Adding subjects that aren't showing/Physics Chatter

[Blakhitman]

Yea nah sorry for my previous, I did the test on tuesday and it's kinda foggy in my head no lol. You were right first it's mg(a+A) which is the same as mga +mgA (I don't why I complained up there lol).

yea part c is 1/2kA^2 but you can't have k, so you need to substitute f/x which is (mg/(b-a))

[schnappy]

Yeah I've done that, I accidentally put it in the next box so I only have 2 tries left on it It's just equating the previous two answers though. Algebra... I can't do it anymore. I smell a quadratic.

[Elnino_Gerrard]

Theres a lot of unneccasrry maths thats the only difference in physics advanced and the normal one.Plus lecturer quality is crap :/ .I wish i had done the other one.I think next semester is where the difference comes in..I may as well drop down then/

[schnappy]

I heard one of the advanced lecturers was good, but don't know anything about the mainstream ones.

Microsoft math says its a quad equation, and using the quad. formula you apparently get:
A=(-sqrt(2a/(b-a)+1)+1)(b-a)

Well that's wrong, now I really don't know what they're looking for and I wasted my first go by writing the thing in the wrong box. :/

[Blakhitman]

yea you use quad formula and get

but just use the positive.

I can scan the working for it if you want.

[schnappy]

Meh, I just want to get this done. So (b-a)+sqrt(bb-aa)
is what the website will accept?

[Blakhitman]

you can do b^2-a^2 instead of bb-aa but yes it is.

[schnappy]

Thanks. I think what I had first was right, but I put something on the wrong side of parenthesis when typing it in. Your form is a lot nicer though, thanks

MasteringPhysics :thumbsdown:

[Blakhitman]

you're welcome.

yea masteringphysics is annoying, and I think we're the guinea pigs haha.

[Elnino_Gerrard]

Yea the new one is better but we only have him for 3 weeks and hes doing the easy topics ala gravity and oscilliations so yeah :/

[Elnino_Gerrard]

Hey guys in centripetal questions,with banked roads u know how the horizontal component of friction plus th horizontal component of the normal reaction force counts towards centripetal..When adding forces vertically would the vertical component of friction also be counted..Like-N -mgcos(theta) -horintal component of friction=o?  the textbook  only seems to do..N-mgcos(thta)=0?

[simpak]

Guise I'm having some Fundamentals issues, LITERALLY BEEN WORKING ON THIS SHIT HAMMERTHROW PROBLEM FOR LIKE FOUR HOURS TOTAL arghhhh!  Anyone interested in helping me out?  Should I post some queries?

[Blakhitman]

No harm in asking I guess.

[schnappy]

Post the question, sounds like a projectile motion problem. They're straight forward for us

[simpak]

I can do the projectile motion bit it's the circular motion bit, I have an aversion to round things and the theta symbol.
I'll upload the picture now...

http://i56.tinypic.com/2zyzy3o.jpg

Mmkay so, I am good with most of it but I'm just getting myself confused with the stuff specifically to do with circular motion, particularly b and f.
For b, I figured that because the tension force is the only thing pulling the hammer towards the centre of the circle that you would find it using centripetal acceleration.  But then I was like 'wat about the gravities' so now I don't know how that comes into play - is the tension force just found using F = mac or does it involve the mg of the weight as well?  Like, when I first looked at the question on an off day, I calculated the tension force using trig because you know the weight force and the angle at release.  Would you sum the value I got there and the value I got for centripetal?  Or ignore second guessing myself altogether?

And then for f, at first I thought the torque was 0 because the man is at the pivot point but NO then I realised this can't be the case and so I tried to find out what the torque was by finding the tangential acceleration and then finding the 'tangential force' and then solving for torque using the linear torque expression t = Frsinphi with F being the value I got as the tangential force and assuming that phi was 90 degrees but I don't know if that is right at all either...

The more I look at the question the more I confuse myself.

OH MAN I AM SO BAD AT PHYSICS.  Really I just want to be able to understand this because circular motion is what I will 100% screw up in the exam.  Thanks guys.

Ps. the reason I used linear expressions to solve for torque in f was because they don't give us the moment of inertia and Sevior was all 'oh yeah you don't need to know how to work that out lol' so I assumed they just didn't want us to use it but maybe I was meant to reference the textbook values?  I had a look but I wasn't sure if the hammerthrow could be thought of as the same thing as 'cylinder with mass distributed around the rim' since the mass is kind of not distributed all over the rim it is centralised at one specific point.  ...that is all.