- #1
- 979
- 1
We will apply a force of some distance for some period of time.
Let's say the goal is to apply a maximum force over a small distance.
We apply 1 Newton at right angles to a frictionless lever, 1 meter from the pivot point.
At 10 centimeters from that pivot point is an object that weighs 1 kg, and the force it receives is 10 Newtons. Because the force is 10 Newtons, the object gains 10 kg*m/s of momentum per second. Two special situations may be considered:
1) Had the force been higher, the object would gain greater than 10 kg*m/s due to greater acceleration.
2) For the same force, a heavier mass would accelerate more slowly but it would still absorb momentum at the same rate as determined by that same force.
The force on each lever arm is inversely proportional to the radius of each lever arm. Therefore, it can be assumed that this system as currently revealed involves the conservation of angular momentum. It also conserves energy. If one increases the length of the arm of input, the effort needed to be applied to do the same amount of work directly decreases.
The question then is, will the momentum going into the lever equal the momentum going out of the lever? If the force going into the lever is 1 Newton, it is absorbing momentum at a rate of 1 kg*m/s per second. If the force going out of the lever is 10 Newtons, it is applying momentum at a rate of 10 kg*m/s per second. This defies common sense. It implies that a rate of momentum utilization of 9 kg*m/s per second is unaccounted for. The question however is, what is the direction of that force? That force is anti-parallel to the force applied on that lever. If we ignore the lever itself and simply accounted for the input and output forces (which are in opposite directions), common sense tells us that a rate of momentum of 9 kg*m/s is being utilized from whatever medium to which the force of 1 Newton is applied and from which the force of 10 Newtons comes from. The momentum, it seems, would have to come from the lever itself. Incongruous!
Even more shocking is when we consider the Newtons applied over a period of time. A force times time equals a change in momentum. Apparently, the kinetic energy stored in some mass would increase with the square of the force and the square of the time the force is applied, assuming a constant force and mass. For a given untethered mass, we could apply twice the force for half the time or half the force for twice the time and it would result in the same kinetic energy gained. The question is however, how far does the lever arm move in that time period. In the case of a lever arm of twice the length, half the force is applied resulting in the same acceleration of the object with the same torque, and the force is made at twice the velocity, resulting in the same power input (since force*velocity=power; e.g. prony brake). The opposite case occurs for a lever arm half as long with an applied force twice as much. Again, moving the same mass at the same rate does not affect the input power or input torque.
However, starting with the same lever arm, the input force * the input arm radius / the output arm radius = the output force. We return to the conundrum we found where output momentum exceeded input momentum, suggesting that the momentum came from the lever itself. If we think the input lever arm radius * input force as the calculation for torque, and if we recognize that torque is quantified as angular momentum transferred per unit time, we can easily see that angular momentum is conserved.
KE = 1/2 * mv^2
where
KE = Kinetic Energy
m=mass
v=velocity
p = mv
where
p = momentum
m=mass
v=velocity
KE = 1/2 * p^2/m
where
p^2/m = mv^2
F = p/t
where
t=time
p=F*t
KE = 1/2 * (F*t)^2/m
where
KE = F*d
d=1/2 * (F/m)*t^2
(F/m)=acceleration
d=1/2 * a * t^2
v_initial=0
d=change in position
These formulas are consistent with physical knowledge and with the above paragraphs.
Lesson: Conservation of momentum is really the conservation of ANGULAR momentum.
Let's say the goal is to apply a maximum force over a small distance.
We apply 1 Newton at right angles to a frictionless lever, 1 meter from the pivot point.
At 10 centimeters from that pivot point is an object that weighs 1 kg, and the force it receives is 10 Newtons. Because the force is 10 Newtons, the object gains 10 kg*m/s of momentum per second. Two special situations may be considered:
1) Had the force been higher, the object would gain greater than 10 kg*m/s due to greater acceleration.
2) For the same force, a heavier mass would accelerate more slowly but it would still absorb momentum at the same rate as determined by that same force.
The force on each lever arm is inversely proportional to the radius of each lever arm. Therefore, it can be assumed that this system as currently revealed involves the conservation of angular momentum. It also conserves energy. If one increases the length of the arm of input, the effort needed to be applied to do the same amount of work directly decreases.
The question then is, will the momentum going into the lever equal the momentum going out of the lever? If the force going into the lever is 1 Newton, it is absorbing momentum at a rate of 1 kg*m/s per second. If the force going out of the lever is 10 Newtons, it is applying momentum at a rate of 10 kg*m/s per second. This defies common sense. It implies that a rate of momentum utilization of 9 kg*m/s per second is unaccounted for. The question however is, what is the direction of that force? That force is anti-parallel to the force applied on that lever. If we ignore the lever itself and simply accounted for the input and output forces (which are in opposite directions), common sense tells us that a rate of momentum of 9 kg*m/s is being utilized from whatever medium to which the force of 1 Newton is applied and from which the force of 10 Newtons comes from. The momentum, it seems, would have to come from the lever itself. Incongruous!
Even more shocking is when we consider the Newtons applied over a period of time. A force times time equals a change in momentum. Apparently, the kinetic energy stored in some mass would increase with the square of the force and the square of the time the force is applied, assuming a constant force and mass. For a given untethered mass, we could apply twice the force for half the time or half the force for twice the time and it would result in the same kinetic energy gained. The question is however, how far does the lever arm move in that time period. In the case of a lever arm of twice the length, half the force is applied resulting in the same acceleration of the object with the same torque, and the force is made at twice the velocity, resulting in the same power input (since force*velocity=power; e.g. prony brake). The opposite case occurs for a lever arm half as long with an applied force twice as much. Again, moving the same mass at the same rate does not affect the input power or input torque.
However, starting with the same lever arm, the input force * the input arm radius / the output arm radius = the output force. We return to the conundrum we found where output momentum exceeded input momentum, suggesting that the momentum came from the lever itself. If we think the input lever arm radius * input force as the calculation for torque, and if we recognize that torque is quantified as angular momentum transferred per unit time, we can easily see that angular momentum is conserved.
KE = 1/2 * mv^2
where
KE = Kinetic Energy
m=mass
v=velocity
p = mv
where
p = momentum
m=mass
v=velocity
KE = 1/2 * p^2/m
where
p^2/m = mv^2
F = p/t
where
t=time
p=F*t
KE = 1/2 * (F*t)^2/m
where
KE = F*d
d=1/2 * (F/m)*t^2
(F/m)=acceleration
d=1/2 * a * t^2
v_initial=0
d=change in position
These formulas are consistent with physical knowledge and with the above paragraphs.
Lesson: Conservation of momentum is really the conservation of ANGULAR momentum.
Last edited: