## 9.1. Center of mass

**The motion of a rotating ax thrown between two jugglers looks rather complicated, and very different from the standard projectile motion discussed in Chapter 4. Experiments have shown that one point of the ax follows a trajectory described by the standard equations of motion of a projectile. This special point is called the center of mass of the ax.**

The position of the center of mass of a system of two particles with mass m

_{1}and m

_{2}, located at position x

_{1}and x

_{2}, respectively, is defined as

_{ }

Figure 9.1. Position of the center of mass in 1 dimension.

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{1}, m

_{2}, m

_{3}and m

_{4}, located at x

_{1}, x

_{2}, x

_{3}and x

_{4}, respectively. The position of the center of mass of m

_{1}and m

_{2}is given by

_{ }

_{3}and m

_{4}is given by

_{ }

_{ }

_{ }

_{1}and m

_{2}and of m

_{3}and m

_{4}we can express the center of mass of the whole system as follows

_{ }

Figure 9.2. Location of 4 masses.

This shows that the center of mass of a system can be calculated from
the position of the center of mass of all objects that make up the system. For
example, the position of the center of mass of a system consisting out of
several spheres can be calculated by assuming that the mass of each sphere is
concentrated in the center of that sphere (its center of mass).**Note**:

- The center of mass of an object always lies on a point/line/plane of symmetry (for homogeneous objects).
- The center of mass of an object does not need to lie within the body of that object (for example: the center of a donut is its center of mass even though there is no mass at that point).

**Sample Problem 9-3**

Figure 9.3a shows a circular metal plate of radius 2R from which a disk if radius R has been removed. Let us call it object X. Locate the center of mass of object X.

Symmetry arguments immediately tell us that the center of mass of object X is located on the x-axis. Suppose the hole in object X is filled with a disk of radius R. The new object (object C, Figure 9.3b) is symmetric around the origin of our coordinate system, and that point is therefore the center of mass of object C. However, object C consist out of object X and a disk with radius R centered on the x-axis at x = - R (this disk is called object D). The center of mass of this system (consisting out of object X and object D) can be easily calculated:

_{ }

_{ }

_{ }

_{ }

Figure 9.3. Sample Problem 9-3.

The position of the center of mass of object X is given by_{ }

**Example Problem 9-1**

Figure 9.4 shows a one-dimensional rod. The density of the rod is position dependent : [rho](x) = a - bx + cx

^{2}. Determine the location of the center-of-mass of the rod.

Figure 9.4. Position Dependent Density.

The mass of a fraction of the rod (length dx) is given by
dm = [rho](x) dx

The position of the center of mass of the rod can be determined as
follows_{ }

_{ }

_{ }

## 9.2. Motion of the Center of Mass

**The definition of the center of mass of a system of particles can be rewritten as**

_{ }

_{ }

_{cm}is the

**velocity of the center of mass**and v

_{i}is the velocity of mass m

_{i}. The acceleration of the center of mass can be obtained by once again differentiating this expression with respect to time

_{ }

_{cm}is the

**acceleration of the center of mass**and a

_{i}is the acceleration of mass m

_{i}. Using Newton's second law we can identify m

_{i}a

_{i}with the force acting on mass m

_{i}. This shows that

_{ }

**the motion of the center of mass is only determined by the external forces**. Forces exerted by one part of the system on other parts of the system are called internal forces. According to Newton's third law, the sum of all internal forces cancel out (for each interaction there are two forces acting on two parts: they are equal in magnitude but pointing in an opposite direction and cancel if we take the vector sum of all internal forces). See Figure 9.5.

Figure 9.5. Internal and External Forces acting on a System of
Particles.

The previous equations show that **the center of mass of a system of particles acts like a particle of mass M**, and reacts like a particle when the system is exposed to external forces. They also show that when the net external force acting on the system is zero, the velocity of the center of mass will be constant.

**Example Problem 9-2**

The center of mass of an exploding rocket will follow the trajectory of a projectile. although its individual pieces can follow a quit complex trajectory. The forces of the explosion are internal to the system, and the only external force acting on the system is the gravitational force.

**Example Problem 9-3**

A ball of mass m and radius R is placed inside a spherical shell of the same mass m and inner radius 2R (see Figure 9.6a). The ball is released and moves back and forth before coming to rest at the bottom of the shell (see Figure 9.6b). What is the displacement of the system ?

Figure 9.6. Example Problem 9-3.

The only external forces acting on the system are the gravitational force and
the normal force. Both act in the y-direction. The x-component of the total
external force acting on the system is zero. The x-component of the
acceleration of the center of mass is therefore equal to zero. The velocity of
the center of mass in the x-direction is initially equal to zero, and will
therefore remain zero. We conclude that the position of the center of mass
along the x-axis will not change. In the initial configuration (Figure 9.6a)
the x-position of the center of mass is given by_{ }

## 9.3. Linear Momentum

**The linear momentum p of an object with mass m and velocity v is defined as**

_{ }

Under certain circumstances the linear momentum of a system is conserved. The linear momentum of a particle is related to the net force acting on that object:

_{ }

The total linear momentum p of a system of particles is defined as the vector sum of the individual linear momenta

_{ }

_{ }

" The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass. "

If we differentiate linear momentum of the center-of-mass with respect to time we obtain

_{ }

_{ext}= 0 N), the linear momentum of the system is conserved.

**Example Problem 9-4**

A stream of bullets with mass m is fired horizontally with speed v into a large wooden block with mass M that is initially at rest on a horizontal table. If the block is free to slide across the table (without friction), what speed will it acquire after it has absorbed n bullets ?

Figure 9.7. Example Problem 9-4.

Consider the closed system shown in Figure 9.7. This is an isolated
system; no particles leave or enter the system. The rate of change of its
linear momentum is therefore equal to the net external force. In this system,
all external forces (normal and gravitational force) act in the y-direction,
and we can conclude that the linear momentum in the x-direction is conserved.
The system shown in Figure 9.7 consist initially out of n bullets, each moving
with speed v, and the wooden block which is at rest. The total linear momentum
in the x-direction is therefore_{ }

_{ }

_{ }

_{ }

**Note**: we did not have to consider what happened when the bullets hit the block since these forces are internal forces.

**Sample Problem 9-10**

Two blocks with mass m

_{1}and mass m

_{2}are connected by a spring and are free to slide on a frictionless horizontal surface. The blocks are pulled apart and then released from rest. What fraction of the total kinetic energy will each block have at any later time ?

Figure 9.8. Sample Problem 9-10.

Figure 9.8 shows a schematic of the system. The velocities of mass
m_{1}and mass m

_{2}are defined to be positive when they are directed towards the right in Figure 9.8 (in Figure 9.8 the velocity of m

_{2}is negative).

Consider the system consisting of the two masses and the spring. This is a closed system. The only external forces acting on the system are the gravitational force and the normal force. Both these forces are directed vertically. The net force along the x-axis is zero, and therefore, linear momentum is conserved along the x-axis.

Initially, both masses are at rest, and the total linear momentum along the x-axis is zero. Suppose at a later time mass m

_{1}has a velocity equal to v

_{1}and mass m

_{2}has a velocity equal to v

_{2}. The total linear momentum at that time is then given by

_{ }

_{f}must be equal to 0. The velocity v

_{2}of mass m

_{2}can now be expressed in terms of m

_{1}and v

_{1}:

_{ }

_{1}and of mass m

_{2}always have opposite sign. The kinetic energy of mass m

_{1}and mass m

_{2}can now be calculated

_{ }

_{ }

_{ }

_{1}is the fraction of the total kinetic energy that is carried by mass m

_{1}we obtain the following equation for f

_{1}:

_{ }

**Problem 43P**

A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of 30 m/s. The third piece has three times the mass of each of the other pieces. What is the direction and magnitude of its velocity immediately after the explosion ?

The vessel is an isolated system on which no external forces are acting. This implies that the total linear momentum of the system is conserved. Since the vessel is initially at rest, the initial linear momentum of the system is zero. Since the total linear momentum is conserved, the final linear momentum of the system must also be zero. Figure 9.9 shows schematically the direction of the three fragments in which the vessel explodes. The problem states that m

_{1}= m

_{2}and that m

_{3}= 3 m

_{1}. Assuming that the total mass of the system is conserved we conclude that

Figure 9.9. Problem 39P.

_{ }

_{ }

_{1}= v

_{2}= 30 m/s. Conservation of linear momentum along the x-axis and along the y-axis requires

_{ }

_{ }

_{ }

_{ }

tan ([theta]) = 1

or
[theta] = 45deg.

The velocity of the third fragment can now be obtained easily_{ }

**Problem 48P**

A 1400 kg cannon, which fires a 70 kg shell with a muzzle speed of 556 m/s, is set at an elevation of 39deg. above the horizontal. The cannon is mounted on frictionless rails, so that it recoils freely. (a) What is the speed of the shell with respect to the earth ? (b) At what angle with the ground is the shell projected ?

The mass of the cannon is M, and the mass of the shell is m. The firing angle is [theta] and the muzzle speed is v

_{0}. The velocity of the cannon and the shell in with respect to the earth is v

_{c}and v

_{s}, respectively. The angle of projection of the shell with respect to the earth is a.

The external forces acting on the shell and cannon are the gravitational force and the normal force. These forces are directed along the y-axis. Since there is no external force acting on the shell along the x-axis, the linear momentum of the system along the x-axis is conserved. The total linear momentum of the system along the x-axis (the horizontal axis) is given by (see figure 9.10)

Figure 9.10. Velocity diagram of shell and canon.

_{ }

_{ }

Figure 9.11. Velocity diagram of shell.

_{ }

_{ }

_{c}

_{ }

_{ }

_{ }

_{ }

## 9.4. The rocket

**The motion of a rocket is a nice example of a system with a variable mass in which nevertheless conservation of linear momentum can be applied. Suppose a rocket is flying through deep space (no friction force and no gravitational force). It is burning fuel. Suppose at some time t the mass of the rocket is M. During a time interval [Delta]t, the mass of the rocket changes by [Delta]M:**

M(t + [Delta]t) = M(t) + [Delta]M

Since the rocket is burning fuel, [Delta]M is negative. The mass of
the exhaust products is - [Delta]M. The result of the burning of fuel is a
change in the velocity of the rocket:
v(t + [Delta]t) = v(t) + [Delta]v

If we consider our system to consist of the rocket and the exhaust
generated during the time interval [Delta]t, we are dealing with a closed
system. Since there are no external forces acting on the system, the total
linear momentum of the system is conserved. The initial linear momentum of the
system (at time t) is given by
p

The final linear momentum of the system is given by_{i}= M(t) v(t)
p

where U is the velocity of the exhaust. Conservation of linear
momentum therefore requires that_{f}= (M(t) + [Delta]M) (v(t) + [Delta]v) + (- [Delta]M) U
M(t) v(t) = (M(t) + [Delta]M) (v(t) + [Delta]v ) + (- [Delta]M) U

The exhaust velocity of the rocket depends on the design of the rocket
engine. Suppose that for the engine used the velocity of the exhaust relative
to the engine is measured to be U_{0}. In the frame of reference in which the rocket is moving, the exhaust velocity is a function of both U

_{0}and the velocity of the rocket

U - U

Using this expression we can rewrite the expression for conservation of
linear momentum as follows_{0}= v(t) + [Delta]v
M(t) v(t) = (M(t) + [Delta]M) (v(t) + [Delta]v) + (- [Delta]M) (v(t)
+ [Delta]v + U

or_{0})
M(t) v(t) = M(t) (v(t) + [Delta]v) - [Delta]M U

We conclude_{0}
[Delta]M U

Dividing each side by [Delta]t gives_{0}= M(t) [Delta]v_{ }

- dM/dt = - R where R is the rate of fuel consumption.
- U
_{0}= - u where u is the (positive) velocity of the exhaust gasses relative to the rocket. - dv/dt is the acceleration of the rocket.

**first rocket equation**"

R u = M a

The mass used in the "first rocket equation" is of course time
dependent (related to R). In order to find the velocity of the rocket (after
burning some fuel) we return to the differential equation previously
discussed_{ }

_{ }

_{ }

_{ }

**second rocket equation**".

## Comments

## Post a Comment