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Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density.. The terms center of mass and center of gravity are used synonymously.The center of mass is the point at which all particles can be considered to be concentrated.It is the point of a body at whi

Systems of Particles

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**1. **Systems of Particles

**2. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. The terms center of mass and center of gravity are used synonymously.
The center of mass is the point at which all particles can be considered to be concentrated.
It is the point of a body at which the force of gravity can be considered to act and which undergoes no internal motion.

**3. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. The center of mass of uniform object is located at its geometric center.
Non-uniform objects which are composed of uniform sections or individual particles, the following equation can be used.

**4. **mi = mass of the individual particle
M = total mass of the system
xi = i component of the position vector for the location of the particle
yi = j component of the position vector for the location of the particle
zi = k component of the position vector for the location of the particle
xCM, yCM, and zCM are the x, y, and z coordinates of the center of mass

**5. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. Three particles, each of mass 2.5 kg are located at the corners of a right triangle whose sides are 2 m and 1.5 m long as shown below. Locate the center of mass.

**6. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. For non uniform objects that are made up of a continuous distribution of matter, the this equation can be used

**7. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. Find the center of mass of a uniform cone of height h and radius R made of copper which has a density of 8930 kg/m3. Hint: Set up your coordinate axis so that the z axis is going upward from the tip of the cone to the top.

**8. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. A tiny slice horizontally through the cone would be a tiny cylinder with height dz.
Using the definition of density and the equation for the volume of a cylinder, we can find that the mass of that tiny cylinder would be dm = 8930(p)(r2)dz

**9. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density.

**11. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. So the location of the center of mass would be
(0,0, )

**12. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. When Newton’s second law is applied to the center of mass, the vector sum of all the forces acting on the system is equal to the total mass of the system times the acceleration of the center of mass.

**13. **Define the center of mass of a system of discrete particles or rigid bodies of uniform mass or density. A rocket is fired into the air. At the moment it reaches it’s highest point, a horizontal distance D from its starting point, it separates into two parts of equal mass. Part 1 falls vertically to earth. Where does part II land? Assume the acceleration due to gravity is constant.

**14. **For projectiles, even if the object itself is rotating, the center of mass will follow a parabolic path.

**18. **A fisherman stands at the back of a perfectly symmetrical boat of length L. The boat is at rest in the middle of a perfectly still and peaceful lake, and the fisherman has a mass 1/4 that of the boat. If the fisherman walks to the front of the boat, by how much is the boat displaced?

**19. **If you’ve ever tried to walk from one end of a small boat to the other, you may have noticed that the boat moves backward as you move forward.
That’s because there are no external forces acting on the system, so the system as a whole experiences no net force.
The center of mass of the system cannot move since there is no net force acting on the system.
The fisherman can move, the boat can move, but the system as a whole must maintain the same center of mass.
Thus, as the fisherman moves forward, the boat must move backward to compensate for his movement.

**20. **Because the boat is symmetrical, we know that the center of mass of the boat is at its geometrical center, at x = L/2.
We can calculate the center of mass of the system containing the fisherman and the boat:

**21. **In the figure below, the center of mass of the boat is marked by a dot, while the center of mass of the fisherman-boat system is marked by an x.

**22. **At the front end of the boat, the fisherman is now at position L, so the center of mass of the fisherman-boat system relative to the boat is

**23. **The center of mass of the system is now 3 / 5 from the back of the boat.
But we know the center of mass hasn’t moved, which means the boat has moved backward a distance of 1/5 L, so that the point 3/ 5 L is now located where the point 2 /5 L was before the fisherman began to move.

**24. **Define linear momentum and rewrite Newton’s Second Law in terms of momentum. Linear Momentum is the product of mass and velocity.
Momentum is a vector since it is the product of a scalar (mass) and a vector (velocity).
The momentum vector is in the direction of the velocity vector.
Momentum is measured in kg·m/s

**25. **Define linear momentum and rewrite Newton’s Second Law in terms of momentum. p = linear momentum
m = mass
v = velocity

**26. **Define linear momentum and rewrite Newton’s Second Law in terms of momentum. A fast moving car has more momentum than a slow moving car of the same mass.
An loaded eighteen wheeler has more momentum than a volkswagon beetle traveling at the same speed.
The more momentum an object has the harder it is to stop and the greater effect it will have if brought to a stop by collision or impact.

**27. **Define linear momentum and rewrite Newton’s Second Law in terms of momentum. A force is required to change the momentum of an object whether it is to increase the momentum, decrease it, or to change the direction.
Newton’s second law could be rewritten as the rate of change of momentum of a particle is proportional to the net force applied to it.
Remember the rate of change of velocity is acceleration.

**28. **Define linear momentum and rewrite Newton’s Second Law in terms of momentum.

**29. **Define linear momentum and rewrite Newton’s Second Law in terms of momentum. Water leaves a hose at a rate of 5 kg / s with a speed of 50 m / s. It strikes a wall, which stops it. (We are ignoring any splashing back). What is the force exerted by the water on the wall?

**30. **State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems. A closed system is one in which objects can neither enter nor leave the system.
An isolated system is one in which there is no net external force acting on the system.
For example billiard balls have the force due to gravity acting on them but it is balanced by the table pushing up on them. So in that case there is no net external force.
There can be internal forces acting.

**31. **State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems. The law of conservation of linear momentum only holds true in a closed and isolated system.
The net external force is zero
The total mass of the system does not change

**32. **State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems. For dP / dt to be equal to zero, the momentum (P) must be constant (not changing over time).
So when there in a closed (mass of system doesn’t change) and isolated (no net external forces) system the total momentum of the system does not change (is conserved)

**33. **State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems. The Law of conservation of momentum states the total momentum before = the total momentum after.

**34. **State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems. The law of conservation of momentum does NOT say that each particle within the system retains its same momentum.
The amount of momentum one particle in the system gains is equal to the amount of momentum lost by another.

**35. **State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems. A 10,000 kg railroad car traveling at a speed of 24 m / s strikes an identical car at rest. If the cars lock together as a result of the collisions, what is their common speed afterward?

**36. **State the law of conservation of linear momentum for a closed, isolated system and apply this law to the solution of problems. Calculate the recoil velocity of a 4 kg rifle which shoots a 0.050 kg bullet at a speed of 280 m / s.

**37. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems. A collision is an interaction between two bodies if the interaction occurs over a short time interval and is so strong that other forces acting are insignificant compared to the forces the two objects are exerting on each other.

**38. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems. An impulse is the product of a net force and time over which it acts.
Remember the net force acting is proportional to the rate of change of momentum of an object or system.
So impulses must cause changes in momentum as well
Beginning with the Newton’s second law definition of momentum, we can derive the impulse momentum theorem.

**40. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems. J = impulse (N·s)
m = mass (kg)
?v = change in velocity (m/s)

**41. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems. Dimensionally, is N·s = kg·m/s? Hint: Remember the base units for a Newton.
1 N = 1 kg·m/s2

**42. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems. Calculate the impulse suffered when a 70 kg person lands on firm ground after jumping from a height of 5 m. Then estimate the average force on the person’s leg if the landing is stiff legged where the body moves 1 cm during impact.
Hint:
Find the impact velocity using linear motion equations.
Find the Impulse
Find the average velocity during impact
Find the time of impact using the definition of average velocity
Find the average force acting during impact

**43. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

**44. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

**45. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

**46. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

**47. **Define collision and state the impulse-linear momentum theorem; apply this theorem to the solution of problems.

**48. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. Elastic collisions are collisions in which both momentum and kinetic energy are conserved.
Inelastic collisions are collisions in which only momentum is conserved.

**49. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems.

**50. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. For elastic collisions, you will typically have to
use the conservation of momentum equation to solve for one of your two unknown variables in terms of the other
substitute that into the conservation of kinetic energy equation
solve for your unknown
substitute that value back into your first derived equation
solve for the other unknown

**51. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. There are some common special cases where it is unnecessary to do the extensive algebra. Those special cases are on the slides that follow.

**52. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. In an elastic collision
If the masses are equal, they will exchange velocities.
For example:
Before the collision
Ball A is going 5 m/s and ball B is going 3 m/s
After the collision
Ball A is going 3 m/s and ball B is going 5 m/s

**53. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. In an elastic collision
If object two is initially at rest (v2 = 0)
After the collision, the objects will have the following velocities.

**54. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. In an elastic collision
If object 2 is at rest (v2 = 0) and object 1 is much more massive than object 2
In other words, object 1 slows down very little and object 2 ends up going almost twice the velocity of object 1

**55. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. In an elastic collision
If object 2 is at rest (v2 = 0) and object 1 is much less massive than object 2

**57. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. Think of the interaction of the basketball and the floor as the case where object 2 (the earth) is at rest and much less massive object 1 (the basketball) hits it.
The basketball will bounce off with approximately the same speed it hit with

**58. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. Think of the interaction between the tennis ball and the basketball as object 2 (the tennis ball) at rest when it is hit by much more massive object 1 (the basketball) so the speed of the basket ball is almost unchanged but object 2 (the tennis ball) takes off with approximately twice the speed of the basketball.
The same situation is true for the interaction between the tennis ball and the ping pong ball.

**59. **Distinguish between elastic and inelastic collisions in one and two dimensions; solve problems involving these types of problems. The conservation of momentum equation can be used in 2 and 3 dimensions as well.
You will just solve for each dimension separately.
Keep in mind that if an object is moving with velocity v
vx = v·cosT
vy = v·sinT
vz = vcosF