- #1
Legaldose
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1. Suppose Earth is a perfect sphere of radius 6,370km. If a person weighs exactly 600N at the north pole, how much will the person weigh at the equator?
Givens:
r = 6,370km = 6,370,000m
Weight at north pole = mg = 600N
ω[itex]_{av}[/itex] = θ/t
ƩF[itex]_{net}[/itex] = ma
a[itex]_{c}[/itex] = v[itex]^{2}[/itex]/r
Okay, so the first thing I did was to draw a picture. Then I noticed that ω[itex]_{av}[/itex] would be 0 at the North Pole because you don't have any rotational velocity! Then I set up a Free Body Diagram for a location at the equator and see that the normal force (N) is acting up perpendicular to the ground, and gravity is acting downward towards the center of the Earths gravity.
I get
ƩF[itex]_{y}[/itex] = ma[itex]_{y}[/itex]
where a[itex]_{y}[/itex] is the centripetal acceleration a[itex]_{c}[/itex]
so
ƩF[itex]_{y}[/itex] = mv[itex]^{2}[/itex]/r = mrω[itex]^{2}[/itex]
and substituting in N and mg for ƩF[itex]_{y}[/itex]
N-mg = mrω[itex]^{2}[/itex]
since we are looking for N
N = mrω[itex]^{2}[/itex]+mg
and so when I substitute the values of m, r, ω, and g I get
N = 61.16kg(6.37 x 10[itex]^{6}[/itex]m)(7.27 x 10[itex]^{-5}[/itex]rad/s)[itex]^{2}[/itex] + 61.16kg(9.81m/s[itex]^{2}[/itex])
solving for N I get 602.06N
But I thought that you weighed less at the equator? I believe I missed something fundamental, can someone please help me out here? Thank you for your time.
Givens:
r = 6,370km = 6,370,000m
Weight at north pole = mg = 600N
Homework Equations
ω[itex]_{av}[/itex] = θ/t
ƩF[itex]_{net}[/itex] = ma
a[itex]_{c}[/itex] = v[itex]^{2}[/itex]/r
The Attempt at a Solution
Okay, so the first thing I did was to draw a picture. Then I noticed that ω[itex]_{av}[/itex] would be 0 at the North Pole because you don't have any rotational velocity! Then I set up a Free Body Diagram for a location at the equator and see that the normal force (N) is acting up perpendicular to the ground, and gravity is acting downward towards the center of the Earths gravity.
I get
ƩF[itex]_{y}[/itex] = ma[itex]_{y}[/itex]
where a[itex]_{y}[/itex] is the centripetal acceleration a[itex]_{c}[/itex]
so
ƩF[itex]_{y}[/itex] = mv[itex]^{2}[/itex]/r = mrω[itex]^{2}[/itex]
and substituting in N and mg for ƩF[itex]_{y}[/itex]
N-mg = mrω[itex]^{2}[/itex]
since we are looking for N
N = mrω[itex]^{2}[/itex]+mg
and so when I substitute the values of m, r, ω, and g I get
N = 61.16kg(6.37 x 10[itex]^{6}[/itex]m)(7.27 x 10[itex]^{-5}[/itex]rad/s)[itex]^{2}[/itex] + 61.16kg(9.81m/s[itex]^{2}[/itex])
solving for N I get 602.06N
But I thought that you weighed less at the equator? I believe I missed something fundamental, can someone please help me out here? Thank you for your time.