Simple physics question

[theNEOone]

I didn't know where else to ask this - I'm sure someone here can help...

It's been a long time since I took physics (or any class for that matter), but this just popped into my head while I was in the bathroom.

From what I remember, F=m*a.

Why does it seem that a bowling ball dropped from one foot in the air will do alot less damage than a bowling ball dropped from 100 feet in the air? If a is constant at 9.8 m/s and the mass doesn't change, F would be the same in both cases. Am I just visualizing this incorrectly, and the damage (i.e. force) will actually be the same from a falling bowling ball, regardless of whether it's dropped from 10 feet or 100 feet in the air?

What am I missing?


=|

 

[Whitecloak]

more momentum and energy.

 

[Uhtrinity]

9.8 m/s^2, so in other words it accelerates, at least until it hits terminal velocity.

 

[rgwalt]

It isn't only the force of impact that needs to be considered... You also need to consider the energy of the object at impact. KE=1/2mv^2

 

[Vageetasjn]

The rate of acceleration is 9.8m/sec/sec. And it gains speed (and force) the longer it falls/travels. Force is really equal to mass*speed at impact.

 

[IronWing]

Originally posted by: theNEOone
What am I missing?

You are missing the point that physics must include experimentation. Otherwise it is just mathematical masturbation. I suggest ten bowling balls and a sturdy ladder. Targets of your own choosing though grad students are the commonly accepted standard target.

 

[theNEOone]

Originally posted by: Vageetasjn
The rate of acceleration is 9.8m/sec/sec. And it gains speed (and force) the longer it falls/travels. Force is really equal to mass*speed at impact.

I realize that a = 9/8 m/s^2. Your explanation still doesn't answer the question. And no, F != mass*speed (i.e. velocity).

It is mass*acceleration.

Originally posted by: rgwalt
It isn't only the force of impact that needs to be considered... You also need to consider the energy of the object at impact. KE=1/2mv^2

Hmm...can you explain why using F=m*a is not a valid way to solve the problem of a falling bowling ball?


=|

 

[CrazyLazy]

Your visualization assumes that it's on a treadmill trying and the bowling ball is trying to fly.

 

[speg]

Another way to think of it is:

At impact the ball has to (de)accelerate from a high speed to zero in a very short period of time. Thus, the faster it's going the greater it (de)accelerates, and thus, a greater force results.

 

[ElFenix]

force is lbs. yes, the weight of a bowling ball is pretty much the same 1 foot from the ground and 100 feet from the ground.

what you're really interested in is work/energy.

 

[Howard]

Do you understand that the bowling ball has a downward velocity which increases as it stays airborne until it reaches its terminal velocity? Do you know how long it takes in standard conditions? You'd think it'd be longer than the time it takes to fall 1 foot from zero velocity, right?

And you are not applying the correct equation. F=ma only gives you the force of gravity from the gravitational acceleration.

 

[Born2bwire]

Originally posted by: speg
Another way to think of it is:

At impact the ball has to (de)accelerate from a high speed to zero in a very short period of time. Thus, the faster it's going the greater it (de)accelerates, and thus, a greater force results.

This. It's the impulse that matters here.

 

[sciwizam]

Originally posted by: rgwalt
It isn't only the force of impact that needs to be considered... You also need to consider the energy of the object at impact. KE=1/2mv^2

Isn't it usually calculated as potential energy which is mass*acceleration due to gravity*height?

PE = mgh

 

[theNEOone]

Originally posted by: Howard
Do you understand that the bowling ball has a downward velocity which increases as it stays airborne until it reaches its terminal velocity? Do you know how long it takes in standard conditions? You'd think it'd be longer than the time it takes to fall 1 foot from zero velocity, right?

And you are not applying the correct equation. F=ma only gives you the force of gravity from the gravitational acceleration.

Ok. So it seems as if I'm applying the wrong formula. I'm using "force" in the colloquial sense - it seems as if "force" means something different in physics. In my head I'm asking myself, "what's the force of the bowling ball as it hits the ground?" Naturally, I choose F=m*a to try and figure it out....but apparently this is not correct.

It's still not clear to me why it's not valid to us F=ma. Please explain in a different way.


=|

 

[LostUte]

You are correct that F=ma. However, when the ball strikes the ground, a != 9.8m/s^2. When the ball strikes the ground, a is the negative acceleration bringing the ball to a stop. The larger velocity from a longer fall means a much greater negative acceleration is required to stop the ball. Therefore, the force was much greater.

Edit:
An equivalent equation to F=ma is F=dp/dt, where dp/dt is the rate of change of the momentum. The ball from with the longer fall has much more momentum, therefore the force is much greater.

 

[Random Variable]

The force you are interested in is the physical force of the the ball on the earth (and vice versa), not the force of gravity on the ball.

 

[TecHNooB]

Work = F x D

That's what you're missing

 

[Jeff7]

F = ma.
Gravity exerts a force on the bowling ball. That force does not change for as long as the bowling ball is within Earth's field of gravity. If it's sitting on a shelf, that force is the same as when the ball is falling. When it's on the shelf though, the shelf exerts a normal force on the ball, which is equal and opposite to that exerted by gravity.

If you drop the bowling ball from two points, one low and one high, the force will act to accelerate the ball for a longer period of time when it is dropped from high up.

If you were to exert a force of 1 Newton on something for 1 second, versus exerting that same 1 Newton on something for 10 seconds, you'll store more momentum in the bowling ball.

 

[mageslayer]

Originally posted by: sciwizam

Originally posted by: rgwalt
It isn't only the force of impact that needs to be considered... You also need to consider the energy of the object at impact. KE=1/2mv^2

Isn't it usually calculated as potential energy which is mass*acceleration due to gravity*height?

PE = mgh

You are both right. Total Energy= GPE + KE
As the ball hits the target, this energy is transferred, but the energy varies with speed and height. This is why different heights produce different impacts.

 

[Hacp]

The acceleration you are looking for is the acceleration of the ball as it hits the ground and bounces back up. At least that's what I gather. I could be completely wrong.

So you measure velocity a few milliseconds before it hits the ground, measure it a few milliseconds after it hits the ground, add the two absolute values, and divide by the time in between to get the accelleration.

 

[Howard]

Originally posted by: theNEOone

Originally posted by: Howard
Do you understand that the bowling ball has a downward velocity which increases as it stays airborne until it reaches its terminal velocity? Do you know how long it takes in standard conditions? You'd think it'd be longer than the time it takes to fall 1 foot from zero velocity, right?

And you are not applying the correct equation. F=ma only gives you the force of gravity from the gravitational acceleration.

Ok. So it seems as if I'm applying the wrong formula. I'm using "force" in the colloquial sense - it seems as if "force" means something different in physics. In my head I'm asking myself, "what's the force of the bowling ball as it hits the ground?"

Not sure how you would figure that out. Maybe if you set up a tough strain gage at the drop site and plotted it on a graph, since the "force" lasts longer than an instant.

I think you should just use KE = 0.5*mv^2, as rgwalt suggested.

 

[speg]

Originally posted by: theNEOone
Ok. So it seems as if I'm applying the wrong formula. I'm using "force" in the colloquial sense - it seems as if "force" means something different in physics. In my head I'm asking myself, "what's the force of the bowling ball as it hits the ground?" Naturally, I choose F=m*a to try and figure it out....but apparently this is not correct.

It's still not clear to me why it's not valid to us F=ma. Please explain in a different way.


=|

You are fine with F=ma. Again, at impact the ball has to (de)accelerate from a high speed to zero in a very short period of time. Thus, the faster it's going the greater it (de)accelerates, and thus, a greater force results.

Let's work through your example:

Dropping a ball from 1m will give it a velocity of:

v = sqrt(2ad) = sqrt(2*9.8*1) = 4.42 m/s

Lets say the impact of hitting the ground takes a tenth of a second. The acceleration during that period will be
~4.42/0.1 = 44.2m/s^2

Thus the force of the impact will be F= m44.2

Similarly for a 10m drop, the velocity at impact will be 14m/s - and the acceleration during impact will have to be ~140m/s^2

Thus the force will be F=m140.

 

[TecHNooB]

Eezee should explain this for fun, since he's got that PhD and all

He's probably single handedly repairing the LHC

 

[RGUN]

Originally posted by: speg

Originally posted by: theNEOone
Ok. So it seems as if I'm applying the wrong formula. I'm using "force" in the colloquial sense - it seems as if "force" means something different in physics. In my head I'm asking myself, "what's the force of the bowling ball as it hits the ground?" Naturally, I choose F=m*a to try and figure it out....but apparently this is not correct.

It's still not clear to me why it's not valid to us F=ma. Please explain in a different way.


=|

You are fine with F=ma. Again, at impact the ball has to (de)accelerate from a high speed to zero in a very short period of time. Thus, the faster it's going the greater it (de)accelerates, and thus, a greater force results.

Let's work through your example:

Dropping a ball from 1m will give it a velocity of:

v = sqrt(2ad) = sqrt(2*9.8*1) = 4.42 m/s

Lets say the impact of hitting the ground takes a tenth of a second. The acceleration during that period will be
~4.42/0.1 = 44.2m/s^2

Thus the force of the impact will be F= m44.2

Similarly for a 10m drop, the velocity at impact will be 14m/s - and the acceleration during impact will have to be ~140m/s^2

Thus the force will be F=m140.

And theres your answer. You are just looking through the wrong frame of reference so to speak. Your reasoning is correct in the sense that at 100 or 1 ft the ball will be subject to the same force due to gravity, but that is not the force causing damage at the impact. That would be a new acceleration that would result in stopping the ball over an interval of time. Of course its not this simple because of elastic and plastic effects, but it would answer your question.

 

[Evadman]

Expand it out and add units and you will figure it out. You are missing the crucial 'how many seconds' is pivotal in acceleration. The equation is the number of newtons PER SECOND that are being applied.

force = mass * acceleration
force (in newtons) = mass (in kilograms) * acceleration (in meters per second per second)
force (in newtons) = mass (in kilograms) * acceleration (in meters per second^2) * (how many seconds)

Example 1: How much force does a 5 kg bowling ball have after falling for 1 second assuming 9.81 m/s gravitational rate?
Substitution: force (in newtons) = 5 (in kilograms) * 9.81 (in meters per second^2) * 1 (how many seconds)
Result: 481.1805 newtons = 5 kg * (9.81^2) * 1

Example 2: How much force does a 5 kg bowling ball have after falling for 2 seconds assuming 9.81 m/s gravitational rate?
Substitution: force (in newtons) = 5 (in kilograms) * 9.81 (in meters per second^2) * 2 (how many seconds)
Result: 962.361 newtons = 5 kg * (9.81^2) * 2

Explanation: A bowling ball will need to dissipate 2 times the amount of energy when falling for 2 seconds than falling for 1 second. That energy is exchanged from the potential energy you gave the ball as you lifted it up a ladder tot he kenetic energy gained as it falls.
Further explanation: Just how high are the bowling balls dropped from in order to fall for 1 second or 2 seconds? Assuming no air resistance:
1 second = 9.81 m/s = 9.81m from impact point.
2 seconds = 9.81m + (9.81*2) = 29.43m from impact point. The reason the distance is 3x as far as 1 second, is because the 9.81m/s acceleration rate is applied per second. So after second 1, the ball is moving at 9.81 m/s. At the start of second #2, the ball already has 9.81 m/s acceleration, but another 9.81m/s^2 of acceleration is attached to it. So during second #2, the ball moves 9.81m + another 9.81m. So the total movement in second #2 is 9.81*2. If there were a second #3, the ball would move 9.81*3 durign that second, since at the end of second #2, the ball was already moving at 9.81m/s. So to fall for 3 seconds, the ball would have to be 9.81m + (9.81m*2) + (9.81m*3) = 58.86m from impact point.

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