threewingedfury

I need some help on some concept questions in a calculus-based physics class, its quite a few so I wont post them here. If anyone is willing to help, please email me at threewingedfury@hotmail.com THANKS!

pig

I'd post 'em here. You'll get better responses, and people can and will correct each other.

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threewingedfury

ok cool - I just didnt think people would want to deal with so many questions lol


The displacement of an object oscillating on a spring is given by x(t)=xmcos(wt+p). If the initial displacement is zero and the initial velocity is in the negative x direction, then the phase constant p is:
a. 0 rad
b. pi/2 rad
c. pi rad
d. 3pi/2 rad
e. 2pi rad
I figured this one was d.


Two uniform circular disks have the same mass and the same thickness are made from different materials. The disk with the smaller rotational inertia is:
a. the one made from the more dense material
b. the one made from the less dense material
c. neither - both are the same
d. the disk with the larger angular velocity
e. the disk with the larger torque
I thought this one was b.


A hoop (I=MR^2) rolls with constant velocity and without sliding along level ground. Its rotational kinetic energy is:
a. half its translational kinetic energy
b. the same as its translational kinetic energy
c. twice its translational kinetic energy
d. four times its translational kinetic energy
e. one third its translational kinetic energy
This one, I have no clue.


When a certain rubber band is stretched a distance x, it exerts a restroing force of magnitude F = ax+bx^2 where a and b are constants. The work done in stretching this rubber band from x = 0 to x = L is:
a. aL^2 + bLx^3
b. aL + 2bL^2
c. a + 2bL
d. bL
e. aL^2/2 +bL^3/3
No clue on this one either.


Vectors A and B have magnitude L. When drawn with their tails at the same point, the angle between them is 30 degrees. The value of A dot B is
a. zero
b. L^2
c. square root(3)L^2/2
d. 2L^2
e. none of these


Vectors A and B have magnitude L. When drawn with their tails at the same point, the angle between them is 60 degrees. The value of A x B is
a. zero
b. L^2
c. square root(3)L^2/2
d. 2L^2
e. none of these

I really hate this stuff - I just cant get a grasp of it.

Silvy

On this forum we tend not to do anybodies homework outright, however we will try to help you understand the stuff.

Please put some work in explaining how you got the answers you put above, and also at what point the other questions elude you or where your specific troubles lie.

Personally I can't help you, these questions go over my head. But if you explain in more detail what concepts you're having trouble with I'm sure other members can be of much more assistance.

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pig

Really quick - on 1. the xmcos(wt+p)...define more clearly xm.

For the work question: W=integral(F.dx)|L-0. It's pretty straight forward.

dot product = mag(a)mag(b)cos(theta).

cross product is another vector. you can calculate its magintude to be a scalar quantity, but how you interpret the answers you have given is up to you. ie. none of them is a vector - so what does "value" of AxB mean?

the other two i'll have to think about - it's been a while since i've looked at some of these formulae.

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threewingedfury

The problem just has x of m - Im thinking its between 0 and 2pi, maybe 0 - but all I have in my book is : The value of the phase constant depends on the displacement and the velocity of the particle at time zero. So I dunno.

So are you just saying for the cross product, just multiply?

pig

. so you're saying x(m)? I am assuming that the x part of xm supposed to be same variable as the x in x(t)?

nope - i'm saying that the cross product is not a scalar quantity, but is instead another vector is that is orthogonal to both A and B. It's magnitude can be calculated to be a scalar quantity. Go to www.mathworld.com and look up "cross product." its a pretty useful site for a lot of mathematical concepts.

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threewingedfury

actually its like m is a subscript - lol I might have said it wrong - sorry so x(m as subscript)

and for the cross product - i forgot it was ab x sin(theta) - so thats no prob

pig

well, in the first one - so that quantity x_m is a scalar constant, not a variable. that's pretty important when calculating the derivative. i get a different answer on the first one from yours, but i suspect that we're very close in reasoning. what are you using for the derivative function?

note, for the cross product - the formula you gave is for the magnitude of the cross product, but that is not the same as the answer for A x B. A x B is a vector quantity.

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threewingedfury

well - I didnt use an integral - I was thinking along the lines of a cosine function and how its considered negative and it has a phase constant of 0.

pig

the way i approached this problem is essentially as follows:

you are given the function for the displacement : x(t) = x_m*cos(wt+p).

you have the initial conditions. x(0)=0, dx/dt=-k, where k is some unknown negative quantity. you can deduce two values of p (not considering periodicity) from the x(0)=0 condition. To determine which value it is, you consider the derivative of the displacement function, dx/dt. the derivative of the displacment function is the velocity. it is positive at one point, and negative at the other. therefore, you need to determine the sign of the derivative of x(t) in order to determine which value is the correct one for your problem.

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threewingedfury

ok I get what you're saying - and for the work question you would say the answer is e. integral of ax+bx^2 is aL^2/2+bL^3/3?

pig

3wing:

well, take it worth a grain of salt, and so forth...but word. that's what i would come to for the work problem.

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shakran

I'm no math guy, so the conceptual one is the only one I can answer. . .


Correct me if I'm wrong but if they're uniform and they have the same mass and the same thickness, then the material they're made from is a red herring. The answer is C because rotational inertia is determined by the mass and where the mass is concentrated. In other words, if they weren't uniform then one disk could have most of its mass on the inside and the other have most of its mass on the outside, and therefore the smaller rotational inertia would be the disk with more mass on the inside. But if they're uniform then as far as rotation is concerned their physical properties are identical.

threewingedfury

^ but they are made of different materials, so the density would be different in them - does the density not matter?

So I'm getting the idea of either half or twice the translational energy - any ideas?

shakran

Ahhh I made a mistake. You are correct, it's b. I=1/2 mr^2 (this works for hoops and discs), so the less dense material, which has a larger radius, will have a greater rotational inertia.

filtherton

If rotational kinetic energy is .5 I w^2 (it seems like it should be)
translational velocity is v
translational kinetic energy is .5 m v^2 (v = wr)
w = angular velocity = v/r (r = radius)
rotational k.e. is then 0.5 mr^2 * (v/r)^2 = the key to your answer (provided i didn't mess things up)

threewingedfury

are you trying to say they are the same?

because I get for the translational - .5m(v/r)r^2 and for the rotational .5mr^2(v/r)^2, which arent the same

Ok, so I found this exact question on the net and it said the answer was between pi and 3pi/2 rads - but it cant be both answers

filtherton

This is kinetic energy right? .5*m*v^2. I'm not sure where the r^2 is coming from in your kinetic energy formula.

threewingedfury

I dunno, I think I messed up - so are you getting the same thing for both rotational and translational?

pig

you made an unnecessary, and i believe incorrect, substitution in the translational kinetic energy expression. if you write the problem as trans = a*rot - where a is (presumably) some constant unknown factor between the two, you can deduce a value of a.

for the displacement, check this: problem

i think you'll find something suspiciously similar to your problem there.

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Aseel

hello eveyone,
am a new mamber and i like to Introduce myself. my name is aseel and i live in new zealand. i am studying at universtiy and doing my fisrt year now..