Semester 2, 2012-13

Mini-Project: Dynamics in a Product

N
ewton Cradle Analysis

In this report, we

a) Identify the requirement of dynamics of the Newton Cradle.

b) Carry a detailed analysis of the dynamics of Newton Cradle.

c) Suggest improvement of the dynamics of the Newton Cradle.

Newton’s cradle

This is a device that can show the conservation of momentum and energy by swinging a series of spheres. When one of the end spheres lifted and released, the resulting force will travels through the line spheres and the fore will act on the last spheres. The last one was pushed upward.

Detail action

The last ball is pulled away and going to fall, it will hit on the series of balls and comes to nearly a dead stop. The ball at opposite side will accept almost all of the velocity and will swings instantly in an arc. The upward height of the ball at opposite side may nearly the same as the height of the ball is pulled. This show the final ball receives most of the energy and momentum that was in the first ball.

The impact produces a shock wave that propagates through the intermediate balls. All the elastic material like steel will repeat this until the kinetic energy is temporarily stored as potential energy in the compression of the material rather than being lost as heat and sound.

Physics explanation

The conservation of momentum (mass x velocity) and kinetic energy (0.5 x mass x velocity^2) can be used to find the resulting velocities.

Consider the two-ball case where each ball has mass m and velocity v at the time of collision. The kinetic energy will be E=mv2 and the momentum p=2mv. So suppose n balls were to fly off the other side with velocity u. The energy would be n2mu2 and the momentum nmu. So by conservation we must have mv2=n2mu2 and 2mv=nmu. It is not difficult to see that the only solution to these equation together is n=2 and u=v.

Observation of Newton’s cradle

There are three situations in this system that can prove the conservation of momentum and energy.

Newton’s cradle 2-ball system

One side of the ball is pulled away and fall. It will hit the other ball and it will nearly dead stop. The other ball accepts almost all the velocity and swings in an arc to the height almost as the height of the ball pulled. This shows the other ball receives most of the energy and momentum.

Newton’s cradle 5-ball system

The same operation is used on the first step. All the energy will transfer to the last ball of the opposite side. After the dead stop of the first ball, the three middle balls will not move because there is no space for the swing upward. Therefore the energy only can transfer to other ball until the last ball.

Newton's cradle 3-ball swing in a 5-ball

The only different is initially pulled up three balls and release. After hitting the rest of balls, the central ball and the rest of two balls are swinging upward. This situation can easy to see that all the energy and momentum will not be los

Newton Cradle Analysis

The cradle

As you see know what is Newton Cradle. 5 balls are suspended and each one of them is holding by 2 wires. They are in equilibrium stage. Before we go to analysis the energy conversation within this system working, we have to make some assumptions. First, each of the ball should be same mass. Second they should be suspended in the same level from the floor by same length of wires. On other thing to note is that we will be neglecting force such as drag and friction. In reality, these play an important role in the motion, but we will only consider a “perfect system”.

Conservation of energy

Energy for the system is separated into two types of energy. They are Potential energy “PE” and Kinetic Energy “KE”. Potential energy is the mass(m) X the acceleration due to gravity(g) X the vertical height that the object is from a datum that we get to set(h). And the KE is equal to 1/2 X the mass(m) X the velocity squared(v2). When you put all this together, it simply stated that all that all the initial energy is equal to all the final energy.

Scenarios

The physics behind the Newton’s Cradle and just about any process in the world can be extremely excessive. To analysis the system, we will look at few different scenarios of a simple motion. The first scenario is when the system is just sitting on a table top, and none of the balls are moving. The velocity is equal to “0” in this case

Sc1) No ball being displaced and released.

The first scenario is when the system is just sitting on a table top, and none of the balls are moving. The velocity is equal to “0” in this case, so there is no momentum or kinetic energy. Moreover, we have set our datum for potential energy on the same axis on which the balls lie when they are at rest. Because they indeed are at rest, the height in our energy equation is “0” and thus potential energy is “0”. Therefore both the momentum and the energy equation are “0” equal to “0”. It doesn’t mean that no forces are acting on the balls. Take a look at the free body diagram to the right. The force due to gravity is balance by the vertical component of the string. The ball is in equilibrium. The horizontal components are also equal. And therefore the ball doesn’t move side to side.

Sc2) 1 ball being displaced

This is the simplest performance of this device. When one of the balls is displaced from its equilibrium position by external force, human hand in this case. Refer to the photos on the right. There is a free body diagram and also the ball is immediately after it is released. Now the forces shown are not balancing. And there is acceleration. This acceleration leads to velocity. The vertical component of the force of gravity outweighs that of the combined tensions in the wires. As the ball progresses through its path, the forces will balance to keep it on the path.

The momentum is there is an external force from gravity acting on the body, o equations pertaining to this will not help us much. At the moment, momentum is changing. Initially, there is no velocity, so the initially Kinetic energy is equal to “0”. The potential energy is governed by “mgh”(mass X gravity acceleration X height). All of which are nonzero quantities. So the momentum is not yet be considered. Initial energy is PE+KE, but the velocity is “0” so the KE is “0” too.

Sc3) The ball is being released

As the ball velocity increases, the momentum and KE increase but the PE is decreasing because the height “h” from the datum is decreasing when “m”(mass of the ball) and “g”(gravity) are remaining constant. PE is converting to KE. Now the ball is at the position it was before. It is literally at the moment of impact with the rest of the balls. This time, the ball has momentum and energy. It is at maximum velocity.

For momentum, we want to know what will happen immediately after impact is occurred. We know that the initial momentum is going to equal the final. And the momentum is at a maximum, the velocity is also at maximum. The equation tells us that the product of the mass and the velocity of the single ball is equal to the product of the mass and velocity of whatever is moving after the collision. Looking at the device, this leaves 2 options; when the ball strikes, the rest of the balls will move away at a much slower velocity or one ball of equal mass will move away at an equal velocity. As the ball accelerates towards the site of impact, it potential energy “PE” is being converted to kinetic energy “KE”. By the time the ball reaches the datum, the final PE is equal to “0” and the final KE is at its max. If we put these results into our equation, we will find the initial PE is equal to the final KE. We can calculate the original PE and therefore find the final KE and the velocity. This velocity is the same as the velocity for the momentum. So, the ball was at rest with PE stored. After the release, an imbalance forces gave it acceleration on a curved linear path. It gains KE and it is losing PE at the bottom of its path at the same time. It has a velocity that can be used to used to determine momentum.

Sc4) The energy has dissipated into other balls

The ball’s energy has dissipated into the other balls, because the balls are rigid. So each of the ball is carrying the same amount of energy travels through the line of balls until reaching the last ball of equal weight. The last ball then leaves the rest and proceeds towards the equal height of the initial. During this time I is decelerating to achieve a velocity of zero at the apex. Although the energy is not needed and not considered for calculation while energy is being transported through the ball. Momentum will tell us what we need to know about last ball when it leaves. However anytime after that, an imbalanced force, gravity are acting on it. To follow whats going on till the apex, we use the conservation of energy.

By using energy, we found the velocity and the momentum. In order to conserve this, we need to have a balanced reaction. The first ball throws all of its momentum into the line of balls. Because the balls are not all attached, the momentum travels to an object of equal mass of the last ball and gives it an equal velocity. This is predicted by the equation, and the velocity of another different sized ball could be calculated by using the same equation. The energy takes back control of the system as soon as

the last ball leaves the chain. The energy takes the reverse path of the starting ball. KE begins to be converted back to PE until the velocity equal to “0” and the height “h” is the same as the starting ball.

Sc5) NEWTON'S LAW OF RESTITUTION

Newton’s law of restitution is good demonstrated by Newton’s cradle in ideal case.

When two particles approach each other and collide directly. The speed they separate is usually less than the speed at which approached each other.

(Separation speed) = e (approach speed

Working Examples

We let each ball weight 0.1kg, and initial we take tha ball 0.1m up.

By conversation of energy, the ball will convert the potential energy to kenitic energy.

mgh = 1/2mv²

10(0.1) = 1/2v²

V = 1.41

For the ideal case, we assume that there is no loss in kinetic energy, it is perfectly elastic, therefore e=1.

e= (V₂-V₁)/(U₁-U₂), U₁=1.41, U₂=0,

1= (V₂-V₁)/1.41

V₂-V₁=1.41m/s---------(1)

By Conversation of momentum,

M₁U₁+M₂U₂=M₁V₁+M₂V₂

0.1(1.41)+0.1(0)=0.1(V₁+V₂)

V₁+V₂=1.41m/s---------(2)

Combine (1) and (2),

V₂=1.41/s

As the action and reaction pair, the second and the third ball will not have the movement, V₂=0 and V₃=0; and the fourth ball V₄ should be equal to 1.41m/s by the Conversation of momentum.

By conversation of energy,

mgh = 1/2mv²

gh = 1/2v²

10(h) = 1/2(1.41)²

h= 0.1m

In ideal case, the fourth ball will raise up to 0.1m and the motion reverse.

Improvement

Since we discover that the Newton’s cradle we use in this project would slowing down over a period of time, we know our Newton’s cradle still not perfect enough, because in ideal case, the system of Newton’s cradle should swing forever if no external force acting on it.

To improve the dynamic of Newton’s cradle, in other words, suggest how to maintain the system swings a longer period:

1. The coefficient of restitution of balls used in Newton’ cradle should be 1.

Reason:

By the Newton’s law of restitution,

Experimental evidence suggests that, for two particular colliding particles, the separation speed is always the same fraction of the approach speed, which states that

(Separation speed) = e (approach speed)

The quantity represented by e is called the coefficient of restitution and it is constant for any two particular objects; its value depends upon the material of which the two objects are made.

If the coefficient of restitution of balls used in Newton’ cradle is e =1, a perfect elastic impact occur in Newton’s cradle, and the relative speeds before and after impact should be equal, which means there is no loss in kinetic energy because of the perfectly elastic impact. Therefore, less energy is loss and the system can maintain longer.

2. The surface of balls in Newton’s cradle should very smooth.

Reason:

To reduce friction between air and the ball surface, reduce the energy loss as thermal energy.

3. Function the Newton’s cradle inside a vacuum container.

Reason:

Reduce the air resistance. Since no particle inside the vacuum container, except the Newton’s cradle, the friction between ball surface and air will be reduced, so less energy loss to the surrounding.

Conclusion

In conclude that, Conservation of energy and momentum are satisfied in Newton’s cradle, the ball slams into the balls at rest. Energy is transfer from one to the next until the momentum of the first ball equals to the mass and then instantly the velocity of another. The last ball decelerates to the apex while losing momentum and KE. Eventually, the PE at the apex is equal to that of the original PE.

Moreover, all balls are in contact with neighbors just after impact. Compression occurs at the interfaces. This stores potential energy and gives rise to interface elastic forces. Momentum conservation applies, and hence, line of balls is now moved at the same speed.

The net impulse on each ball due to interface forces is zero, because those balls that end up going the same speed, hence they experience no net change of momentum, and therefore no change in speed.

Finally, in a perfect system, the scenarios and results would be valid. However, considering the drag, friction, energy lost to the surrounding.