Friends at Mt Kilimanjaro
https://isaacphysics.org/questions/friends_at_kilimanjaro?board=b30ca120-4de2-4640-ac3d-c3efa1c33657&stage=a_level
For part a of this question, I thought the relative velocity of the friend would just be the difference in their velocities with omega*r, but I don't seem to be getting anywhere.
If anyone could take a look, help is greatly appreciated.
For part a of this question, I thought the relative velocity of the friend would just be the difference in their velocities with omega*r, but I don't seem to be getting anywhere.
If anyone could take a look, help is greatly appreciated.
Reply 1
10
Working:
I got theta ~ 1.1x10^-3 rad
and worked out h as ~ 3m
and calculated omega as 2pi/24*60^2
so the relative velocity = omega*6,370,000 - omega*(6,370,000-h) = 2.8x10^-4m/s but this was wrong
I got theta ~ 1.1x10^-3 rad
and worked out h as ~ 3m
and calculated omega as 2pi/24*60^2
so the relative velocity = omega*6,370,000 - omega*(6,370,000-h) = 2.8x10^-4m/s but this was wrong
Reply 2
21
Original post by mosaurlodon
Working:
I got theta ~ 1.1x10^-3 rad
and worked out h as ~ 3m
and calculated omega as 2pi/24*60^2
so the relative velocity = omega*6,370,000 - omega*(6,370,000-h) = 2.8x10^-4m/s but this was wrong
I got theta ~ 1.1x10^-3 rad
and worked out h as ~ 3m
and calculated omega as 2pi/24*60^2
so the relative velocity = omega*6,370,000 - omega*(6,370,000-h) = 2.8x10^-4m/s but this was wrong
Not sure what youre doing with the final relative velocity calculation. The rest looks ok, but it may be easier to think of the triangle(s) as isosceles rather than right. Agree with theta and omega, then if you draw the velocity triangle, it should be similar to the position isosceles triangle and ...
(edited 1 year ago)
Reply 3
10
Wait now im slightly confused - I thought the Earth rotates about poles so the greater your perpendicular distance to that axis, the greater your velocity ie. if your on the poles, you go nowhere and the equator you have v_max - so I thought the aim of the question was to find the perpendicular distance to that axis and use omega*r to get velocities and subtract them to get the relative velocity.
Also, I dont really know how an isosceles triangle plays a role in this - I thought the velocities for each person would be in the same direction acting tangentially into/out of the diagram I drew? Maybe because my reasoning is completely flawed.
I'll have a try rn and see what I get/reason with.
Also, I dont really know how an isosceles triangle plays a role in this - I thought the velocities for each person would be in the same direction acting tangentially into/out of the diagram I drew? Maybe because my reasoning is completely flawed.
I'll have a try rn and see what I get/reason with.
(edited 1 year ago)
Reply 4
21
Original post by mosaurlodon
Wait now im slightly confused - I thought the Earth rotates about poles so the greater your perpendicular distance to that axis, the greater your velocity ie. if your on the poles, you go nowhere and the equator you have v_max - so I thought the aim of the question was to find the perpendicular distance to that axis and use omega*r to get velocities and subtract them to get the relative velocity.
Also, I dont really know how an isosceles triangle plays a role in this - I thought the velocities for each person would be in the same direction acting tangentially into/out of the diagram I drew? Maybe because my reasoning is completely flawed.
I'll have a try rn and see what I get/reason with.
Also, I dont really know how an isosceles triangle plays a role in this - I thought the velocities for each person would be in the same direction acting tangentially into/out of the diagram I drew? Maybe because my reasoning is completely flawed.
I'll have a try rn and see what I get/reason with.
The two points are east - west of each other, so are rotating with the same rotational speed, but their relative (linear) velocity is because the two tangents (linear velocities) are at an angle.
(edited 1 year ago)
Reply 5
10
Oh I think I see what you mean? So does the diagram below roughly make sense?
so the friend has some tangential velocity v_f at an angle to the 7km, and I (me) also have a tangential velocity v_m, both the velocities are same magnitude but it is the difference in the angle that creates the relative velocity?
Oh and the velocities are perpendicular to the center since they are tangents?
so the friend has some tangential velocity v_f at an angle to the 7km, and I (me) also have a tangential velocity v_m, both the velocities are same magnitude but it is the difference in the angle that creates the relative velocity?
Oh and the velocities are perpendicular to the center since they are tangents?
(edited 1 year ago)
Reply 6
21
Original post by mosaurlodon
Oh I think I see what you mean? So does the diagram below roughly make sense?
so the friend has some tangential velocity v_f at an angle to the 7km, and I (me) also have a tangential velocity v_m, both the velocities are same magnitude but it is the difference in the angle that creates the relative velocity?
Oh and the velocities are perpendicular to the center since they are tangents?
so the friend has some tangential velocity v_f at an angle to the 7km, and I (me) also have a tangential velocity v_m, both the velocities are same magnitude but it is the difference in the angle that creates the relative velocity?
Oh and the velocities are perpendicular to the center since they are tangents?
Pretty much, if I understand you correctly.
Reply 7
10
Oh I got it! The other methods that the "correct" text gives are actually pretty genius.
I guess you were supposed to use cosine rule? I kind of just resolved and did 3d pyth.
I didnt even think about a circle of radius 7km that solution surprised me a lot.
Thanks a lot
edit: also cool thing I noticed for part b, if you use that genius method, then the new radius thats traced out is just Pythagoras of the altitude and distance away.
I guess you were supposed to use cosine rule? I kind of just resolved and did 3d pyth.
I didnt even think about a circle of radius 7km that solution surprised me a lot.
Thanks a lot
edit: also cool thing I noticed for part b, if you use that genius method, then the new radius thats traced out is just Pythagoras of the altitude and distance away.
(edited 1 year ago)
Reply 8
21
Original post by mosaurlodon
Oh I got it! The other methods that the "correct" text gives are actually pretty genius.
I guess you were supposed to use cosine rule? I kind of just resolved and did 3d pyth.
I didnt even think about a circle of radius 7km that solution surprised me a lot.
Thanks a lot
I guess you were supposed to use cosine rule? I kind of just resolved and did 3d pyth.
I didnt even think about a circle of radius 7km that solution surprised me a lot.
Thanks a lot
There was little / no trig necessary. So once you have the linear speed and angle, then the relative velocity triangle is isosceles which is similar to the original posiiton triangle as it doesnt really matter whether the 7km is an arc or a straight line. You can use cos rule or bisect the isosceles base to get (two) right triangles and do simple sin or ... but the angle is so small that really its just a simple ratio.
Reply 9
10
Actually I've been trying to understand the other method and im not really getting anywhere, do you mind explaining the alternative method slightly - the one where "you can avoid thinking about the radius of the Earth, and simply note that from your point of view your friend rotates around you once a day, tracing a circle of radius 7.00km."
and then for part b you can do the same thing but use the hypotenuse of the 2 lengths, but I don't really get how they trace the circumference of this circle with that radius when the Earth does one rotation - I guess im struggling to visualize how they relatively do that.
and then for part b you can do the same thing but use the hypotenuse of the 2 lengths, but I don't really get how they trace the circumference of this circle with that radius when the Earth does one rotation - I guess im struggling to visualize how they relatively do that.
(edited 1 year ago)
Reply 10
21
Original post by mosaurlodon
Actually I've been trying to understand the other method and im not really getting anywhere, do you mind explaining the alternative method slightly - the one where "you can avoid thinking about the radius of the Earth, and simply note that from your point of view your friend rotates around you once a day, tracing a circle of radius 7.00km."
and then for part b you can do the same thing but use the hypotenuse of the 2 lengths, but I don't really get how they trace the circumference of this circle with that radius when the Earth does one rotation - I guess im struggling to visualize how they relatively do that.
and then for part b you can do the same thing but use the hypotenuse of the 2 lengths, but I don't really get how they trace the circumference of this circle with that radius when the Earth does one rotation - I guess im struggling to visualize how they relatively do that.
Imagine the sun is stationary and is north of earth (for the slice you mentioned earlier). Then
•
at time 0hr youre due east of the center and the other person is 7km due south of you and their face is fully lit by sunlight
•
at time 6hr youre due north of the center and the other person is 7km due east of you and their face is half lit by sunlight
•
at 12hr youre due west of the centre ...
So in 24 hrs, the other person has rotated once around you and the radius is 7km.
(edited 1 year ago)
Reply 11
10
So smth like this?
I've been trying to play it out a lot in my mind to try to develop some intuition/sense for this.
So basically since I am completely still in my frame of reference they appear to be going around a full circle around me when the Earth does one rotation.
I've been trying to play it out a lot in my mind to try to develop some intuition/sense for this.
So basically since I am completely still in my frame of reference they appear to be going around a full circle around me when the Earth does one rotation.
Reply 12
21
Original post by mosaurlodon
So smth like this?
I've been trying to play it out a lot in my mind to try to develop some intuition/sense for this.
So basically since I am completely still in my frame of reference they appear to be going around a full circle around me when the Earth does one rotation.
I've been trying to play it out a lot in my mind to try to develop some intuition/sense for this.
So basically since I am completely still in my frame of reference they appear to be going around a full circle around me when the Earth does one rotation.
Yes, assuming youre the orange dot, though I dont fully follow your lines. If you start at 3 o clock and go anticlockwise, then at 12 o clock you should have a trace which starts due south of the orange dot (old red position) and does quarter of a turn anticlockwise to the new red position. When you move to 9 o clock, then the red dot is due north and the trace goes from due south to due north anticlockwise ...
Its not hard, but there are famous examples of similar incorrect questions being asked
https://www.scientificamerican.com/article/the-sat-problem-that-everybody-got-wrong/
Reply 13
10
Oh I think I see what you mean..
So because the red dot is initially behind the orange dot it has some relative "upwards" and anticlockwise velocity which makes it appear to "jump" over the orange dot every quarter turn forming a quarter circle each time?
Oh also thank you for the link - I remember watching a veritasium video on the topic but never realised it applied to celestial bodies orbiting,
So because the red dot is initially behind the orange dot it has some relative "upwards" and anticlockwise velocity which makes it appear to "jump" over the orange dot every quarter turn forming a quarter circle each time?
Oh also thank you for the link - I remember watching a veritasium video on the topic but never realised it applied to celestial bodies orbiting,
(edited 1 year ago)
Reply 14
21
Original post by mosaurlodon
Oh I think I see what you mean..
So because the red dot is initially behind the orange dot it has some relative "upwards" and anticlockwise velocity which makes it appear to "jump" over the orange dot every quarter turn forming a quarter circle each time?
Oh also thank you for the link - I remember watching a veritasium video on the topic but never realised it applied to celestial bodies orbiting,
So because the red dot is initially behind the orange dot it has some relative "upwards" and anticlockwise velocity which makes it appear to "jump" over the orange dot every quarter turn forming a quarter circle each time?
Oh also thank you for the link - I remember watching a veritasium video on the topic but never realised it applied to celestial bodies orbiting,
Looks about right, with a bit of artistic licence about your arcs. While the orange dot moves around the circle (x,y) position, the orientation is fixed south (for instance, starting at 3 o clock). So the red dot goes from being ahead, to being to the left, to being behind, to being to the right ... So orbiting the orange dot every 24 hrs.
Reply 15
10
Yeah hah just try to ignore the artistic flair I added.
Ohhhhh making the orientation of the orange dot fixed actually makes the rotation a lot easier to understand.
I was thinking of it as the red dot would always be on the right side of the orange dot and I wasnt really getting anywhere with that though process.
So I guess with relative motion, you have to always consider the orientation to be fixed of the person you are taking the reference frame from, even if they are standing on a rotating object.
Thank you so much
Ohhhhh making the orientation of the orange dot fixed actually makes the rotation a lot easier to understand.
I was thinking of it as the red dot would always be on the right side of the orange dot and I wasnt really getting anywhere with that though process.
So I guess with relative motion, you have to always consider the orientation to be fixed of the person you are taking the reference frame from, even if they are standing on a rotating object.
Thank you so much
(edited 1 year ago)
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