# How do I calculate the acceleration of the C.O.M. of a pulley system?

1. Oct 23, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass $m_1=\frac{3}{5}kg$, and its center is initially at xy coordinates $(-\frac{1}{2}m,0)$; the block has mass $m_2=\frac{2}{5}kg$, and its center is initially at xy coordinates $(0,-\frac{1}{10}m)$. The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart-block system? (b) What is the velocity of the com as a function of time $t$?

2. Relevant equations
$m_1=\frac{3}{5}kg$
$s_1=<-\frac{1}{2}m,0>$
$m_2=\frac{2}{5}kg$
$s_2=<0,-\frac{1}{10}m>$

$s_{com}=<\frac{Σmx}{Σm},\frac{Σmy}{Σm}>$
$F_{pulley-system}=(m_2-m_1)g=(m_1+m_2)a$
$a_{pulley-system}=\frac{(m_2-m_1)g}{(m_1+m_2)}$

3. The attempt at a solution
$x_{com}=\frac{(\frac{3}{5}kg)(-\frac{1}{2}m)}{1kg}=-\frac{3}{10}m$
$y_{com}=\frac{(\frac{2}{5}kg)(-\frac{1}{10}m)}{1kg}=-\frac{1}{25}m$

$s_{com}=<-\frac{3}{10}m,-\frac{1}{25}m>$
$a_y=(g)(\frac{m_2-m_1}{m_2})=(\frac{5}{2}kg^{-1})(\frac{49}{5}\frac{m}{s^2})(\frac{2}{5}kg-\frac{3}{5}kg)=-\frac{49}{10}\frac{m}{s^2}$
$a_x=(g)(\frac{m_1-m_2}{m_1})=(\frac{5}{3}kg^{-1})(\frac{49}{5}\frac{m}{s^2})(\frac{3}{5}kg-\frac{2}{5}kg)=\frac{49}{15}\frac{m}{s^2}$

I don't really know how to do this problem properly.

Last edited: Oct 23, 2016
2. Oct 23, 2016

### Simon Bridge

Explain how you are reasoning your way through this.
(Note: "m" stands for "meters" so it should not appear in your equations.)

If you have the position of each mass at time t, can you find the coordinate of the center of mass at time t?
OR you can look for a lecture on how to find acceleration of the center of mass from the accelerations of the individual particles... ie.

Last edited: Oct 23, 2016
3. Oct 24, 2016

### Eclair_de_XII

Let's see. Let me redo this... I don't think I had the right idea before about the gravitational forces of the masses cancelling each other; it would only apply if they were both suspended from ledges on a pulley system.

$x_{com}(t)=\frac{3}{5}(\frac{1}{2}+\frac{1}{2}gt^2)$
$y_{com}(t)=-\frac{2}{5}(\frac{1}{10}+\frac{1}{2}gt^2)$

So: $s_{com}(t)=\langle\frac{3}{10}(1+gt^2),-\frac{2}{50}(1+5gt^2)\rangle$? I was a bit ambivalent about using the calculated com's as the $x_0$ and the $y_0$, but I figured it would make more sense for the position of the com to vary as time passes, instead of $com_0$ being a constant.

4. Oct 24, 2016

### kuruman

You are assuming that the acceleration is g. This is not correct because it says that the masses are in free fall when they are not. You need to find the common acceleration of the masses.

5. Oct 25, 2016

### Eclair_de_XII

So I figure it's something like this: (1) gravity acts on the suspended mass, but not on the cart; (2) the tension of the spring transmits a force from the cart to oppose the gravity of the suspended mass; (3) the resulting acceleration is the difference in the forces divided by both masses. Even if I assume that the opposing force is the normal force, I still end up with the acceleration proposed in the first post, assuming that the normal force has the same acceleration coefficient as the gravitational force: $a_{pulley−system}=(g)(\frac{m_2−m_1}{m_1+m_2})$. I'm missing something here...

6. Oct 25, 2016

### haruspex

This is counting the inertia of the cart twice, once in subtracting it from the gravitational force, and again in dividing by the sum of masses.
There is one driving force, one acceleration, two masses.

7. Oct 25, 2016

### Eclair_de_XII

Okay, so I know there's gravity that's acting on $m_2$ and indirectly on $m_1$ because the tension in the string transmits it. The force on $m_2$ must pull not only itself, but $m_1$ as well. If $m_1$ were suspended (which it's not), only then would the net force be $F=(m_1-m_2)g$... But I have no idea what the case would be if it were just on solid ground like the problem describes.

8. Oct 25, 2016

### haruspex

There is a standard way to approach statics and kinetics problems. Trying to shortcut that is leading you astray.
Draw a free body diagram for each rigid body separately. Consider the forces and accelerations of each, and write down the corresponding equations. Solve.
What are the forces acting on the suspended body? Invent variables for unknowns as necessary.
Do the same for the cart. Since there is no friction, you can omit the vertical forces on that; the gravitational dorce will just cancel the normal force.

9. Oct 26, 2016

### Eclair_de_XII

Okay, I got it. It's on an air track, so there's no friction; and normal force is always perpendicular to the motion, so it doesn't affect the horizontal motion of $m_1$, and in turn, the vertical motion of $m_2$. It was so simple. The gravity of $m_2$ provides all the force for the acceleration.

$m_1$

$m_2$

$F=m_2g=(m_1+m_2)a$
$a=(g)(\frac{m_2}{m_1+m_2})=(\frac{49}{5}\frac{m}{s^2})(\frac{2}{5}kg)(1kg^{-1})=\frac{98}{25}\frac{m}{s^2}$

$s_{com}(t)=\langle\frac{3}{10}(1+at^2),-\frac{2}{50}(1+5at^2)\rangle$
$v_{com}(t)=\langle\frac{3}{5}(at),-\frac{2}{5}(at)\rangle$
$a_{com}(t)=\langle\frac{3}{5}a,-\frac{2}{5}a\rangle=\langle\frac{294}{125},-\frac{196}{125}\rangle\frac{m}{s^2}$

Thanks, everyone.