Welcome. I dislike physics problems. Here's one I've been working on.




First section seems easy. It should just be the sum of the interias, right?

for a disk with axis through the center
I=(m)(r²)(.5)
so that would lead me to do
(.5)(.82)(2.61²)+(.5)(1.68)(5.03²)

Except the units of the problem want it in m, not cm, so it would look more like this
(.82*(0.0261^3)+1.68*(0.0503^2))/2)
Which gives 0.00213256522

Incorrect answer. What did i do wrong?
Edit: Typo in the calculation had me cubing instead of squaring one of the radii. Shoot me now. Will post other problems i run across in this topic.

Post has been edited 1 time(s), last time on Mar 25 2013, 3:07 am by TiKels.

So I leave you with this next problem, since vrael solved my last issue so quickly...

for clarity:
T1 = tension on the left side
T2 = tension on the right side
m1 = mass on the left side
m2 = mass on the right side


I started to draw free body diagrams for each of the things, with tension as well as mass*gravity for each of the weights. But when I got to the wheel I was slightly unsure as how to represent it. How do I set that up? What do i do to represent my intertia as a force? I'd imagine it'd look like T1-T2-Intertialforce=torque ?

I'm close. I got some other formulas floating in my head but I'm not 100% sure. Also tired.

m1*g-T1=m1a
m2*g-T2=m2a

I'm just going to give you a hint first. T1 and T2 are both the tension in the same rope, right?

Have you tried using the syrup conjecture?

Yes but they should be different tensions.

It says I should use energy methods, so I'm guessing I don't even need to use the tensions... ? You can give me a bit more of a hint than that.

You may want to double check this conjecture.

I'm losing credit for this homework being late (meh).

K.

Since it asks me to use energy methods. I'd assume that means that I'd describe the energy of the left block as mgh, the energy of the right block would be herp derp edit ...and the pulley... How does one describe the energy "lost" through moving the pulley though? Or the energy lost through having to lift the right block? Would it just be like mah (mgh with acceleration of the system instead of g?) I mean obviously it'd be like

Energies:
Lblock-pulley-rblock=lblockend

Fuck Masteringphysics, stupid website.

Reasons why throwing formulas at problems in hoping they stick is a bad strategy.

"The energy method," clearly means using potential/kinetic energies.

Okay? My question is how do I account to energy lost spinning the wheel.

.....

Post has been edited 1 time(s), last time on Apr 1 2013, 7:32 pm by Sacrieur.

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Post has been edited 1 time(s), last time on Apr 1 2013, 7:32 pm by Sacrieur.

Sacrieur, your answer is both not helpful due to lack of explanation, and also wrong. Calling him a numbskull when you posted wrong to begin with isn't very nice.

So idk if you still are working on this or not, but what I was driving at in my hint was T1 = T2. Since its the same rope connected to both boxes, the tension is constant throughout the entire rope. You can't pull one end of a rope without pulling on the whole thing (unless it isn't taut, but that's not the case here), so the tension in the rope is the same at both ends.

Since we're assuming the pulley has mass, or inertia, that also has to factor into the problem. We know all parts of the system along the rope will move in unison, so we have for the wheel:
Torque = I*alpha, where I is inertia and alpha is the rotational acceleration.

alpha = r*a, where r is the radius of the wheel and a is the acceleration of the two blocks, which must be equal since the system moves in unison with the wheel (no slip condition). The torque is going to be equal to the residual force that causes the acceleration of the system times the radius.

Since the rope does not slip, there must be a force of static friction. If we "unbend" the system, basically pretending its one long chain instead of bent due to gravity, we end with:
(M1)g - T - F + T - (M2)g = (M1 + M2)a
In this equation the inertia of the pulley is represented by the static friction force it will cause on the rope. Since the tensions are equal, they cancel each other out.

We follow up with this by noting that the frictional force by the pulley on the rope is the same as the torque by the rope on the pulley, in the opposite direction:
Torque = I*alpha, F = I*r*a
Solving for F from earlier:
F = (M1 - M2)g - (M1 + M2)a = I*r*a
Rearrange for a:
(M1 - M2)g / [ I*r + M1 + M2] = a


Use the acceleration in your basic distance formula to get speed
X = V_0*t + 1/2at^2
V_0 = 0
5 = .5*a*t^2
sqrt(10/a) = t
V_f = V_0 + at
V_f = 0 + sqrt(10a) and thats the answer.

By energy methods:
All the potential energy is going to be converted into kinetic energy. Some will go into the two boxes and some into the wheel:
(M1)gh - (M2)gh = .5(M1+M2)(V^2) + .5Iw^2
we know w in terms of v because the rope doesn't slip: V = w*r, w = V/r
(M1)gh - (M2)gh = .5(M1+M2)(V^2) + .5I(V/r)^2
10g(M1-M2) / [ M1 + M2 + I/r^2] = V^2
Plug in the numbers and take the square root, much simpler by energy methods but I figured I'd go both ways to hopefully help you understand.

Energy is never "lost" in these problems, just transfered from one state to another. In this case, you would need to know that the energy of a rotating body is (1/2)*Inertia*Rotational_Velocity^2, where Rotational_Velocity is measured in Radians/second. In the instant before the box hits the floor, all the potential energy of the two boxes has been transferred into kinetic energy. The inertia is measured around the axis of rotation, its pretty obvious for a wheel where that is but if you had a generalized body it would typically rotate about its center of mass, or for weird problems it could be elsewhere.

Post has been edited 7 time(s), last time on Apr 1 2013, 5:28 am by Vrael.