<p>Can anyone explain to me simple harmonic motion? I'm still not getting it...</p>
<p>And rotation please.</p>
<p>I have a few notes on this, hold on...</p>
<p>Thanks a lot. :)</p>
<p>In simple harmonic motion, an object oscillates between spatial positions for an indefinite period of time with no loss of mechanical energy.</p>
<p>x = Acos(wt+theta) or x = Asin(wt+theta), depending on initial conditions for x and v; if objects starts from x = 0, then the 2nd equation applies. A = amplitude (how far objects goes from equilibrium during oscillation)</p>
<p>Max velocity = wA
Max acceleration = (w^2)A</p>
<p>For spring mass system, w = (k/m)^1/2
For pendulum (massless), w = (g/L)^1/2
For physical pendulum (one with moment of inertia), w = (mgd/I)^1/2</p>
<p>Period = T = (2pi)/w
Frequency = 1/T</p>
<p>Energy:</p>
<p>U = 1/2(k)(x^2)
K + U = 1/2(k)(A^2) (k = force of spring)</p>
<p>w is angular velocity, right?</p>
<p>Thanks for those notes. I couldn't understand what my review book was saying.</p>
<p>Rotation:</p>
<p>v = rw
a = r(alpha)</p>
<p>K = 1/2(I)w^2, I = moment of intertia</p>
<p>Torque = rFsin(theta), r = distance between point of application and pivot point; torque = RxF</p>
<p>Torque = I(alpha)
w = integral of torque
P = (Torque)(w)</p>
<p>w = angular velocity</p>
<p>^^ in addition to that:</p>
<p>Angular momentum = Iw (with I being inertia)</p>
<p>for the "disk rolling down the incline" problem:</p>
<p>always remember that if you are setting up the KE/PE equation the KE final for the disk is (1/2) mv^2 + (1/2) Iw^2</p>
<p>Bump.</p>
<p>Also, what about all the moments of inertia? (like for a disk, hoop, etc). I think those are worth knowing.</p>
<p>And the disk thing is "Translational KE + Rotational KE" when it doesn't slip?</p>
<p>hmm...........i think wut confuses ppl is that w right there. it means angular frequency, kinda abstract idea</p>
<p>hmm for moment of inertia.. we probably have to know the basic ones like a thin rod with rotation axis through center(probably most important?) and solid cylinder.</p>
<p>long thin rod w/ rotation axis thru center
I=(1/12)ML^2</p>
<p>solid sylinder
I=(1/2)MR^2</p>
<p>but im pretty sure we don't have to know everything-meaning hollow cylinder, thin spherical shell and stuff like that. it's juss wayy to much.</p>
<p>ahh one more.
long thin rod w/ rotation axis thru end
I=(1/3)ML^2</p>
<p>but they dont give u I for special shape, or they do??</p>
<p>they probably don't. so i doubt we need to know I for special shape...</p>
<p>Physical pendulum T = 2pi sqrt(I/mgl) which mass on a bob = 2pi sqrt(l/g)</p>
<p>Period for pendulum is independent for mass.
For small angles sin theta = theta and so the acceleration back to the eq position is like kx, its mgLsintheta = mgL * theta for small angles. THis is the torque which acts as a restoring force
Springs add like resistors (was on MC a few years ago)
x= A cos(wt)
v = -A w sin(wt)
a = -A w^2 cos(wt)
kx = ma
-k(Acos(wt)) = m (-A w^2 cos(wt))
w^2 = k/m
w = sqrt(k/m)
T = 2pi/w = 2pi sqrt(m/k) </p>
<p>Kepler laws of equal area because area swept = 1/2 v * r and r*mv = constant because there are no external torques on the planet system. and so Area swept out can be witten as L / 2m.
Satellites in cicular orbit mv^2/r = GmM(e) / r^2 and so v= sqrt(GM(e)/r)
T^2 porpotional to r^3 from this</p>
<p>Momentum always conserved
perfectly inelastic = objects stick together
inelastic collision - only momentum conserved, objects move apart at 90degrees (maybe)
elastic = KE conserved</p>
<p>Moments of inertia = integral x^2 dm
This rod one end = 1/3
Thin rod in middle = 1/2
thin hoop = 1
solid circle = 1/2
solid sphere = 2/5
hollow sphere = 2/3</p>
<p>I = Icm + M(d)^2 where d = distance from that point so for getting thin rod middle from one end
I = 1/12Ml^2 + M(l/2)^2 = 1/12 + 1/4 = 4/12 = 1/3Ml^2 </p>
<p>if a thing rolls down an incline with friction, the one that has a lower MOI will get to the bottom first. If there is no friction then they must have the same velocity at the bottom, (no torque) and it doesnt matter their MOI. Theres something Ill try to find it about like mg/(1+B^-1 ) or something like that.</p>
<p>damn thats like **** load of them.............even my teacher, who nvr gives us any formula, when encount the I thingy, will give us the formula......</p>
<p>but u can put them into ur calculator, they dont make u erase it. but u cannot use it during MC......</p>
<p>why is angular momentum sometimes</p>
<p>mvr?
what's that mean.. like when you equate it</p>
<p>i know Iw
but what's mvr mean?</p>
<p>regular momentum is just mv</p>
<p>where'd the r come from?</p>
<p>not torque.. cause torque is mgr..</p>
<p>r is radius. mvr is angular momentum. mv is linear momentum.</p>
<p>L (angular momentum) = r x mv</p>
<p>what's that x (cross) again? Sin or Cos?</p>
<p>i think its sin, but it depends on the set up of the problem</p>