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Consider an incline, which is one part of a circle, radius 'r' m. From one end of an incline(A), to the center O, there is a frictionless plane inclined at an angle of theta. A mass (A) slided down the striaght. slope. Another mass (B) which has the same mass of A, slides down the circular incline from the other end. Determine which mass reaches the center O first. It's better if you show your calculations...You're not guessing, are you? :)
 one year ago
 one year ago
Here's a fun one: Consider an incline, which is one part of a circle, radius 'r' m. From one end of an incline(A), to the center O, there is a frictionless plane inclined at an angle of theta. A mass (A) slided down the striaght. slope. Another mass (B) which has the same mass of A, slides down the circular incline from the other end. Determine which mass reaches the center O first. It's better if you show your calculations...You're not guessing, are you? :)
 one year ago
 one year ago

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ShadowysBest ResponseYou've already chosen the best response.1
dw:1350647588577:dw hint: theta will vanish.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
\[2\sqrt{\frac{ r }{ g }}\]vs\[\frac{ \pi }{ 2 } \sqrt{\frac{ r }{ g }} \] \[...\huge ?\]
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
lol calc then one of them is the right one.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
you sure about the sqrt for the g?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
yup. doesn't the g during the derivation has a square?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
don't think so. I see mine is wrong for a different reason though: used the wrong angle for chord length.... so what's the simple way to do it?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
hmm..I'm used the workenergy theorem and striaghtline motion equations but let's see if others have more elegant answers.
 one year ago

vannayenBest ResponseYou've already chosen the best response.0
dw:1350732149436:dw compare v1 and v2 when it reach center O dw:1350733431967:dw
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
Would you mind labelling the graph? lol Didn't get 'l' because I can't the see the whole pic. And who is 1 and 2?\ Hint: Find the time taken. The velocity is just part of it. Because the distance taken might(might!!) just cancel them out.
 one year ago
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