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answer to Q4 (multiple choice) (1 Viewer)

rent

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what is the answer to Q4? the rollercoaster and g force one. Only reply to me if you are certain. But feel free to give me your opinions too. thanks.
 

Ace-Pilot

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Not sure about this, i think i might be wrong.

Even though alot of ppl chose B for Q4, about the rollecoster - i still agree with what i selected which is C.

This is the reason:

At point "Q" the rollecoster just came from a negative gravity force. At the Exact point Q u should only feel the standard gravity, because if u closely analyze the rollecoster diagram u can see "Q" is exactly between the curve, thus implying at that EXACT point g=9.8.

The question asks at what POINT will the occupant experience maximum g which logically seems at the EXACT point "R" the rider experiences maximum g, because he is traveling up the slope.

So i strongly believe the correct answer for Q.4.) is C

However i could be wrong, but if u think about it logically and exactly what the question asks, I dunno.

Laz u out there ? What do u think man ?
 

spice girl

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g force = (g + a)/9.8 where a is the upward acceleration.

Lets dodgily suppose that the roller coaster moves towards the right at a constant velocity. So the curve PQRS is like a graph: the horizontal axis being time, and the vertical axis being displacement. Differentiating this curve gives us a velocity-time graph. Differentiating it again gives us an acceleration-time graph.

At Q, S we have v=0, at P we have minimum v, and at R we have maximum v.

Drawing the v-t graph, we now find that the curve is stationary at P, R, but steepest at Q, S. Thus acceleration is greatest at Q, S.

Acceleration at Q is upwards, and acceleration at S is downwards, and since velocity is at max/min at P, R, acceleration is at 0 at P, R

This answer is Q, as acceleration is max at Q, maximising g force.
 
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ace -pilot u the man i agree its c
because at c the rider experiences 1.their weight mg
2.their normal reaction force acting up i thought it was b at the start of the qustion but looked more closely and came up with c

i could be wrong
 

macca202

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damm, i put C(R), cos at q u are only just being affected by change in concavityy. Its a stupid question anyway, cos wuld it make a difference if the stupid car was tyravelling at constant speed? because then R would have a better chance.

The reason i put R was because i went skiing for the first time in July holidays, and i was at the bottom of a double black diamond run at Mt Hotham (Mary's slide if anyones been there and know wot im talking about), and at the end it dips down, and onto a bridge(over a creek) and back up again, just like the diagram. Well i made the Q part of this quite easily, but when i came up the dip near R and S, the g-forces pulling me down were absolutely massive!!! my knees just gave way causing my to fall backwards in my boots, still attached the the skiis. trying to stop it kinda lent sideways and my skiis pulled me around to the right, and i started sliding back down the bank, only to the side of the bridge, to the CREEK!. as i was desperately trying to dig my hands into the snow to slow down i was looking as a snowboarder already in the creek trying to get out his board. you should have seen his face when he saw me comming. With one last desperate stab at the snow i was able to stop, only half a foot from a drop into the creek. There went all my pride.
Anyway my point is, i had a PRACTICAL kind of experimental experiance of this question, and i probably got it wrong. maybe i should have gone to Australias wonderland instead!!!! did anyone go there and answer their question according to their experiance?
Damn pyhsics, it just doen't make sense!!
 

macca202

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i think they should put another point in. i shall call it.... QR. if u move the page so the line PQ is horizontal, at Q there is a fair bit of force, but look at QR, g is massive!!! oh, damn gravity....
 

Bob

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macca i would put your experience down to bad skiiing technique. Your skiies tend to pull you around at high speed if you haven't got proper control and that sideways force of an edge biting is often thought to be your skies getting bogged because in the stiff boots it almost feels like its pulling down.
Having skiied for 7 years and snowboarded for 4 i generally know what i'm talking about, being on a double black run on your first year is nothing short of a skiing hazard.
But none the less as you go over the crest of hills at speed there is a noticable drop in G-force which causes you to get slightly airbourne and you partially loose control. whereas in a sharp bottom dip it you tend to turn slightly, bend you knees or speed up to avoid getting to heavy in the snow (because of the extra G's) and stacking :]


Hope you enjoyed your first ski trip :] best sport in the world.
 

-=«MÄLÅÇhïtÊ»=-

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im having some doubts about c, coz i asked my dad and he says max g force isnt max acceleration. He says if the ques is askn for max acceleration, den it's B, but if its max g force, he wreckons it's C.

And he's an expert wiv g force coz he drags races at traffic lights all the time
jk


i dunno, maybe he's gotten his g force mixed up wiv sumfin else.
 
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spice girl

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if g force = (g + a)/9.8 like it says in all the texts, if max g force isn't max downward acceleration, then ??
 

-=«MÄLÅÇhïtÊ»=-

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i dunno
i put B!

my dad juz reckons its C. And ppl at ANSTO and siemens bow down to him

dun wanna discuss this ques right now
i have chem 2moro
 

Indefinite

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its at point Q- it says in the excel book that people experience the most g forces during a 'dip' or turn in a rollercoaster. this is due to centripidal force
 

Felix: I

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play rollercoaster tycoon and you'll see the G is at max at the bottom of a drop.........excellent source of physics answers that game.

even riding on a real rollercoaster the force is greatest at the bottom of the drop..so i put B
 

Minai

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hahahah I love that, rollercoaster tycoon helped u answer that question

anyway yes I put Q, because I knew max G force = max acceleration, and I used simple calculus to confirm it...like spice's answer
 

saladsurgery

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Just think... if you were to sprint along that rollercoaster track where is the most likely place of putting out your knee?
- the answer is logical:D
 

Mr Chicken

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Q!

Fellas G-force can be approximated as N/mg, where N is the normal reaction force acting on the person at that time.
Clearly, at Q this person who's been accelerating down the hill under the influence of 1g must how undergo a normal force well above 1g to change their motion from down to up. The person will actually be accelerating upwards at Q, so this further adds to the g-forces as they're fighting against gravity.
At R the person is experiencing 1g, so hey, that's less than Q.
 

Ace-Pilot

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Originally posted by saladsugery
Just think... if you were to sprint along that rollercoaster track where is the most likely place of putting out your knee?
- the answer is logical:D
Damn i hate to admit it, but thats a pretty good point. Geez, well theres 1 mark.
 

afnya

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we had the exact same question in our trial exam and i put the point which was the same as Q which was correct SO THE ANSWER IS 100% WITHOUT A DOUBT Q
 

rent

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Important argument

anfya guy is right - that question was in the catholic trials.

A g force is defined as apparent weight/normal true weight.
And this is where the formula g force = (g + a)/9.8
BUT "a" is acceleration upwards or downwards, by definition!! (weight forces are only on the vertical plane!) and so acceleration on the horizontal plane has no effect, right?

so that's why my answer was C, at R.

I just think this syllabus point (comparing g forces of a rollercoaster and rocket stages).

someone reply to me!
 

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