Kinetic Energy

Get a flywheel. Any flywheel, a bicycle wheel will do. Get it spinning at a constant known RPM. Let it freewheel and count the revolutions until it comes to a stop.

Next, get it spinning at twice the original RPM and repeat the experiment. If it completes 4 times as many free revolutions you've proven the equation.

slo

Except…..friction losses increase with speed!

I am a black belt in TaeKwonDo. I break bricks with my fist. Trust me, eight times as fast is 64 times the impact force. That's about as "hands on" as it gets. It's also why there is a very big difference between crashing your car at 55mph and 110mph. I hear they have some really good ones on the Autobon.

I wana make one thing clear, 2 times speed isn't equal to 4 times "ENERGY", rather it is 4 times Moment Of Inertia (P=MV). The equation of kinetic energy actually tells that:-

For an object moving with a certain speed, if u double its speed keeping the mass constant, then its moment of inertia gets 4 times. But if you double the mass keeping the speed constant, then moment of inertia gets 2 times.

So if the object is made to collide with something, it will cause 4 times much damage in case you double its speed but cause 2 times much damage if u double its mass keeping the speed constant.

This principle is used in gun bullets.

"I wana make one thing clear, 2 times speed isn't equal to 4 times "ENERGY", rather it is 4 times Moment Of Inertia (P=MV). The equation of kinetic energy actually tells that:-"

I think you have got it 100% arse about.

Momentum = Velocity x Mass

Kinetic Energy = ½ Mass x Velocity²

twice the velocity

Momentum = 2 x Velocity x Mass = 2 fold increase

Kinetic Energy = ½ Mass x (2 x Velocity)² = 2² = 4 fold increase

Look at it from the energy conversion point of view. Say the kinetic energy is completely converted into potential energy when a mass m is thrown vertically upward with an initial velocity v until it comes to rest.The mass m will travel a height h that will give an equivalent energy (to the initial kinetic energy) in the form of potential energy. If the velocity v is doubled, you will see that the height the mass travels will be 4 times h. This is ofcourse with the assumption that frictional effects are neglected. So all that you have to do is throw a stone upward and measure the height!

I like the spinning weel idea, sloco! Ill give it a try.

Impact force examplez make me think that the formula iz rong: If twice the speed = 4 timez the energy, than an impact shoud have 8 timez the force, bekuz you are disapating 4 timez the energy in haf the time.

I'm gessing that friction iz the same az a slow impact, so the weel experiment will end up being inconclusive, but maybe not.

Hi Z man, you wrote: "I like the spinning wheel idea, sloco! I'll give it a try."

Put an electric motor on your wheel to spin it up. If you keep a constant voltage and can measure the electrical current over time (i.e., get the integrated watt-minutes), you can verify the kinetic energy equation approximately. The required watt-minutes should be roughly 4 times when you double the r.p.m. Air resistance and bearing friction need to be accounted for if you want to be accurate, of course.

I say you drop something from 3ft and drop something from 12 feet.

one should have twice the speed of the other, and you can infer the rest. Video tape with a grid background to prove it. Do the math, it should work out.

Takes less effort than building a rig if you have a video camera, and dropping something is fairly efficient transfer of potential to kinetic energy, considering there are very dense things like chunks of metal or a rock that will suffer small wind losses.

Good luck.

Quote: "I say you drop something from 3ft and drop something from 12 feet."

Yep, nice and simple if you have a way to measure the speed - the video camera might work for the relatively slow speeds. I made a quick calc and the speeds will be ~27.712 ft/sec for a 12 ft drop and ~13.856 ft/sec for a 3 ft drop.

Out of the calculation comes an even simpler method. Just time the two drops! Since v = at, the times are, like the speeds, in the relation 2:1. I get ~0.866 and ~0.433 seconds respectively. Within the experimental error, this will prove that the speeds are ~ 2:1 while the energies are ~ 4:1.

I already did sumthing like this.

I made 1/2 and 1 pound brass wates with a holez in the center, a base with a long rod and a spring loaded impact absorber.

The absorber haz 2 latchez that stop it frum bouncing back up wen the absorber plate haz been pushed down at least 5mm. it haz 8 pockets to retain springz, allowing 2, 4, 6, or 8 springz to be installed.

The problem for me iz that the rezults made sense frum 2 different perspectivez. Twice the hite = twice az many springz can be compressed AND less than twice the energy compressez twice az many springz.

Zman wrote: "....The problem for me iz that the rezults made sense frum 2 different perspectivez. Twice the hite = twice az many springz can be compressed AND less than twice the energy compressez twice az many springz."

Dropping identical weights from different heights and timing or measuring speed (no springs!) will give a unique answer.

Try it!

The anser will be the acceleration uv gravity. (9.8 meterz per second iz it?) Why duz this help?

It seemz to me that this only verifiez the objection to the formula. Since gravity iz accelerating the wate for less time in the bottom haf uv the fall, it iznt adding az much kenetic energy.

Zman wrote: "The anser will be the acceleration uv gravity. (9.8 meterz per second iz it?) Why duz this help?"

You misunderstand! If you measure the speed at the bottom of a fall, you observe a direct conversion from potential energy into kinetic energy. If you do it from different heights and use ratios, the value of g cancels out and plays no role whatsoever.

Sure, you have to accept the universal law of gravity as correct. If you don't, you are wasting your time!

I am assuming I misunderstand. I'll take a few dayz to try to digest wut you mean by the rest.

Z: It sounds like you may be confusing momentum and kinetic energy. See if this link helps:

http://hyperphysics.phy-astr.gsu.edu/hbase/truckc2.html

Good link.

Duznt help tho. It still leavez the 'disapating twice the energy in haf the time = 4 X the force' idea intact.

A recurring problem in the teaching of physics comes from simplifications, over simplifications, and not so obvious assumptions.

Consider a car that will not start, being pushed by its driver, alongside a perfectly flat road. The car has some energy by virtue of its rest mass. That energy, E, equals the mass of the car 1000 kg * c squared. That's enough energy to end life as we know it, everywhere. So let's make a simplification. We'll ignore that energy entirely. Newton did, and he was pretty smart, and smart engineers do it every day.

The car has a half tank of gas, also containing a large amount of potential energy, even in the Newtonian sense. We'll ignore that too. We'll also ignore the fact that the car could be torched, and that many of its materials also contain energy in the Newtonian sense. We'll also ignore the energy content of the driver, which is the only energy source that is causing the car to move: A major bit of ignoring, because how else is the car going to get its kinetic energy? But will ignore it anyways, just for kicks.

The car rolls on tires that have a coefficient of friction of .01 (just a tad lower than for real tires). This friction is constant for speeds up to 50 km/h (very close to real world, too). The driver had hoped he could apply a force to the car for a while to accelerate it, and then simply have the car coast to a repair station. However, he finds that every time he stops pushing, the car slows down and stops. Its kinetic energy is being converted to heat in the tires.

So let's say he gets the car up to a speed of 1 meter per second, and stops pushing. [Kinetic energy: (1 / 2) * (1000 kg) * ((1 (m / s)) squared) = 500 joules] The car will come to a stop in some time, so we want to find the deceleration rate. F=MA, or A=F/M. If we use 10 m/s/s as the acceleration due to gravity to find the force the car exerts on the ground, we get 10,000 newtons. Therefore, the retarding force is 100 newtons. A=100/1000 or .1m/s/s. So, the time required for deceleration will be 10 seconds. The average speed will be .5 m/s. therefore the distance traveled will be 5 meters. (Work done by tires = 5M * 100 N = 500 NM = 500 joules.)

Copying the above paragraph, but doubling the speed:

So let's say he gets the car up to a speed of 2 meters per second, and stops pushing. [Kinetic energy: (1 / 2) * (1000 kg) * ((2 (m / s)) squared) = 2000 joules] The car will come to a stop in some time, so we want to find the deceleration rate. F=MA, or A=F/M. If we use 10 m/s/s as the acceleration due to gravity to find the force the car exerts on the ground, we get 10,000 newtons. Therefore, the retarding force is 100 newtons. A=100/1000 or .1m/s/s. So, the time required for deceleration will be 20 seconds. The average speed will be 1 m/s. therefore the distance traveled will be 20 meters. (Work done by tires = 20M * 100 N = 2000 NM = 2000 joules.)

This is an experiment you could do, with a friend driving (or pushing, if it is a really good friend). Make sure the engine is running so you have brakes. Make sure the road is flat and there is no wind. You could measure speed by counting off one meter paces, one per second, versus two per second.

Wouldn't it be a lot easier with something less massive like a push bike. Then again the harder you work the more likely you are to remember the lesson.