Centripetal Acceleration - General Physics - Solved Past Paper, Exams of Physics

Download Centripetal Acceleration - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 4. (30 pts) a) (10 pts) Given that the earth rotates once in 24 hours, what is the centripetal acceleration required to keep a 70 kg person rotating with the earth if they are standing at the equator (6.37×106 m from the center of the earth)? How fast does a point at the surface of the earth move? 2π × 6.32 × 106 m day × 1 day 24 h × 1 h 3600 s = 463 m/s 2 Then a = v2/r, so a = 4632 6.37 × 106 = 3.37 × 10−2 m/s 2 b) (10 pts) What is the magnitude of the force required to keep this person accelerating in this manner? F = ma = (70)(3.37 × 10−2) = 2.36 N c) (10 pts) How fast would the earth have to be spinning in order for the person to feel weight-less (the centripetal force equal to the weight of the person)? This would require that all the gravitational acceleration of the Earth be used to keep the person mov- ing in a circle leaving none to press them against the surface, thus giving the feeling of weightlessness. So the required acceleration would be 9.8 m/s2. Putting this back into the equation for centripetal acceleration gives 9.8 = v2 6.36 × 106 v = 7.9 × 103 m/s This is the speed of a point on the surface of the earth. We can convert this to revolutions per day as was given in part (a): 1 rev 4 × 107 m × 7.9 × 103 m s × 3600 s h × 24 h day = 17.1 rev/day