a) When the earth rotates an object(man) on the equator undergoes distance equal to the circuference of earth at equator in 24 hours. So, its velocity = distance undergone /seconds in 24 hrs = 2pi*64000*1000 metrs/(24*60*60) =465.4,211339meters/second is speed of any object on the earth due to the earth's rotation.
b)
The person along the earth moves with a constant speed along the earth's circumference. In other words his constant relative speed along the earth wrt the earth is zero. So to accelerate the 95 kg man by 1meter/s^2 you require 95kg*1m/s^2 newtons =95N
c)The weight of the person is due to the inherent gravitational attraction between the person and the earth. So gravitational force on the person =GMm/R^2 where G is the gravitational constant and its value = 6.67258*10^-11Nm^2/kg^2, M is the mass of the earth = 5.978*10^24 kg and m is the mass of the person =95 kg , given. So the weight due to gravitation = 6.67258*10^-11)(5.978*10^24)(95)/(6400000)^2 = 925.1526 N
d)The apparent weight of the person, here in this question due to earths rotation about its axis alone, is reduced by mv^2/R = 95*(465.4211339)^2/(6400*1000) = 95*(0.033846379) =3.2154N which is due to the rotation of the earth. So the apparent weight of the person = weight due to gravitation - force on the person due to the earth's rotation = (925.1526-3.2154)N = 921.9372 N.
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