<p>^
So yea, I also need help on a physics problem. It includes a drawing, but Im pretty sure I can illustrate it good enough. Here it is:</p>
<p>A system consists of two blocks, each of mass M, connected by a spring of force constant k. The system is initially shoved against a wall so that the spring is compressed a distance D from its original uncompressed length. The floor is frictionless. The system is now released with no initial velocity.</p>
<p>[Picture of a block of mass M on a corner. Imagine a pythagorean triangle, and imagine that block to be the 90-degree mark. Connected to that block is a spring that has another block of mass M attached to the other side of it. I think the problem illustrates it enough actually :-/ ]</p>
<p>b. Determine the speed of the center of mass of the system when the left-hand block is no longer in contact with the wall.</p>
<p>This problem may sound familiar. It is a Physics C Mechanics problem from 1974. Thanks a lot in advanced!</p>
<p>a) (1/2)kD^2 = (1/2)mv^2
Dsqrt(k/m) = v</p>
<p>b) Conservation of Momentum
m(0) + m(v) = 2m(v center of mass)
v/2 = v center of mass
(you can sub in part a for v if you wish)</p>
<p>c) period is easy: T = 2(pi)sqrt(m/k)
masss = reduced mass = 1/((1/m) + (1/m)) = 2m
so T = 2(pi)sqrt(2m/k)</p>
<p>yay</p>
<p>m(0) + m(v) = 2m(v center of mass)</p>
<p>I do not understand that part. Can you please describe how you incorporated the velocity of the system's center of mass into that conservation of momentum equation? Thanks alot. :D</p>
<p>BUMP. Along with this question, how do you find the direction of the Torque on a spinning bicycle wheel [wheel is vertical, perpendicular to the floor]?</p>
<p>Here it is.</p>
<p>After the right block reaches its maximum KE, it begins to slow down, as per Hooke's law. But momentum must be conserved at this stage, since the wall is no longer providing a normal force to the left block and no net force acts on the system from the outside. The system weighs 2M, so the momentum of its center of mass must equal that of the sum of the momenta of the left and right blocks <em>exactly</em> when the right block achieves its maximum. This occurs when</p>
<p>*(1/2) k D² = (1/2) m v² => v = D sqrt(k / m) = D omega.</p>
<p>Thus the momentum is mv = D sqrt(km) = D omega m.</p>
<p>This is the maximum momentum of the right block before the left block leaves the wall. When it does, the center of mass maintains the same momentum...</p>
<p>2m (v COM) = D omega m => (v COM) = (D omega) / 2 = v / 2.</p>
<p>Thus the velocity of the COM is 1/2 that of maximum velocity attained by the right guy.</p>
<p>Now as far as torque is concerned... you would use the right hand rule to determine its direction, if that's what you're asking. Simply curly the fingers of your right hand in the direction of rotation and stick out your thumb. It should be along the axis of rotation, horizontal with respect to the floor. If you're asking about the torque which causes it to rotate, however, it has to do with friction providing the perpendicular force.</p>
<p>thanks alot for the responses. I think I was being rather vague for the torque question. Im really just curious to know how a gyroscope works. I know that the angular momentum vector sorta "chases" the torque vector, causing it to spin around. Anyone know of a good site where this can be explained well?</p>