## Section 7.5 Center of Mass Motion

### Subsection 7.5.1 Center of Mass Velocity, Acceleration, and Momentum

Center of mass of an object provides an important way to follow the overall motion of the object. We have already seen the formula for the position of the center of mass. For a body that is a collection of discrete masses \(m_1\) at \(\vec r_1\text{,}\) \(m_2\) at \(\vec r_2\text{,}\) \(\cdots\text{,}\) \(m_N\) at \(\vec r_N\text{,}\) the center of mass will be at \(\vec R_\text{CM}\text{,}\) given by

where total mass \(M=m_1+m_2+\cdots+m_N\text{.}\) Now, studying the rate at which position \(\vec R_\text{CM}\) changes will give us the velocity of center of mass, \(\vec V_{\text{CM}}\text{.}\)

Similarly, when we look at the rate which \(\vec V_{\text{CM}}\) changes, we get the acceleration of the center of mass, \(\vec A_{\text{CM}}\text{.}\)

The total momentum of all particles will be

Now, this is just \(M \vec V_{\text{CM}}\) as you can see from Eq. (7.5.2) when you multiply it by \(M\text{.}\)

We call the product \(M \vec V_{\text{CM}}\) the center of mass momentum, \(\vec P_{\text{CM}}\text{,}\) which is actually total momentum of all particles in the body.

### Subsection 7.5.2 Center of Mass Motion - Constant Mass Case

Let us consider a system with \(N \) parts of masses \(m_1 \text{,}\) \(m_2 \text{,}\) \(\cdots \text{,}\) \(m_N \text{.}\) Let \(\vec F_1 \text{,}\) \(\vec F_2 \text{,}\) \(\cdots \text{,}\) \(\vec F_N \) be forces acting on those parts. For the sake of simplicity, let us look at the constant mass system, i.e., where masss of each part is fixed in time.

Now, we write Newton's equation of motion for each part for constant mass.

Summing them, we find that the left side can be written as sum over all the forces from objects outside the \(N \) parts making up the system, which we call the *external forces*, and a sum over all the forces that one part applies on another part.

By Newton's third law, internal forces, say by part 1 on 2 and by part 2 on 1 come in pairs whose sum is zero. Therefore,

That gives us the following simplified equation after summing the equations of all \(N \) parts.

whose right side is just the total mass \(M \) times the center-of-mass acceleration \(\vec A_{\text{CM}} \text{.}\)

It is interesting that the overall motion of a malti-part system can be captured essentially by the motion of a single (fictitious) particle of total mass placed at the special point of center of mass. This is the equation we solve whenever we study the overall translation motion of finite-size objects since they are inherently mutiparticle systems.

*General Case (Calculus):*

We have seen before that the general case of mass varying as well as constant mass systems is given in terms of rate of change of momentum.

Now, if you apply this to the \(N\)-particle system, you will get

Similarly, the integrated form in terms of impulse and change in momentum will be

### Subsection 7.5.3 (Calculus) Center of Mass Velocity, Acceleration, and Momentum

(You can safely skip this section)

To study the motion of the center of mass (CM) of a multi-part system, we need to define velocity and acceleration of the CM. We use Calculus to do that. Recall that velocity is defined as the time derivative of position and acceleration the time derivative of velocity. We do the same with CM-position in Eq. (7.5.1).

where \(\vec v_1 \text{,}\) \(\vec v_2 \text{,}\) \(\cdots \text{,}\) \(\vec v_N \text{,}\) are the velocities of the individual parts, or more precisely, the velocities of the centers of masses of the parts. Therefore, we find that the total momentum, i.e., the sum of the momenta of parts, is equal to the total mass times the momentum of the center of mass.

We call the \(M\vec V_{\text{CM}} \) the center-of-mass momentum, \(\vec P_{\text{CM}}\text{.}\) Writing momentum of the parts as \(\vec p_1\text{,}\) \(\vec p_2\text{,}\) etc, we get

Taking a derivative of \(\vec V_{\text{CM}}\) with respect to \(t \) will give the acceleration of the CM of a system whose mass is constant in time,

where \(\vec a_1 \text{,}\) \(\vec a_2 \text{,}\) \(\cdots \text{,}\) \(\vec a_N \text{,}\) are the accelerations of the individual parts.

### Subsection 7.5.4 (Calculus) Center of Mass Motion - General Case

(You can safely skip this section)

For the general case of \(N \) part system, we will have to start with equations of motion of individual parts in the momentum form.

Now, summing them, we get the left side to equal the \(F_{\text{net}}^{\text{ext}} \) as above, and the right side equals the rate of change of total momentum.

But by Eq. (7.5.10), we can replace the sum of momenta of the parts by the center-of-mass momentum, or the total momentum.

where \(\vec P_{\text{CM}} = M \vec V_{\text{CM}}\text{.}\) This equation takes care of even those situations in which the total mass of the system is changing, regardless of which part that may be gaining or losing mass. Of course, if a system is not gaining or losing mass, we would use much simpler analysis based on Eq. (7.5.5).

Eq. (7.5.13) can be integrated to yield the following important relation between the change in total momentum and impulse by external forces.

where

###### Example 7.5.1. CM of an exploding shell.

As an example consider the motion of pieces of an exploded shell in free-fall.

Although the pieces fly away in various directions, the CM falls as if no explosion had taken place since during the explosion all forces are internal to the system. The motion of CM is that of a single particle with total mass of the system. The fictitious particle at the CM falls freely with acceleration equal to g, the acceleration due to gravity, *until one of the pieces hits the ground.*

###### Checkpoint 7.5.2. Speed When Jumping Off Vertically.

Suppose you are trying to practice a vertical jump. You crouch down and push against the ground.

The magnitude of the normal force from the ground on you varies with time as shown in Figure 7.5.3 of \(F\) versus \(t\text{.}\) The initial normal force is just your weight. As you push down, the normal force increases. You leave the ground at the point normal force is zero.

As a net result of normal force from the ground, and weight, your center of mass has acceleration upward. What will be your speed at the instant you are no longer in contact with the ground?

Add the area under the curve to get impulse from the normal force. Then, subtract the impulse from weight since it is in opposite direction. Use the net impulse to deduce the change in momentum of the center of mass.

\(1.47\text{ m/s}.\)

Let positive \(y\) axis be pointed up. We will implement the following law for the center of mass since CM momentum is equal to \(p_\text{tot}\text{.}\)

To get the net impulse, we need to vectorially add up impulses from all forces on the body. We have two forces on the body here - the weight and the normal force. From \(t=0\) point in Figure 7.5.3, we note that magnitude of weight is \(500\text{ N}\text{.}\) Since weight has negative \(y\) component, its \(J_y\) will be negative.

The impulse from the normal force is the area under the given plot.

which is obtained by slicing the area into triangles and rectangle. Therefore,

The mass of the person is obtained from the weight.

Initially, the momentum was zero. Therefore, the velocity at the end is

###### Checkpoint 7.5.4. Impulse for a Vertical Jump.

A \(78.0\text{-kg}\) athlete jumps straight up so that his CM moves by a height of \(1.2\text{ m}\) after he leaves the ground. Find the magnitude of impulse from the ground.

Impulse will equal \(mv_0\) where \(v_0\) is the velocity of CM just before the athlete leaves ground.

\(378.3\text{ N.s}\text{.}\)

The jump is due to push by the athlete on the ground which causes a normal force that exceeds his weight and gives upward momentum when he leaves the ground. This force from the ground during the jumping process is responsible for imparting initial upward momentum.

Let \(v_0\) be the velocity pointed upwards at the instant athlete leaves the ground and \(m\) mass of the athlete. Then, the change in momentum from zero to \(mv_0\) will equal the the impulse by ground. Let us work with \(y\) axis pointed up. Then, \(y\)-component of impulse will be

The CM starts with velocity \(v_0\) and has constant acceleration under the external force of gravity so that its velocity is zero at maximum height \(h\text{.}\) Using constant acceleration of the motion of CM along \(y\) axis we get

Hence, impulse by ground is

###### Checkpoint 7.5.5. The Motion of the Center of Mass of a Diver.

A diver jumps off a board with initial speed \(12\text{ m/s}\) at an angle \(60^\circ\) with the horizontal direction. She falls a total height of \(8\text{ m}\) before entering water. Find the speed of her center of mass when she enters water?

The CM moves as if it were a particle in free fall.

\(17.4\text{ m/s}\text{.}\)

Since the only force on her body after she takes off is her weight. The CM will obey the free fall equations. Therefore, we start with components of her initial velocity.

The final \(v_x\) is same as the initial, and the final \(v_y\) is obtained from

where \(\Delta y = - 8\text{ m}\text{,}\) which is negative since the final location is below the initial point, and \(v_{fy}\) is negative since it is pointed down while positive \(y\) axis is pointed up.

From \(v_{fx}\) and \(v_{fy}\) we get the speed just before she enters water.

###### Checkpoint 7.5.6. Motion of Center of Mass of Two Blocks Attached by a Spring.

Two blocks of mass \(m\) each is attached by a spring of spring constant \(k\) and unstretched length \(l_0\text{.}\) The blocks are placed on a horizontal frrictionless track.

While holding one of the blocks, say B in the figure, the other block is pushed in so that spring compresses by \(D\text{,}\) and then everything is released from rest at \(t=0\text{.}\) Use \(x\)-axis so that at \(t=0\text{,}\) \(x_A=D\) and \(x_B=l_0\text{.}\)

Find the position of the center of mass, \(x_\text{cm}\text{,}\) at an arbitrary instant \(t\text{.}\)

Since there is no external force, \(x_\text{cm}\) will remain fixed in time. You can show this by a detailed calculation.

\(x_\text{cm} (t)= \frac{D+l_0}{2}\text{.}\)

Since there is no external force, \(x_\text{cm}\) will remain fixed in time. Since the center of mass is at \(x=(D+l_0)/2\) at \(t=0\text{,}\) it will remain there. This does not mean A and B do not move - they do, but their CM remains at the same point in space. We can also show this by a detailed calculation as follows.

Figure 7.5.8 shows situation at an arbitrary instant when the spring is stretched. The spring forces on the two blocks are shown.

The \(x\)-equations of motion for the two blocks will be

Here \(m_A=m_B=m\) and \(x_B \gt x_A\text{.}\) Since center of mass is give by

we can take two derivatives with respect to time to get

Using the equations of motions for \(x_A\) and \(x_B\text{,}\) we immediately get

This has a simple solution.

where \(a\) and \(b\) will be fixed from the initial conditions.

Writing Eq. (7.5.18) in terms of \(x_A\) and \(x_B\) and also writing their derivative we have

Evaluating these at \(t=0\) we get

Therefore, we have the following answer for the center of mass at arbitrary instant \(t\text{.}\)

That is, CM remains fixed at \(x= (D+l_0)/2\text{.}\)

###### Checkpoint 7.5.9. (Calculus) Motion of Center of Mass of Two Blocks Attached by a Spring with One Block Piushed Against a Wall.

In Checkpoint 7.5.6, the two blocks moved freely on the frictionless horizontal track.

Now, suppose you were pushing block A, you kept block B against a wall as shown in Figure 7.5.10.

After compressing the spring by amount \(D\text{,}\) you release both blocks at rest. What will be the motion of the center of mass if \(l_0\) is the natural length of the spring?

First solve the situation when block B is in contact with the wall. Then solve the situation after B loses contact with the wall. When in contact, set up equation of motion for the CM and another equation of motion for B. Also, use the relation between \(x_A\) and \(x_\text{cm}\text{.}\)

Overall strategy:

The motion of the center of mass will be controlled by external forces on the two-block (and spring) system. When block B is in contact with the wall, the net external force on the two-block system comes from the normal force of the wall, which will vary with the varying length of the spring. When block B is no longer in contact with the wall, the normal force from the wall will be zero, which will make net external force zero on the two block system. Thus, after block B is no longer in contact with the wall, the center of mass will just move with a constant velocity.

*Block B in Contact with the Wall:*

Figure 7.5.11 shows situation at an arbitrary instant when block B is still in contact with the wall. The equation of motion of the center of mass will be

We need an expression for \(N(t)\) to solve this equation. We get that by looking at force on B as shown in Figure 7.5.12. Balancing spring and wall forces on B gives

We also note that when B is at the wall \(x_\text{cm}\) and \(x_A\) are related by

Using the three equations above, we get the following equation of motion for \(x_\text{cm}\text{.}\)

This has the following general solution.

The velocity of CM will be

Using the initial conditions on A and B, we have the following intial conditions on CM.

Using these in Eqs. (7.5.20) and (7.5.21), we find

Therefore, the position of CM and its velocity are

*Block B not in Contact with the Wall:*

At the instant, block B loses contact with the wall, the CM will be at \(l_0/2\text{.}\) At that instant onward the velocity of the CM will remain unchanged since there are net external force on the two-block system is zero now. We can find the time when this happens from Eq. (7.5.22). Let this instant be at \(t=T\text{.}\)

Therefore,

From Eq. (7.5.23), the velocity of CM after this instant will be constat at

This expression can be simplified by noting that

To prove this try \(y=\cosh^{-1}(x)\) and solve for \(e^y = x + \sqrt{x^2-1}\) and \(e^{-y} = x - \sqrt{x^2-1}\text{.}\) Using this, we simplify CM velocity for \(t\gt T\) to be

The velocity is \(x\)-component and pointed towards negative \(x\)-axis. That is why there is a negative sign in the front.