<p>you guys scared me badly. T-T I’m going to write the late exam this Friday.</p>
<p>@DataBox: <a href=“http://talk.collegeconfidential.com/1062552133-post338.html[/url]”>http://talk.collegeconfidential.com/1062552133-post338.html</a></p>
<p>could someone post their answer for FRQ#1 on the Mechanics exam i want to compare answers to mine. thnks.</p>
<p>Mech 1:</p>
<p>a) E=K+U
E=m(v^2)/2+U(-.5)
E= 7.0J (i think, I don’t want to calculate anything right now)</p>
<p>b) K=0 when U(x)=7.0J
4.0x^2=7 then solve for x.</p>
<p>c) p=mv
E=7.0J=U(.60)+m(v^2)/2 solve for v and plug it into p to get momentum</p>
<p>d) |a|=|F/m|
F=-dU/dx=-8.0x
|a|=8.0x/m then plug and chug</p>
<p>e) ok, I will do my best to explain this one.</p>
<pre><code>x(t) should be a sine wave
K(t) should be a cos^2 wave (because v=dx/dt and K(t)=1/2mv^2) so therefore it should start with maximum KE, b/c v(t) has a max, it should equal 0 when x has its max and mins b/c v=0 (this is also a relative min) and then it should have a max every point after where x=0 and a min (0) whenever x has a max or min. I hope that was clear.
</code></pre>
<p>Hey chong, thanks a lot for posting up your answers for everyone. I was wondering if anyone could post the FINAL correct answers for Mechanics because I can’t keep track of all the corrections made. If anyone has the Final correct answers for only the mechanics questions, that would be extremely helpful for me. Thanks a lot!</p>
<p>This is, hopefully, a 100% answer to Mechanics 3.</p>
<p>a.) v = sqrt(gh). Use orangebase’s explanation. He describes it very thoroughly.</p>
<p>b.) F = (M/L)yg</p>
<p>F of Gravity = Mg</p>
<p>The hanging part of the rope is represented by the linear density of the rod (M/L) times the amount of distance over (y). Therefore the mass of the distance over the edge can be represented by (M/L)y. Plug it into the Force of Gravity equation and voila.</p>
<p>c.) W = Integral of Force. </p>
<p>Therefore, W = Integral of (M/L)yg dy. Because (M/L)g is constant you can pull it outside the integrand and then have the integral of y, which equals (y^2)/2. Multiply that by the constant (M/L)y and then</p>
<p>W = g(M/L)(y^2)/2</p>
<p>d.) This one can be tricky. Use the same kinematic idea presented in part a.</p>
<p>(1/2)(Mass of the entire rope)v^2 = (Center of Mass of y)gh.</p>
<p>(Mass of the entire rope) = M
h = y
(Center of Mass of y) = Mass over the edge divided by two. Therefore,
(Center of Mass of y) = (M/L)(y/2)gy = g(M/L)(y^2)/2</p>
<p>So,
KE = PE
(1/2)Mv^2 = g(M/L)(y^2)/2, solve for v and you should get</p>
<p>v = sqrt((g/L)y^2)</p>
<p>e.) v(block) = sqrt(gh) </p>
<p>The distance traveled is L therefore h = L, so,</p>
<p>v(block) = sqrt(gL)</p>
<p>v(rope) = sqrt((g/L)y^2)</p>
<p>The distance traveled is L therefore y = h, so,
v(rope) = sqrt(gL)</p>
<p>The speeds are equal. Although I believe this is correct I very well have made an error.</p>
<p>The speeds are the same. Those are the same exact answers I put. </p>
<p>What did people put for 2c and 2d?</p>
<p>Are you talking about the mechanics? There is no 2d.</p>
<p>sorry,</p>
<p>2b and c, I forgot that a was in two parts.</p>
<p>Thanks for posting. I am fairly certain I did well on this from reading the answers, along with the fact that like 60% is a 5.</p>
<p>2b. The question is asking how to determine the moment of inertia of the bar. To find the answer we can use the result found in part a.</p>
<p>T = 2pi*sqrt(I/mgx)</p>
<p>Solve for I and you get,</p>
<p>I = ((T/(2pi))^2)*mgx</p>
<p>Because the variables m, g, x are already given, if we can find the period T of the rod, the moment of inertia I can be obtained. </p>
<p>The process for finding the period can vary as to how you approach the problem. </p>
<p>Possibilities include using laser timing to find the frequency (T = 1/f) or use a radar system to measure the linear speed which can used to find the angular speed and then the period
(T = (2pi)/w).</p>
<p>Errror is individual to the type of process you used to find T.</p>
<p>2c.) I rolled this one off at the last minute and im sure there are more appropiate ways then what I used.</p>
<p>I understood that at the center of mass, a pivot of some sort whether located below it with the bar sitting on it or with the bar being held by a rope, should be parallel to the ground. The position could be adjusted accordingly to try to find this value. And at this position the center at mass could be determined by a meterstick.</p>
<p>A little shaky but it should work nonetheless.</p>
<p>O.K. i put something similar for part b, i just forgot to talk about error.</p>
<p>c, i didn’t write anything, ran out of time.
how many points do you think those two sections will be?</p>
<p>I’m a bit late, but if anyone still wants to discuss the answers I had the exact same answers except for a problem in 3d.</p>
<p>KE = PE</p>
<p>I don’t think this is exactly true… Even after the rope has left the table, all the potential energy isn’t converted into kinetic energy. The kinetic energy will continue to rise until it has hit the ground.</p>
<p>Also, I don’t think you can say PE = (Center of Mass of y)gy either. Wouldn’t that mean that as the rope falls the potential energy increases? What I did was solve for the acceleration a = gy/L and then differentiated that to find v. I could be wrong though.</p>
<p>Oh, I love discussing these problems.</p>
<p>PE is a relative thing, you can measure height from anywhere, because you can say that a book resting on a table has 0 PE or you could say that it does have PE because it is above the floor. And how did you do the second degree differential equation that a=gy/L gives you (d2y/dt2=gy/L), Going through the problem, you were suppose to find the work done to get the whole entire string off, (antiderivative of the force of gravity (Myg/L) which gives you My^2g/2L) The work done on the string equals the change in KE, so you have Mgy^2/2L=1/2Mv^2 solving for v you get sqrt(gy^2/L) since the total distance fallen is L, you plug L into y and get v=sqrt(gL) which is the same as the answer in part A.</p>
<p>Yeah, I just wanted to make sure I understood it fully. I actually got the same answer.</p>
<p>What I did was took the acceleration a= gy/L and then plugged it into the equation v^2 = 2a(sf-s1), but since acceleration is variable I used an infinetesimal distance dy</p>
<p>d(v^2) = 2(gy/L)dy </p>
<p>which integrates to</p>
<p>v^2 = (g/L)y^2 + C where C=0 so</p>
<p>v = sqrt((g/L)y^2).</p>
<p>But I see now, taking the work equal to kinetic energy makes sense.</p>