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**4.3.1 Definition of the angular impulse of a force**

The angular impulse of a force is the time integral of the moment exerted by the force.

To make the concept precise, consider a particle that is subjected to a time varying force F(t) , with components in a fixed basis {i,j,k} , then

Let

denote the position vector of the particle, and

M(t)=r(t)xF(t) = (y(t)F_{z}(t)-z(t)F_{y}(t))i+(z(t)F_{x}(t)-x(t)F_{z}(t))j+(x(t)F_{y}(t)-y(t)F_{z}(t))k

the moment of the force about the origin.

The **Angular Impulse** exerted by the force about O during a time interval is defined as

The angular impulse is a vector, and can be expressed as components in a basis

If you know the moment as a function of time, you can calculate its angular impulse using simple calculus. For example for a constant moment, with vector value M_{0} , the impulse is

**4.3.2 Definition of the angular momentum of a particle**

The angular momentum of a particle is simply the cross product of the particleâ€™s position vector with its linear momentum

The angular momentum is a vector â€“ the direction of the vector is perpendicular to its velocity and its position vectors.

**4.3.3 Angular impulse â€“ Angular Momentum relation for a single particle**

Consider a particle that is subjected to a force F(t) for a time interval

Let r(t) denote the position vector of the particle

Let denote the moment of F about the origin

Let denote the angular impulse exerted on the particle

Let denote the change in angular momentum during the time interval Î”t.

The momentum conservation equation can be expressed either in differential or integral form.

1. In differential form

2. In integral form

Although itâ€™s not obvious, these are just another way of writing Newtonâ€™s laws of motion. To show this, weâ€™ll derive the differential form. Start with Newtonâ€™s law

Take the cross product of both sides with r

Note that since the cross product of two parallel vectors is zero. We can add this to the right hand side, which shows that

This yields the required relation.

Angular momentum conservation: For the special case where the force is parallel to r, the moment of the force acting on the particle is zero and angular momentum is constant

**4.3.4 Examples using Angular Impulse â€“ Angular Momentum relations for a single particle**

The angular impulse-angular momentum equations are particularly helpful when you need to solve problems where particles are subjected to a single force, which acts through a fixed point. They can also be used to analyze rotational motion of a massless frame containing one or more particles.

**Example 1: Orbital motion.** A satellite is launched into a geostationary transfer orbit by the ARIANE V launch facility. At its perigee (the point where the satellite is closest to the earth) the satellite has speed 10.2km/sec and altitude 250km. At apogee (the point where the satellite is furthest from the earth) the satellite has altitude 35950 km. Calculate the speed of the satellite at apogee.

**Assumptions: **

1. We assume that the only force acting on the satellite is the gravitational attraction of the earth

2. The earthâ€™s radius is 6378.145km

**Calculation:**

1. Since the gravitational force on the satellite always acts towards the center of the earth, angular momentum about the earthâ€™s center is conserved.

2. At both perigee and apogee, the velocity vector of the satellite must be perpendicular to its position vector. To see this, note that at the point where the satellite is closest and furthest from the earth, the distance to the earth is a max or min, and so the derivative of the distance of the satellite from the earth must vanish, i.e.

where we have used the chain rule to evaluate the time derivative of Recall that if the dot product of two vectors vanishes, they are mutually perpendicular. We take a coordinate system with i and j in the plane of the orbit, and k perpendicular to the orbit.

3. We take the satellite orbit to lie in the i , j plane with k perpendicular to the orbit. The angular momentum at apogee and perigee is then

where r_{a} , r_{p} are the distance of the satellite from the center of the earth at apogee and perigee, and v_{a} , v_{p} are the corresponding speeds.

4. Since angular momentum is conserved it follows that

5. Substituting numbers yields 1.6 km/s

Example 2: More orbital motion. The orbit for a satellite is normally specified by a set of angles specifying the inclination of the orbit, and by quoting the distance of the satellite from the earths center at apogee and perigee r_{a} , r_{p} It is possible to calculate the speed of the satellite at perigee and apogee from this information.

**Calculation**

1. From the previous example, we know that the distances and velocities are related by

2. The system is conservative, so the total energy of the system is conserved.

3. The potential energies when the satellite is at perigee and apogee are

4. The kinetic energies of the satellite at perigee and apogee are

5. The kinetic energy of the earth can be assumed to be constant. Energy conservation therefore shows that

6. The results of (1) and (5) give two equations that can be solved for v_{a} , v_{p }in terms of known parameters. For example, (1) shows that - this can be substituted into (5) to see that

Similarly, at apogee

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