2
$\begingroup$

I am trying to model a mass being driven by a motor (Angular acceleration $\dot{\omega}$ and shaft-polar modulus $J$) with belts and pulleys. This is then being compared to numerical simulations. I find a 20-30% difference in results.

The system is so:

enter image description here

Here, $\tau$ is the torque provided by the motor whose shaft has moment of inertia $J$. Torque may be calculated as $\tau = J \dot{\omega}$ where the pulley (or sheave) has a diameter of $b$ and hence a radius of $b/2$. The mass being driven is $m$. The effect of gravity and any slack in the belt is neglected. A viscous force acts with a coefficient of $\nu$ (N/m/s).

The belt is actually 3 x ropes. However, this is not showing up in my force balance and I am asking what went wrong.

The horizontal direction is the $x$ direction and the corresponding displacement, velocity ($v$) and acceleration ($a$) of the mass are $x, \dot{x}, \ddot{x}$ respectively.

Force balance, where tension in single rope is given by $T$:

$$F = m\,a$$ $$3T - \nu \, v = m\,a$$ $$3\frac{\tau}{b/2} = m \ddot{x} + \nu \dot{x} $$

$\because$ the angular acceleration may be given as $\dot{\omega} = a/(b/2) = \ddot{x}/(b/2)$:

$$ 3\frac{\tau}{b/2} = 3 \frac{J \dot{\omega}}{b/2} = 3 \frac{J \ddot{x}}{b^2/4}$$

$$\therefore m \ddot{x} + \nu \dot{x} = 3 \frac{J \ddot{x}}{b^2/4}$$

or

$$ \frac{m}{3}\ddot{x} + \frac{\nu}{3} \dot{x} = \frac{4 J\ddot{x}}{b^2}$$

$$ \left[\frac{m}{3} - \frac{4 J}{b^2}\right] \ddot{x} + \frac{\nu}{3} \dot{x} = 0$$

The initial conditions for this differential equation are for position and velocity:

$x(0) = 0, \dot{x}(0) = \omega b/2$

For the following physical parameters: $m=2000$kg, $\nu=1000$N/m/s, $\omega = 2$ rev per sec, $b=0.1$m, $J=1$kg/sq.m, I use Mathematica to solve the 2nd order ODE and plot the position wrt time.

**My model**

Simulations run by a proprietary software however, returns the following response:

Proprietary model

I can have my model get about 1-2% close to the proprietary model, through the following ODE

$$ \frac{1}{3}\left[m - \frac{4 J}{b^2}\right] \ddot{x} + \nu \dot{x} = 0$$

This ODE is different from what I derived from a force balance. What gives? What went wrong?

For those interested, Mathematica code to solve my model:

Clear[m, Ir, b, \[Nu], s, t, \[Omega]];
m = 2000.;(*mass in kg*)

Ir = 1.;(*moment of inertia of shaft in kg-m^2*)

b = 100*10^-3 (*sheave diameter in meter*);
\[Nu] = 1000.;(*Viscous drag in N/m/s*)
\[Omega] = 
  N[120/60];(*Rev per second of shaft in 1/s*)

\!\(TraditionalForm\`sVal = DSolveValue[{\((m/3 - 
\*FractionBox[\(4\ Ir\), \(\(b\)\(\ \)\(b\)\(\ \)\)])\)\ \(\*
SuperscriptBox["s", "\[Prime]\[Prime]",
MultilineFunction->None](t)\) + \[Nu]/3\ \(\*
SuperscriptBox["s", "\[Prime]",
MultilineFunction->None](t)\) == 0, s(0) == 0, \*
SuperscriptBox["s", "\[Prime]",
MultilineFunction->None](0) == 
\*FractionBox[\(b\ \[Omega]\), \(2\)]}, s(t), t]\) // Expand
(*
sVal=DSolveValue[{(1/3)(m-(4 Ir)/(b b )) s^\[Prime]\[Prime](t)+\
\[Nu] s^\[Prime](t)\[LongEqual]0,s(0)\[LongEqual]0,s^\[Prime](0)\
\[LongEqual](b \[Omega])/2},s(t),t]//Expand *)

Plot[sVal, {t, 0, 20}, PlotRange -> {{0, 20}, {0, 0.25}}, 
 ImageSize -> Medium, PlotStyle -> {Thick, Black}, 
 BaseStyle -> {FontSize -> 15}, Frame -> True, 
 FrameLabel -> {"Time,t", "Position, x(t)"}, GridLines -> All]
$\endgroup$
  • $\begingroup$ From the figure one can see that $\omega$ is closer to 0.2 rev/s than 2 rev/s. Is there a typo? $\endgroup$ – user26872 Jul 19 '17 at 22:28
  • $\begingroup$ No it should be 2rps. Which figure suggests 0.2? $\endgroup$ – dearN Jul 20 '17 at 1:35
  • $\begingroup$ From the second figure, $v_0$ is roughly $0.03/0.5$. Since $b = 0.1$, $\omega_0$ is roughly $(0.03/0.5)/(0.1/2) = 1.2$. But $2\cdot 2\pi\simeq 12.6$, not $1.2$. $\endgroup$ – user26872 Jul 20 '17 at 2:00
  • $\begingroup$ I am using the fact that $\omega_0 \simeq \Delta\theta/\Delta t \simeq (\Delta x/\Delta t)/r$, where $r$ is the radius of the pulley. $\endgroup$ – user26872 Jul 20 '17 at 2:20
0
$\begingroup$

Please note that I may have totally misinterpreted your problem?

I was surprised that as a result of your analysis the displacement of the mass tends to a constant value.
I would have expected that the velocity of the mass should have tended to a constant value when the net force $(3T - \nu \dot x)$ on the mass became zero cf terminal velocity of a falling mass in air.

Another surprise was that your differential equation did not include the driving torque $\tau$.

Is there not more than one torque acting on the driving pulley?

Torque may be calculated as $\tau = J \dot{\omega}$

should be $\tau - 3T \, \frac b 2=J\, \dot \omega$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.