximum centripetal force that can act on the penny is the maximum static frictional force, we have MAX cs Ff = . Since F c = mv 2 / r (Equation 5.3), it follows that MAX 2 s / f mv r = . Substituting this expression into Equation (1) yields 2 MAX s s N N mv f r FF = (2) The speed of the penny can be determined from the period T
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HW5 solutions - HomeWork 5 solutions Physics 203 1. Chapter...

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