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Edexcel - M3 - 18th May 2016

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Original post by BBeyond
U mean about the y-axis?


Yes I do, sorry
Original post by wr123
How would you find the volume generated by rotating the area bounded by a curve about the y-axis? Is it by intergrating x^2 dy and x^2y dy?


Yeah that's right, it uses the same integrals as for finding the x-bar when it is rotated about the x-axis but the x's and y's are switched
Original post by oShahpo
How can max height occur when V is less than zero? Max height by definition occurs when V=0 as v = dX/dT . Do you mean max tension? If you look at the equation of T, you'll find it decreases as theta decreases. Thus minimum T occurs when minimum Theta occurs. Minimum theta occurs when V = 0, which is 19.5 degrees.
This is all based on the assumption that the string remains taut, which it does.


LOL. i said we get a contradiction if we assume min tension is at the top since we get v^2 is less then zero which is nonsense.
I don't even know what your talking about.
what did i say that was wrong? I am just explaining why this is a special case since it doesnt go above the horizontal and why the min tension occurs at max height(min theta). by max height i mean the maximum height above the bottom of the circular motion. if you got it right stick to your methods ill stick to mine lol. btw the implications are the other way round we find minimum theta because of the restriction v^2 greater or equal to zero. not the other way
Original post by Alby1234
Yes I do, sorry


Yh potentially all u do is switch x values and y values to it would be integral of pi(x^2)y dy
Original post by wr123
How would you find the volume generated by rotating the area bounded by a curve about the y-axis? Is it by intergrating x^2 dy and x^2y dy?


Yea
Reply 705
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wr123
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Original post by Alby1234
Yeah that's right, it uses the same integrals as for finding the x-bar when it is rotated about the x-axis but the x's and y's are switched


Thanks, it was irritating me that I didn't know if that was right ahah
Original post by physicsmaths
LOL. i said we get a contradiction if we assume min tension is at the top since we get v^2 is less then zero which is nonsense.
I don't even know what your talking about.
what did i say that was wrong? I am just explaining why this is a special case since it doesnt go above the horizontal and why the min tension occurs at max height(min theta). by max height i mean the maximum height above the bottom of the circular motion. if you got it right stick to your methods ill stick to mine lol. btw the implications are the other way round we find minimum theta because of the restriction v^2 greater or equal to zero. not the other way


I understand you said it's a contradiction, but I don't understand why you were solving for V^2 at max height when it is by definition = 0. I am not saying you're wrong, but I just didn't understand why you're doing it the other way around.

I think the misunderstanding comes from what you mean by max height, I am talking about the actual max height of the ball, which occurs at theta = 19.something not the max height that is at the top.

I guess you were trying to check if it reaches the top of the circle, right?
(edited 7 years ago)
Original post by oShahpo
I understand you said it's a contradiction, but I don't understand why you were solving for V^2 at max height when it is by definition = 0. I am not saying you're wrong, but I just didn't understand why you're doing it the other way around.

I think the misunderstanding comes from what you mean by max height, I am talking about the actual max height of the ball, which occurs at theta = 19.something not the max height that is at the top.

I guess you were trying to check if it reaches the top of the circle, right?


that is what i am talking about the max height above the circle. you see it is max since v hits zero then it goes back down trivially. I checked it to get a range of values of theta which were 1/3 and 1 which trivialised the problem since for some reason i found tension in the first part.i didtnt solve at tp of circle since it just doesnt reach there by the expression 3sinx-1 in v^2
Original post by physicsmaths
that is what i am talking about the max height above the circle. you see it is max since v hits zero then it goes back down trivially. I checked it to get a range of values of theta which were 1/3 and 1 which trivialised the problem since for some reason i found tension in the first part.i didtnt solve at tp of circle since it just doesnt reach there by the expression 3sinx-1 in v^2


I thought you were talking about the max height of the motion, but yea you're right you get a negative V^2 if you consider the top of the circle which means it doesn't reach it. By the way T is not necessarily minimum when V=0 had the ball crossed the horizontal but not reached the topic. If it had crossed the horizontal T would go to zero at one point and the string would go slack, right?
Reply 709
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Nerrad
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Just did the Jan 16 IAL paper and got 75/75, anyone else use the similar triangle methods instead of comparing the length? Feeling pretty good about tomorrow's exam(UK). Good luck to everyone!
Original post by Nerrad
Just did the Jan 16 IAL paper and got 75/75, anyone else use the similar triangle methods instead of comparing the length? Feeling pretty good about tomorrow's exam(UK). Good luck to everyone!


yep similar triangles are the way to go with this. :smile:
Reply 711
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Nerrad
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Original post by physicsmaths
yep similar triangles are the way to go with this. :smile:


I don't really understand the main method they had used though. Can you explain it?
Original post by Nerrad
I don't really understand the main method they had used though. Can you explain it?


they have done what you done(well what i think you did) but applied costheta to it so that your deealing with OA etc. the reason for use of lengths is that we know for it to be in equilbrium the com must lie between O and the perpendicular bisector of the base of the hemishpere and the vertical from point of contact. the cos everything to us OA and xbarcos instead of using the lengths on the vertex line.
Reply 713
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jhill98
Original post by Nerrad
Just did the Jan 16 IAL paper and got 75/75, anyone else use the similar triangle methods instead of comparing the length? Feeling pretty good about tomorrow's exam(UK). Good luck to everyone!


Where did you guys find this paper, this question seems interesting?
Reply 714
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Nerrad
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Original post by jhill98
Where did you guys find this paper, this question seems interesting?


Keep going back pages by pages and eventually you'll find a link.
Original post by jhill98
Where did you guys find this paper, this question seems interesting?


The question is basically a cone, with a hemisphere stuck to it, on a horizontal table where the side of the cone coincides with the table, find the limit to where the centre of mass can be.
Hey Is it true that you can find the amplitude by using w=va ? I've never seen this formula before.
Original post by JustDynamite
Hey Is it true that you can find the amplitude by using w=va ? I've never seen this formula before.


The general formula is V^2 = W^2 (A^2 - X^2). When X = 0 this simplifies to V=AW
Original post by oShahpo
The question is basically a cone, with a hemisphere stuck to it, on a horizontal table where the side of the cone coincides with the table, find the limit to where the centre of mass can be.


What is the similar triangle method in this context if you dont mind?

Thank you.
Original post by JustDynamite
Hey Is it true that you can find the amplitude by using w=va ? I've never seen this formula before.


Do you mean vmax = aω?

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