<p>uh from the velocity equation v^2 = 2ad is how I got the velocity.</p>
<p>You’re absolutely right that gravity only accelerates the hanging part of the rope. I’m contending that gravity only accelerates the hanging portion of the mass.</p>
<p>Indeed, which is why you have to create the function (y/L)Mg. The velocity equations only work when there is a constant acceleration.</p>
<p>There is a constant acceleration for part a. I might have done the rest wrong, but I’m only talking about part a.</p>
<p>Hm…well what about tension?</p>
<p>[Sliding</a> Block Accelerated by Hanging Block Problem](<a href=“http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DynamicsProblems/BlockHangingBob/BlockHangingBobMain.html]Sliding”>http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DynamicsProblems/BlockHangingBob/BlockHangingBobMain.html)</p>
<p>There’s a problem and a solution just like it. I’m almost pretty sure I’m right about part a.</p>
<p>The tension initially balances it in that problem. In the free response system, the block is held there initially and then released from rest.</p>
<p>I still want to know why you’re doing sum of the forces and not conservation of mechanical energy. Mgh=1/2mv^2!</p>
<p>I can do it with conservation of energy too.</p>
<p>m/2<em>g</em>d = 1/2<em>m</em>v^2. </p>
<p>v = sqrt(g*d)</p>
<p>Only one of them has a GPE while both of them have the same v so you can combine it as one mass.</p>
<p>There you go. Sqrt(2gd).</p>
<p>But the sliding block has a velocity as well (that is equal). So I combined it together as I stated.</p>
<p>I see what you’re saying. You could be right, but I wrote m/2 because it asked for the velocity of the hanging block. Wouldn’t that v be the velocity of the system and not just of the hanging block?</p>
<p>For those wondering about tension, it is a non-issue because the system starts held at rest, meaning that there is no tension between the two blocks. You can actually ignore that the pulley exists and treat it like a rope with its mass evenly distributed between its two ends. It’s kind of a trick question because typically pulley systems in physics involve a beginning tension. </p>
<p>To make things simpler, the string and rope systems are virtually identical, it’s just that the string system has its mass distributed on its two ends rather than throughout the string. Consequently their ending velocities must also be equal.</p>
<p>That’s interesting headaslpode, but once again even though that’s true, you’re comparing the velocity of the hanging block ONLY to the rope.</p>
<p>I’m trying to remember exactly what I did mathematically to get that the final velocities of both the rope and the hanging rope equal (I probably fudged something somewhere), but in either case the velocity of the hanging block should be equal to the velocity of the sliding block.</p>
<p>Well the final velocity of the rope is sqrt((gy^2)/L). I got sqrt(2gd) for the hanging block. I’m just not sure if I was supposed to do conservation of the whole system or not. I suppose I was and I fudged up. :/</p>
<p>^Got the same answer for the hanging block. </p>
<p>But my dumb-self didnt think about the y-portion of the rope being a part of the whole Mass until after the exam.</p>
<p>Ah! That’s what I didn’t think to explain.</p>
<p>Acceleration is equal to g in both cases, regardless of any mass in the system, so either way you have this sort of equation:</p>
<p>v = at
d (or y) = 1/2at^2 (no initial velocity)
t = sqrt(2d/g)
v = g * sqrt(2d/g)
= sqrt(2dg)</p>
<p>There’s no need to use conservation of energy.</p>
<p>Side note: Acceleration of the hanging block is also perpendicular to the movement of the sliding block, so the sliding block’s presence does not decrease the acceleration of the hanging block…I think :/. I was working on the assumption that since we aren’t told the rope is taut we don’t have to divide between both masses, but that might be entirely wrong.</p>
<p>Here’s How to do mech. #2 A/B)</p>
<p>Recall from SHM that a(t)=-(w^2)x(t) (this is for SHM caused by force). For SHM caused by torque, the equation becomes Alpha(t)=-(w^2)Theta(t). The equation for torque in the case of any physical pendulum(SHM again) is -xmgsin(theta)=I(Alpha) where x is the disance to the center of mass from the point of pivot. Now you had two options for the differential equation. One way is to repace alpha with ((d^2)(theta))/((dt)^2) or you could replace alpha with -(w^2)(theta) and then replace w with d(theta)/dt. That’s part a. For part B, it was asking for you to prove the equation for period in any physical pendulum. This is similar to the way you do it for a simple pendulum. You have to take the equation, -xmgsin(theta)=-I(w^2)(theta), assume that it’s a small angle approximation(sin(theta)=Theta), and solve for w. You get that w=sqrt((mgx)/I). Then you use the equation T=(2(pi))/W to find the period. The solution is T=2(pi)sqrt(I/(mgx)). That’s part B.</p>
<p>so can you use k-effective=-d(torque)/d(theta)? and the period is 2<em>pi</em>(I/k)^.5? and k is mgx? cause cosx=1 when x is small…</p>
<p>Thanks for the complete explanation on Mech #2 bobwrit. That’s a very interesting way to think of it headasplode. I like it with the constant acceleration of g. Good call, good call.</p>
<h1>339</h1>
<p>k-effective? I thought kx was based only on spring or something? O_o What is this you speak of? 
(I realized that you were talking about the idea of kx that it’s restoring force. I like the idea because it’s interesting but I have no idea at all about the theory being true.)</p>
<p>RentonT,
yeah, kx is the restoring force on the center of mass…it can be used for both simple pendelums and physical ones
in barron’s AP Physcis C book, that’s how they derive the I/mgx equation using the torque, which in mgxsin(theta), and then using cos=1 when theta is small</p>