Shower Rod IV? - circular shower rod
A thin rod (mass m, length L) has u0026lt a solid round cross section with radius R (\\ \\ \\ \\ \\ \\ \\ \\ u0026lt \\ \\ \\ \\ \\ \\ \\ \\, L). It is pivotally mounted at one end and horizontally by a rope. The string is sugarcane, which is on the support pin in a vertical plane.
A small drop \\ \\ \\ \\ \\ \\ \\ \\ u0026lt \\ \\ \\ \\ \\ \\ \\ \\ u0026lt, L and mass \\ \\ \\ \\ \\ \\ \\ \\ u0026lt \\ \\ \\ \\ \\ \\ \\ \\ u0026lt, M is free to slide above and below the counter. The coefficient of friction between the heel and the bar is moved. How do you cut the rope, the account is still on the spit. How long and up to the end of the bar?
I had not worked, so that the equations of motion can be uncomfortable and require a numerical solution. If yes, please give only the equations of motion.
And remember people, do you think) please our public and private disputes (or block trolls.
Friday, January 8, 2010
Circular Shower Rod Shower Rod IV?
Subscribe to:
Post Comments (Atom)
2 comments:
I'll give it a shot.
Notes: denotes Theta - steel angle and r = distance account from the edge of the bar.
Forces at the bar:
m * g * g * mass, mass mu * g * cos (theta)
Forces on the account:
Mass * g, n, mu * mass * g * cos (theta), [centrifugal force]
Moments in the bar:
m * g * L * cos (theta) + mass * g * r * cos (theta)
Moment of inertia:
L * m ^ 2 / 3 + 2 mass * r ^
By u0026lt u0026lt \\ \\ \\ \\ \\ \\ \\ \\; mass m I will simplify the first easy to forget the effect of the law on the counter. The tribe is an independent clock which verified:
2/dt d ^ ^ 2 (theta) = 3 / 2 * g / L * cos (theta)
Theta (0) = (d / dt theta) (0) = 0
An EU EOM nonlinear, which is already a problem to solve, the answers fall known as elliptic functions. Suppose we know:) So now we have:
theta = theta (t), or at least without units theta = theta (t * sqrt (g / L)).
The permanent account first to the pin. It feels like a force to go: mass * g * sin (theta) on the other end of the rod. this FOrce be compensated by the static friction to:
Mass * g * sin (theta) mu_s = mass * g * cos (theta)
Ie, until:
theta_start = arctan (mu_s)
At this stage, the account that r = 0 After this point:
2/dt mass * d ^ 2 (r) = mass * g * sin (theta) - mu_dyn * mass * g * cos (theta) + mass * (d (theta) / dt) ^ 2 * r
2/dt d ^ ^ 2 (r) = g * (sin (theta)-mu_dyn * cos (theta)) + (d (theta) / dt) ^ 2 * r
Integrating this, we want to know when r = L. ..
R (0) = 0, dr / dr (0) = 0, theta_start theta (0) =.
In principle, it is now numerically soluble, but I think many more are, through research and analytical approximation of the elliptic functions. Subjects do not survive as the account to a full rotation of the rod must be responsible.
-------------------------------------- ...
2. Expt
From Wikipedia:
2/dt d ^ ^ 2 (theta) = g / L * sin (theta)
This is known as the Mathieu equation
(is the same equation we had, with theta> pi/2-theta g / L of 3 / 2 renormalized)
d / dt (theta) = gRT (2 * g / L * cos (theta))
[In our case, cos (theta (0)) = 0]
I thought in the same direction. But as I developed it, I write my solution.
We have already established relations of acceleration and angular velocity of the rod. It is assumed that the light of accounts HTE no influence on the kinematics of the bar.
3g/2L α = cosθ
ω ^ 2 = 3 g / L sinθ
The problem here is that when we welcome expressions for ω and α, we can take the form of time, a beautiful expression obtained for θ.
ω = dθ / dt = √ (3gsinθ / L)
dθ / √ (sinθ) = √ (3 g / L) dt
Therefore, the use of a numerical procedure T is find () θ.
We ω = Δθ / Dt
and Dt = Δθ / ω
Since θ = 0 can be increased in steps of Δθ = π/500 and can give us the time to θ = π.
While pleased with the kinematics of the bar. We have numerical values of θ, ω and α in terms of T.
***
Well, the kinetics of this account.
It is the weight of Mg, the normal reaction of the N-rail, parallel to the connecting rod bearings, μN.
Use describe cordinated r-θ, the trajectory of the account
A_R = r "(t) - rω ^ 2
a_θ = rα + 2 * r * ω
Mgsinθ - μN = m * A_R ... (1)
Mgcosθ - N = m * a_θ ... (2)
We can use equation (2) to find N to [θ (t)]
to find it with equation (1) r (t).
From the initial conditions
R (0) = 0, r '(0) = 0, N (0) = Mg
we go with r (t) with a numerical method (Euler found works well with Excel).
Some results for a simple case:
M = 1, G = 10, L = 10, μ = 0
http://i37.tinypic.com/k00kt2.jpg
In this case, the account reaches the end of the rod (R / L) = 1 if θ = ~ 115 degrees, and t = 1.72 sec
Post a Comment