dynamics help!! Please!!!! I beg you!

[Deathslayer7]

http://i198.photobucket.com/albums/aa125/death_slayer7/dynamics1.png

The assembly consists of an 8-kg disk and a 10-kg bar which is pin connected to the disk. (a) If the system is released from rest, determine the angular acceleration of the disk. The coefficients of static and kinetic friction between the disk and the inclined plane are μs = 0.6 and μk = 0.4, respectively. Neglect friction at B.
(a) Model the complete system and solve for all unknown variables in Mathcad.
(b) Determine and plot the angular velocity, φ-dot of the disk as a function of time t in the range 0 < t < 2 seconds.


can anyone help me here. I did a Free Body diagram but i have no idea if its right and if it is what the constraint equations would be.

For forces
i got the tangential to the left at C.
the normal upwards at C
weight downwards at A
Friction up at 30 degrees to the left
Normal of C perpendicular to the plane.
normal of B perpendicular to the plane????? (not sure if this matters)
and since the mass of the bar isnt given no weight for it.

from there you use summation of tangential, normal, and the moment about B or C (idk which one).

so any help would be appreciated.

[mucat]

Why do you say the mass of the bar isn't given? You said at the beginning that it was an 8-kg disc and a 10-kg bar.

Here are some ideas, but bear in mind that I'm taking a physicist's approach rather than an engineer's. The substance is the same -- Newton's laws are Newton's laws, regardless of who uses 'em -- but if you are an engineer, we might tend to use the words a little differently.

Throughout this problem, it's probably easiest to use a tilted coordinate system, where x points tangent to the slope, and y points perpendicular to the slope slope.

I would start with a free-body diagram for the bar itself, as a separate object from the disc; that will be simpler, since you said there's no friction at point B.

So the relevant forces are a normal force at B, the bar's weight applied at its center of mass, and a force exerted by the axle of the disc at point A.

So you have four unknowns: the y-component of the normal force (its x component is zero in this coordinate system), the x and y components of the axle force, and the x-component of the acceleration (its y-component is zero.) The weight of the bar has both an x and a y component, but you can easily find both of those.

You can set up three equations:
The x components of the forces add to mass of bar times acceleration
The y components of the forces add to zero.
The torques around the bar's center of mass add up to zero.

(Careful when figuring the moment arm of the axle contact forces. If you're tempted to take the torques around the point A instead of the bar's CM to eliminate this hassle, remember that the bar is accelerating; that makes it tricky to calculate torques about any point other than the CM.)

You now have three equations and four unknowns; can't solve the system yet. So we move on to the disc, and do a second free-body diagram. The relevant forces are:

Weight of the disc, applied at its CM.
The axle contact force from the bar; same force you were using in the bar's equations, but with the direction flipped.
A normal force from the ground at point C. (Not the same as the normal force on the bar at point B.)
Static friction (presumably, unless it slips) tangent to the slope at point C.

The new unknowns are the y-component of this normal force (again, the x-component is zero), and the x-component of static friction (and no, it's not equal to mu times the normal force; that would be the maximum possible static friction.)

The disc's acceleration is the same as the bar's, and its angular acceleration alpha is equal to its linear acceleration divided by its radius.

Again you have three equations:

The x components of the forces add to mass of disc times accel.
The y components of the forces add to zero.
The torques around the disc's center of mass add up to its moment of inertia (which you can calculate) times alpha, which in turn is (a / R).

So that's six equations, and six unknowns (the unknowns are the two normal forces, the two components of the contact force at the axle, the magnitude of the static fricion force, and the acceleration. Alpha is not a separate unknown, as discussed above.)

So you pound out the system of equations, and declare victory.

I wish I could suggest a less algebraically tedious last step. At least they're linear equations, so you can shove them into a giant matrix and let matlab do the ugly part for you.

Good luck!

EDIT: Once you find all the forces, make sure that the force of friction turns out less than or equal to mu-static times the normal force at C. If it doesn't, then the disc slips, and you have to modify your equations to use kinetic friction instead...I don't think that will happen here, though.

[Deathslayer7]

all right must have just read over the mass of the bar. and thank you. I'll try to use what you said,

[Deathslayer7]

i only seem to have 5 unknowns, 6 equations.

Force of the axle (i split it into F_a*cos/sin(beta) for its x and y components.
accelreation in the x direction.
Normal of C
mu static in the x direction
Normal of B

[Xyk]

I thought from the title you meant music-related dynamics.

f means loud
mf means kinda loud
mp means kinda quiet
p means quiet

adding more f's makes it more loud
adding more p's makes it more quiet.

There ya go! :smallbiggrin:

[mucat]

i only seem to have 5 unknowns, 6 equations.

Force of the axle (i split it into F_a*cos/sin(beta) for its x and y components.
accelreation in the x direction.
Normal of C
mu static in the x direction
Normal of B

Count 'em again. That's six unknowns!

(Remember, the axle force counts as two. You can either think of the x and y components -- which is probably the better option for this problem -- or of the force's magnitude and its angle...either way, there are two things about that force you don't know.)