<p>^its 0.24</p>
<p>you DON'T NEED to do "Parallel Axis Theorem"</p>
<p>I = Sigma(mr^2) = (1/12)ML^2 + mr^2 [the first is the given from the question for the rod, the block is just the basic mr^2, where r is the distance from the axis of rotation]; plug and chug gives 0.24 :)</p>
<p>Dexter, I think it's .42. </p>
<p>stanfordrat, v = Mg*sin(θ)[1-e^(-bt/m)] / b, so take the derivative to get</p>
<p>a = Mg<em>sin(θ)</em>e^(-bt/m) / m. t = 0 => a = Mg*sin(θ) / m. As t -> +inf, a -> 0. This gives you the y-intercept and the shape of the graph.</p>
<p>@snipez90: why do you say its 0.42; I think its 0.24 [see post above yours for why]?</p>
<p>The answer is 0.42:</p>
<p>You do NOT use 1/12 M L^2 for the moment of inertia of the rod, you use 1/3 M L^2. 1/12 M L^2 is the moment of inertia with respect to the center while 1/3 M L^2 is with respect to the end.</p>
<p>never mind, you use the parallel axis theorem :)</p>
<p>List</a> of moments of inertia - Wikipedia, the free encyclopedia</p>
<p>go to the bottom, there are two moments of inertia for a rod. I don't think it was as simple as applying the summation given on the formula sheet. Anyways I don't think I can explain the right answer, but that's my justification for why .24 is incorrect.</p>
<p>^got it...the explanation is parallel axis theorem :)</p>
<p>darn, well will i get any points for deriving the moment of inertia for a rod about its hinge?</p>
<p>I know I should have like memorized the parallel axis theorem, well because if kind of makes sense. It takes the mass of the entire object and makes it seem like it's a weight where the center of mass is because it is distance R from the end, so MR^2. But I was still angry that they didn't give it to us.</p>
<p>Hopefully we won't get doubly penalized for correctly substituting the answer in C into D. I did the parallel axis theorem as
I = Icm + h^2, which is obviously retarded, but of course I wasn't thinking straight, especially after the Bio exam.</p>
<p>More over guys, for E/M 1, part c, the graph, the starting point of the last one is less than that of the E inside the sphere because the radius is greater but the charge is the same...</p>
<p>The parallel axis theorem alone isn't enough for the rotational inertia in 2C.</p>
<p>If the question were asking for the inertia of just the rod (from the hinge), then .24 would be the answer. It asks for more, though. It says to calculate the inertia of the rod-block system. The block is included in that. The block on the end of the rod adds to the inertia. The block can be treated as a point on the end of the rod, which means that its individual inertia is mr^2= (.5)(.6^2)=.18 Since the inertia of a system is the sum of the inertias of all the individual inertias in that system, the inertia of rod-block system (around the axis, in our case the hinge) is the inertia of the rod + inertia of the block= .24 + .18=.42.</p>
<p>oh man.. for the moment of inertia, i put 7/12 * MR^2 instead of .42</p>
<p>They are the same, but forgot to substitute in the numbers for M and R.</p>
<p>And for #3B on E&M, i put that the B field would be 0 when there was one loop below and one loop above the point P. Right hand rule says that correct? Someone said it added up instead of canceling out.</p>
<p>For the spring question on mech, i said that the ball would free fall until past equilibrium position, where the eleastic force would act upon it. My mass was like 2.8 or something like that.</p>
<p>haha...then there wouldnt be any magnetic flux for part d and e....lol</p>
<p>I added the block's moment of inertia in part D, because I suppose I misread part C. I correctly calculated the block's moment of inertia, but not the rod's. Partial credit ftw???</p>
<p>I totally got the magnetic field question for E&M wrong >< Bleh, I've not got much experience with Biot-Savart. Well, you only need like 60% to get a 5 so I hope I can still get 5 =(</p>
<p>i have a feeling physics all around is going to have a big curve this year....:)</p>
<p>wait can someone explain how the flux is not 0 for part D?</p>
<p>Okay, here are my answers for M3 from memory. Haven't redone them yet.</p>
<p>a) F=kx. Simple F vs. x and draw best fit line.
b) k is just slope of graph. I think I got 25 N/m.
c) Potential energy of the mass at the top is converted all to spring potential when the string is at 1.5m. The stretch is 1.5 - 0.6 = 0.9m. Set 1/2kx^2=mgh, solving m to be 0.675 if g=10 and 0.689 if g=9.8.
d) i) Equilibrium position. Set mg=kx. x should be solved to 0.27m. Adding 0.27 to 0.6 gets you 0.87m as spring length.
ii) Simple enough. The most spring potential is converted to kinetic energy. There's always some spring potential always in the spring, though.
iii) 1/2kA^2 - 1/2kx^2 = 1/2mv^2. I forgot to subtract out 1/2kx^2 at the equilibrium position, so boo.</p>
<p>
[quote]
wait can someone explain how the flux is not 0 for part D?
[/quote]
</p>
<p>It is not, because the initial flux is Bnet<em>s^2. But because of the dot product, it is Bnet</em>s^2<em>cos(theta), and theta changes with time. So you can substitute theta for omega</em>time and then take the derivative of cos(omega*time) with respect to time.</p>
<p>haha....i only put Bneet *s^2......TOTALLY forgot about the cos part....oh well.....probably still got like 2/3 points for that</p>