A woman living on the equator weighs herself and finds that her apparent mass to be 50 kg. A correction in her mass should be made for the Earth's rotation (centripetal force). What should be her correct mass, assuming that the Earth is a perfect sphere of radius 6.37 x 10^6 meters.
2007-11-29 05:16:16 · 2 answers · asked by labelapark 6 in Science & Mathematics ➔ Physics
_____________________________________-----
Weighing machine records weight in kilogram weight ( kg wt ) unit
Apperent weight =50 kg wt
Apperent mass of woman = 50 kg
A woman living on the equator is moving on cicular path of radius equal to radius of earth
Radius of earth = r = 6.37*10^6 m
Angular velocity of earth = w =2pi / period of rotation of earth
period of rotation of earth = T =86400 s
w = 2pi /86400
The centripetal force required for circular motion of woman = F = m r w ^2=50*6.37*(10^6)(2pi)^2/ ( 86400 )^2
The centripetal force required for circular motion of woman = F = 1.6844 N
Apperent weight = 50 kgwt =50*9.8= 490 N
Correct weght = apperent weight + centripetal force
Correct weght =490 + 1.6844 = 491.6844 N
Correct weight = 491.6844 N
Correct weight = 50.17188 kgwt
Correct mass of woman is 50.17188 kg
______________________________________________
2007-11-29 06:07:53 · answer #1 · answered by ukmudgal 6 · 0⤊ 0⤋
The angular acceleration required to move the woman in a circle of radius r = Re over time 24 hours is
r w^2
where r is the radius of the Earth and w is the angular velocity in radians per second.
w = ( 2 pi ) / (24 hours ) ( 3600 sec / hour )
Calculate the angular acceleration. Her true weight is larger than her measured weight by this value multiplied by her mass.
2007-11-29 05:22:21 · answer #2 · answered by jgoulden 7 · 0⤊ 0⤋