A. Impuls
Often in systems of particles, two particles interact by applying a force to each other over a finite period of time, as in a collision. Now we will look at the general case of forces acting over a period of time. We shall define this concept, force applied over a time period, as impulse. Impulse can be defined mathematically, and is denoted by J :
| J = FΔt |
|
Just as work was a force over a distance, impulse is force over a time. Work applied mostly to forces that would be considered external in a system of particles: gravity, spring force, friction. Impulse, however, applies mostly to interactions finite in time, best seen in particle interactions. A good example of impulse is the action of hitting a ball with a bat. Though the contact may seem instantaneous, there actually is a short period of time in which the bat exerts a force on the ball. The impulse in this situation is the average force exerted by the bat multiplied by the time the bat and ball were in contact. It is also important to note that impulse is a vector quantity, pointing in the same direction as the force applied.
Given the situation of hitting a ball, can we predict the resultant motion of the ball? Let us analyze our equation for impulse more closely, and convert it to a kinematic expression. We first substitute F = ma into our equation:
J = FΔt = (ma)Δt
But the acceleration can also be expressed as a = 
. Thus:
J = m 
Δt = mΔv = Δ(mv) = mv f - mv o
The large impulse applied by the bat actually reverses the direction of the ball, causing a large change in velocity.
Recall that when finding that work caused a change in the quantity 
mv 2 we defined this as kinetic energy. Similarly, we define momentum according to our equation for an impulse.
B. Momentum
From our equation relating impulse and velocity, it is logical to define the momentum of a single particle, denoted by the vector p , as such:
| p = mv |
|
Again, momentum is a vector quantity, pointing in the direction of the velocity of the object. From this definition we can generate two every important equations, the first relating force and acceleration, the second relating impulse and momentum.
Equation 1: Relating Force and Acceleration
The first equation, involving calculus, reverts back to Newton's Laws. If we take a time derivative of our momentum expression we get the following equation:
= 
(mv) = m 
= ma = 
F
Thus
|
|
|
It is this equation, not F = ma that Newton originally used to relate force and acceleration. Though in classical mechanics the two equations are equivalent, one finds in relativity that only the equation involving momentum is valid, as mass becomes a variable quantity. Though this equation is not essential for classical mechanics, it becomes quite useful in higher-level physics.
Equation 2: The Impulse-Momentum Theorem
The second equation we can generate from our definition of momentum comes from our equations for impulse. Recall that:
J = mv f - mv o
Substituting our expression for momentum, we find that:
| J = p f - p o = Δp |
|
This equation is known as the Impulse-Momentum Theorem. Stated verbally, an impulse given to a particle causes a change in momentum of that particle. Keeping this equation in mind, momentum is conceptually quite similar to kinetic energy. Both quantities are defined based on concepts dealing with force: kinetic energy is defined by work, and momentum is defined by impulse. Just as a net work causes a change in kinetic energy, a net impulse causes a change momentum. In addition, both are related to velocity in some way. In fact, combining the two equations K = mv 2 and p = mv we can see that:
| K = |
|
This simple equation can be quite convenient for relating the two different concepts.
This section, dealing exclusively with the momentum of a single particle, might seem out of place after a section on systems of particles. However, when we combine the definition of momentum with our knowledge of systems of particles, we can generate a powerful conservation law: the conservation of momentum.
Momentum Conservation Principle
One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value. This law of momentum conservation will be the focus of the remainder of Lesson 2. To understand the basis of momentum conservation, let's begin with a short logical proof.
Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows.
F1=-F2
The force are equal in magnitude and opposite in direction
The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as
t1=t2
Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as
F1 x t1=-F2 x t2
The impuls are equal in magnitude and opposite in direction.
Angular Momentum
The angular momentum of a rigid object is defined as the product of the moment of inertia and the angular velocity. It is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle if there is no external torque on the object. Angular momentum is a vector quantity. It is derivable from the expression for the angular momentum of a particle
Moment Of Inertia
Moment of inertia is the name given to rotational inertia, the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation. For a point mass the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr2. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses.
COMPARISON OF LINEAR AND ANGULAR MOMENTUM
| Linear Momentum, p | Angular Momentum, L |
| Tendency for a mass to continue moving in a straight line. | Tendency for a mass to continue rotating |
| Parallel to v. | Perpendicular to both v and r |
| A conserved, vector quantity. | A conserved, vector quantity
|
| Magnitude is inertia (mass) times speed. | Magnitude is rotational inertia times angular speed. |
| Net force required to change it. | Net torque required to change it. |
| The greater the mass, the greater the force needed to change momentum. | The greater the moment of inertia, the greater the torque needed to change angular momentum. |
Sample Problems and Solution
1.What is the momentum of of an object with m=2.00 kg and v-40.0 m/s?
Answer:
p = mv
p = (2.00)(40.0) = 80.0 kg*m/s
2. A force of 30000 N j is exerted for 4.00 s, on a 95,000 kg mass.
(a) What is the impulse of the force for this 4.00 s?
(b) What is the mass's change in momentum from this impulse?
(c) What is the mass's change in velocity from this impulse?
(d) Why can't we find the resulting change in kinetic energy of the mass?
Answer:
a. Impulse = Ft = 30000(4) = 120000 N-s
b. change in momentum = impulse = 120000 N-s
c. j = m∆v
120000 = (95000)∆v
∆v = 1.26 m/s
d. We do not know the initial velocity.
3. A particle with momentum mv1 experiences a force which leaves it with momentum mv2. Draw an arrow to represent the impulse the particle experienced.
1. j = mv2 - mv1 = mv2 + (-mv1)
4. A 1000 kg car accidentally drops from a crane and crashes at 30 m/s to the ground below and comes to an abrupt halt. What impulse acts on the car when it crashes?
Answer:
j = mΔv assuming constant mass.
j = 1000(-30) = -30 000 kg-m /s
The impulse is 30000 kg-m/s upwards.
5. A freight car weighing 25 tons runs into another freight car of the same weight. The first was moving at 6 mi/hr (8.8 ft/sec) and the second was at rest. If the cars are coupled together after the collision, what is their final speed?
Solution: We have a collision problem in 1‑dimension. We draw both 'before' and 'after' pictures and select a coordinate system as shown. We have conservation of linear momentum: pAi Å pB i = pA f Å pB f .
|
|
|
We have a perfectly inelastic collision problem. Hence: m1 v1i Å m2 v2i = (m1 + m2) vf
For x‑components this becomes: m1 v1 + 0 = (m1 + m2) vf .
Since m1 = m2 , we have: vf = (1/2) v1 = 3 mi/hr (4.4 ft/sec).
6. A cue ball collides elastically with a red ball and a blue ball which are both initially at rest at 10 m/s. If the cue ball comes to rest and the blue ball begins to move at 2m/s after collision, what is the speed of the blue ball?
Solution:
Because it is an elastic collision, the relative velocity of approach is equal to the relative velocity of recession. Which means that the total magnitude of velocity is conserved in the system and because all balls have equal masses, the third ball, blue one will have the difference of velocities of other two balls.
vblue = 10m/s - 2m/s = 8m/s
7. If the stone does 6 rotations in 1 second find the angular velocity of it.
Solution:
If the stone does four rotations in one second then its frequency becomes 6.
f=6s-1
T=1/f=1/6s/s
ω=2π/T=2.3/1/6s=36radian
http://www.kelvinsewow.blogspot.com/
