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M3 IAL Jan2015 Q5(b)

I am so stuck with this question, I couldnt understand even I read the Mark Scheme. Could anyone help me out please? Many thanks.
Original post by Kamisama
I am so stuck with this question, I couldnt understand even I read the Mark Scheme. Could anyone help me out please? Many thanks.


Can you post/link the question and markscheme?
Reply 2
Avatar for Kamisama
Kamisama
OP
Original post by ghostwalker
Can you post/link the question and markscheme?


Sure
QP: https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202015%20(IAL)%20QP%20-%20M3%20Edexcel.pdf

MS: https://8a40d6c38bafca75cc407741c0f3e1889c8e66b2.googledrive.com/host/0B1ZiqBksUHNYWENTbktxdlk5dEU/January%202015%20(IAL)%20MS%20-%20M3%20Edexcel.pdf
Original post by Kamisama


So, S is suspended from a point on its circumference, and the line OA makes an angle of 12 degrees with the vertical.

Here's a diagram:

Untitled.jpg

Green line is supposed to be vertical. Point on circumference is suppoed to be A, not S - my bad.

So, Tan 12 = "distance of CofM from O" / r

And using the result from part a, the rest follows.
(edited 7 years ago)
Reply 4
Avatar for Kamisama
Kamisama
OP
Original post by ghostwalker
So, S is suspended from a point on its circumference, and the line OA makes an angle of 12 degrees with the vertical.

Here's a diagram:

Untitled.jpg

Green line is supposed to be vertical. Point on circumference is suppoed to be A, not S - my bad.

So, Tan 12 = "distance of CofM from O" / r

And using the result from part a, the rest follows.

Not 5c, is 5b
Thank you
Original post by Kamisama
Not 5c, is 5b
Thank you


Sorry - I must be going mad.

OK. S is on the ground with it's shorter side resting on the ground.

It will remain that way as long as the CofM is vertically above the line V to the circumference. As k gets larger the CofM moves to the right, until the shape tips about the circumference. When k is the greatest possible value and we still have eqiulibrium, the Cof M will be over the circumference.

In diagram the two angles c are equal.

Then tan c = r/4r = x bar / r

Untitled.jpg
Reply 6
Avatar for Kamisama
Kamisama
OP
Original post by ghostwalker
Sorry - I must be going mad.

OK. S is on the ground with it's shorter side resting on the ground.

It will remain that way as long as the CofM is vertically above the line V to the circumference. As k gets larger the CofM moves to the right, until the shape tips about the circumference. When k is the greatest possible value and we still have eqiulibrium, the Cof M will be over the circumference.

In diagram the two angles c are equal.

Then tan c = r/4r = x bar / r

Untitled.jpg


It is much clearer, thank you so much

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