Adding subjects that aren't showing/Physics Chatter

[schnappy]

A centripetal force is the force that keeps the object going in a circle, at a constant speed v (assuming a uniform motion - its not speeding up. It is accelerating due to a change in direction, but its speed is constant).

A centripetal force is made up of other forces. It might be gravity, normal forces or how have you, in this case it is tension.

Centripetal force = Tension force
mvv/r = T

But I'm assuming the circular path is parallel to the Earth. The diagram suggests otherwise, but it doesn't give an angle specifying the angle of the circular path to the Earth, from which to find the component of gravity contributing to the centripetal force.

For f, you know that w_0 = 0, w_f = r/v
You found v in part a. Find alpha (angular accel) using a = (change in v) / t
Torque = I*alpha

It says all the mass of the system is in the 'hammer'. This means I = mrr.

That assignment is just as hard as the Physics 1 assignment, and it looks like you don't get 3 goes :S If this is sort of the equivalent to our angular kinematics question, I'd rank this higher actually.

Sorry my explanation is a bit hurried and I don't know latex

[simpak]

Thank you for that!  Thanks so much, makes sense.  So for b, once I find the centripetal force I don't have to do anything else to find the tension force?  Someone else is suggesting that I find the weight force of the hammer in the same direction as the tension force, so the 'angular' component and subtract it from the centripetal force?  By the way, they said to assume the hammer is released at ground level (but I thought that was only helpful for part a so you know how high above the ground it's released to solve for the initial velocity using projectile motion).  But would that kind if imply that there is a normal force counteracting/cancelling the weight force anyway because it is 'on the ground'?  This is the most ridiculous question of all time.

Arrrrghhh, I know right and we got yelled at for saying they were going to fast.  Three goes?  We have this (there are only four questions, the others are fine I think this one is just the most difficult) and then two written response questions, one on ABS breaks and the other on earthquakes changing moment of inertia and therefore day length.  But the questions are all on kinematics...we're not even doing kinematics anymore now, we're doing sound which I am better at!  Wish the assignment was on that!

[schnappy]

Yes, you do need to add on the weight component contributing to the centripetal force. But there's no angle to allow you to quantify it (Or am I missing it?). Or are you meant to state your answer in terms of an angle?

If it's at ground level, doesn't mean it's in contact with the ground. Questions obviously aren't worded the best, but I'd go with it being just above the ground (so the tension is holding it up, not a normal force). Talking about a normal force holding it there would make it messy, and ruin everything.

If the hammer is released at ground level, wouldn't it just skid along the ground? (No vertical component of launch velocity) I thought it said the angle of launch was 45deg.

Are you meant to model the motion as a circle that's standing on top of the ground? Or as a circle that's bent over like in the picture? If you use the 45deg launch angle to find the weight component, but anything other than that and I'm confused.

[simpak]

That's what I mean, the 45 degree launch angle...BUT I DON'T KNOW IF THAT'S RIGHT.

I actually hate this question.

[schnappy]

See attached. Might have the wrong trig function in the wrong place. I assume you're doing Calc1 and so know what the i-hat and j-hat vector notation all means. It's just breaking it up into vertical and horizontal components.

From what I can tell the weight force 'pulls' the tension into the center of the circle, making a circular motion.

You might get a better response in the VCE Physics section.

[simpak]

I DO NOT DO CALC 1 what is this i-hat j-hat?  Thanks so much for that by the way.

[schnappy]

oooo ok.
Basically the number (including the sign) with the i-hat (It's an 'i', but it's wearing a hat, like a ^) is a horizontal component, j is vertical. A vector has direction, at some angle theta. the i and j hat just breaks the angled vector into a horizontal and vertical parts, so you can add them together easily. If it's confusing just do vector addition however you know too, but it's probably worth learning.

Attached - There's probably some fancy definition for everything, but that'll show you what I mean. It's just a tool to add things together, though mathematicians think it's something to make difficult and silly.

[simpak]

So not this reasoning...  

Ps I fucked that up at the top it should say T = Fc - x but I am lazy.

[schnappy]

Ummm to be honest I'm not entirely sure. I don't have full faith in my own working (Any of it - not sure if I'm looking at it the right way)
EDIT: After subbing numbers in and looking at the result, I either still can't use a calculator OR this really doesn't seam right. What velocity did you work out in part a?
Decompose the centripetal force into its horizontal and vertical components:
Horizontal: (mvv/r)*sin(45)
Vertical: (mvv/r)*cos(45)

This is the resulting force, from the vector addition of the weight and tension.
Weight (only vertically - make it negative): -mg

For Tension,
Vertical: O
Horizontal: P
O-mg = (mvv/r)*cos(45)
O = ( (mvv/r)*cos(45) ) + mg
P = (mvv/r)*sin(45)

O is vertical, P is horizontal. The result is just a pythag problem:
O^2 + P^2 = T^2

Where T is the tension, which we're looking for.

The i-hat and j-hat notation is just doing the above, but having to write 'vertical' and 'horizontal' out goes away and you can just write a single line equation.

Please don't submit this as an aswer, I'm really not confident with it. Try plugging in numbers and see if you get a reasonable answer.

[simpak]

Got 29.16 for velocity!

Thanks for that I'll give it a go!

[happyhappyland]

horrible assignment...

[simpak]

Horrible test!  I must have actually failed.  I didn't realise that 1 b was actually a part b and I spent like 5 minutes trying to figure out how to get an answer without mass.  Good one Simone.

And then I couldn't answer some other three mark question.
So there's 7 marks of blank.

It was too long for 20 minutes >:

[Elnino_Gerrard]

think i got like two marks for my advanced phsycis test l.Test was too long for 20 mins fak..an so little room.I ran out and was franticlly looking fros space all over :/

[simpak]

For us the lecturer was like 'this won't take most of you 20 minutes'.
Nobody I spoke to even finished.

[Elnino_Gerrard]

For us the lecturer was like 'this won't take most of you 20 minutes'.
Nobody I spoke to even finished.

JOke eh.Similiar story here.Except not only did we not finish ,most couldnt even do most of it xD