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A Candy Cane

https://isaacphysics.org/questions/candy_cane?board=3b7f762f-33e0-4987-850b-c6f172244bd9&stage=a_level

Could anyone please check how I went wrong with my following working:

Any help greatly appreciated.

Reply 1

mosaurlodon

I basically said 2a/cos(alpha) = (2r/pi + (2pir - x_com/2cos(alpha)))
But I must have done something wrong because theyre not leading me anywhere.

(edited 9 months ago)

Reply 2

mqb2766

Original post by mosaurlodon

I basically said 2a/cos(alpha) = (2r/pi + (2pir - x_com/2cos(alpha)))
But I must have done something wrong because theyre not leading me anywhere.

Not really sure what you tried to do, but Id work out the com x-y coordinates relative to the point where its suspendend and assuming the cane is horizontal, then it should easily map to the suspended case.

(edited 9 months ago)

Reply 3

mosaurlodon

oh yes!! I see now. thank you very much
I actually found mapping it to the suspended case the hardest part tbh, but that might be because I rushed my diagrams or used a weird method.

Reply 4

mqb2766

Original post by mosaurlodon

oh yes!! I see now. thank you very much
I actually found mapping it to the suspended case the hardest part tbh, but that might be because I rushed my diagrams or used a weird method.

If youd drawn the rotated horizontal and vertical vertical lines to the com and formed a right triangle, that would have been pretty much what was described in the previous post. Tbh, sometimes just thinking about what you can find out, especially with trig / rotation problems, often gives a way to adapt to the actual problem.

Reply 5

mosaurlodon

I believe thats what I did?

I basically then said that tan(theta) = r+2b/2a-2r/pi
but for some reason that last expression took a lot of thinking power

I never wouldve thought to consider the horizontal case but looking back it makes so much more sense

math.jpg96.3KB

Reply 6

mqb2766

Original post by mosaurlodon

I believe thats what I did?

I basically then said that tan(theta) = r+2b/2a-2r/pi
but for some reason that last expression took a lot of thinking power
I never wouldve thought to consider the horizontal case but looking back it makes so much more sense

Id do it about the point of suspension (top) as it directly relates to the original diagram. The ycom value will be flipped, though youve a couple of typos in the xcom, so clearly write down the masses and displacements and do the calc.

(edited 9 months ago)

Reply 7

mosaurlodon

Sorry I dont really know what you mean?
I check my original working to see if there were any typos but I dont think there are any? Unless I just cant find them
According to my diagram (rotated cane):

tan(theta) = r+2b/2a-2r/pi

Reply 8

mqb2766

Original post by mosaurlodon

Sorry I dont really know what you mean?
I check my original working to see if there were any typos but I dont think there are any? Unless I just cant find them
According to my diagram (rotated cane):

tan(theta) = r+2b/2a-2r/pi

I misinterpreted what a and b represented etc and thought you were taking moments about the point of suspension so can do everything in terms of r without needing to introduce a and b. Do you get the right answer now?

Reply 9

mosaurlodon

My bad I shouldve stated what i was doing - you can do evertyhing with just r, I just kept it simple for my brain by using a and b.
I got the right answer since reply #3 when you told me to think about the horizontal case first.