artemisxrpg
Earth revolves on its axis once every 24 hr. Assuming that Earth's radius is 6400 km, find the following.
a. angular speed of earth in radians per day and radian per hours
b. linear speed at the north pole or south pole
c. linear speed at quito,ecuador, a city on the equator
d. linear speed at salem, oregon (halfway from the equator to the north pole)
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ganeshie8
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\(\large \omega = \frac{2 \pi}{T}\)
ganeshie8
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earth is taking 1 day to to rotate around itself.
\(\large \omega = \frac{2 \pi~radians}{1~ day}\)
artemisxrpg
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it's 2 pi because that how much one revolution is?
ganeshie8
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Yup ! to go around circle once, it takes 2pi radians
ganeshie8
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|dw:1380369147272:dw|
artemisxrpg
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so angular speed in 24 hrs would be pi/12
ganeshie8
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Yes, pi/12 radians per hour
ganeshie8
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2 pi radians per day
us same as saying,
pi/12 radians per hour
ganeshie8
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its just the unit conversion
ganeshie8
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*is
artemisxrpg
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v = r*theta/t, so in this case it would be 6400*2pi/1 day or 24 hr?
ganeshie8
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theta/t is the angular speed \(\omega \)
ganeshie8
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linear speed v = r\(\omega \)
ganeshie8
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simply multiply radius wid \(\omega\)
ganeshie8
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whats the radius of earth at north pole ?
artemisxrpg
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6400?
ganeshie8
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let me put it this way :-
whats the distance from axis of rotation to you, when you are at north pole ?
artemisxrpg
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I'm still confused|dw:1380370272714:dw|
artemisxrpg
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this is where the north pole would be?
ganeshie8
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|dw:1380370352008:dw|
ganeshie8
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|dw:1380370410225:dw|
ganeshie8
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thats the axis of rotation, straight line from North pole to Sounth pole
wats the distance ? :)
artemisxrpg
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12800?
ganeshie8
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'point on North pole' , is right 'next to axis of rotation'
ganeshie8
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so, distance between them = r = 0
ganeshie8
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for a person on North / South poles :-
linear velocity \(v = r \omega = 0\times \omega = 0 \)
ganeshie8
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see if that makes more/less sense
ganeshie8
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|dw:1380370749167:dw|
ganeshie8
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starting from equator,
as you move towards the poles, the linear velocity decreases and becomes 0 at poles
artemisxrpg
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so this is where the equator is?|dw:1380371306773:dw|
artemisxrpg
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not at the center of the graph?
ganeshie8
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yup ! thats the equator, and since distance from axis is maximim, equals radius of earth, the speed wud be maximum at equator
artemisxrpg
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so then radius would be 6400 at the equator?
ganeshie8
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lets call it, distance from axis.
at equator,
distance from axis = radius of earth = 6400
ganeshie8
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in ur speed formula,
\(v = r \omega\) ,
\(r\) is distance form axis
artemisxrpg
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then if the linear speed is halfway from the equator to north pole, radius would be 3200?
ganeshie8
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nopes,
use trig to find r, the distance from axis
ganeshie8
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|dw:1380372161083:dw|
ganeshie8
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|dw:1380372288001:dw|
ganeshie8
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|dw:1380372490038:dw|
ganeshie8
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thats the distance from axis, \(r\)
you can find it using proper trig ratio
artemisxrpg
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180-45= 135 - 90= 45. since it's a special triangle, both side are 6400 so use pythagorean theorem and the answer should be 9050.97?
ganeshie8
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Nope, careful 6400 is the hypotenuse
ganeshie8
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|dw:1380442777970:dw|
artemisxrpg
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sin45 = x/6400
= 4525.48?
ganeshie8
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Yup ! thats the \(r\)