Simple physics question

[theNEOone]

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.


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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.

I think this is what I'm looking for....finally, i can now reconcile the mental picture of the force of a falling ball with the mathematics behind the action.


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[theNEOone]

Originally posted by: 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.

<edit>
spelling optional.

No, you're applying the formula incorrectly....acceleration is a constant here. It's just that there are two different accelerations (deceleration) based on the velocity of the ball at impact.

Think about the scenario of when a ball reaches terminal velocity....assume that terminal velocity is achieved after 60 seconds, and that a 60 second drop translates into 5,000 feet. Your explanation implies that a ball dropped from 10,000 feet would have a more violent impact because it's in the air longer, despite the fact that the both balls (5,000 ft. 10,000 ft.) will have the same velocity at impact.

I think speg summarized it nicely.


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[Freshgeardude]

it isnt hard to understand. the bowling ball is at a lower height. energy is conserved. so mgh(PE)=1/2mv^2(KE). less h = less energy= less velocity

 

[theNEOone]

Whoops. Wrong post.


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[Hacp]

Originally posted by: freshgeardude
it isnt hard to understand. the bowling ball is at a lower height. energy is conserved. so mgh(PE)=1/2mv^2(KE). less h = less energy= less velocity

He wanted the problem explained through F=ma.

 

[Fenixgoon]

first of all, F != m*a.

F = dp/dt = v*dm/dt + m*dv/dt

it's dependent on both the rate of change of velocity and mass.

let's assume our mass doesn't change, so now we have F = m*dv/dt = m*a

for impact events, the time required is on the order of dt, so we can say F= m*(v2 - v1) / (t2 - t1)

now let's look at when the ball comes to a momentary rest, v2 = 0...... F = m*(-v1) / (t2 - t1)

so the force felt is dependent on the velocity at impact. the longer the ball accelerates, the harder it hits.

 

[silverpig]

There is so much bloody fail in this thread. It's like, a botched abortion worth of fail. On a failboat. The SS abortaboat.

OP: You can pretty much ignore every post in this thread. There are a few that are pretty close, but you'll have to sift through a mountain full of pure crap to get to it.

If you want something pretty good, Fenixgoon's post just above this one is pretty correct, although the nitpick about F != m*a is a little much (even though he's correct that it's F = dp/dt = mdv/dt + vdm/dt).

 

[Evadman]

Originally posted by: theNEOone
Think about the scenario of when a ball reaches terminal velocity....assume that terminal velocity is achieved after 60 seconds, and that a 60 second drop translates into 5,000 feet. Your explanation implies that a ball dropped from 10,000 feet would have a more violent impact because it's in the air longer, despite the fact that the both balls (5,000 ft. 10,000 ft.) will have the same velocity at impact.

Uh, no. Unless you are taking aerodynamics, or a class way more advanced than it looks like, aerodynamic drag will not be included. However I am making it as easy as possible to understand, since you are trying to wrap your head around f=ma.

Hell, the formula is for a constant acceleration force. Where in the equation is anything about an opposing force (aerodynamic drag), where you need to net the 2 out? Especially considering the aerodynamic drag force is anything but constant. At low speeds the drag is almost nothing, while a falling human body at about 53 m/s will experience a (roughly) 9.81m/s^2 aerodynamic force that directly counters gravity. That means that there will be no further acceleration, and the velocity remains constant.

Long story short, since it's bedtime. 481.1805 newtons or 962.361 newtons are how much "energy" is put into the bowling ball by the acceleration of gravity over a 1 or 2 second drop. However a newton is a measure of force, while a joule is a measure of energy, and they can not be directly converted. to get joules, you need to take the end result of the force, and compute the energy (joules) now in the object as kenetic energy.

I accelerated the bowling ball using 481.1805 or 962.361 newtons. But now that the bowling ball is traveling at 9.81m/s or 29.43m/s the ball must lose that velocity (to 0 in this example), and the ball does that by transferring the kinetic energy into many different kinds of energy at impact (sound, heat, etc). That equation is KE=1/2mv^2. So:

1/2 * 5kg * 9.81m/s^2 = 240.59 joules.
1/2 * 5kg * 29.43m/s^2 = 2165.31 joules.

Now you have units that you can wrap your head around. Even if you don't know what a joule is, you can see that allowing a bowling ball to fall for 2 seconds results in an impact that must dissipate more than 9 times the energy than allowing a bowling ball to fall for 1 second.

Or, conver it into imperial units, and relate it to the bowling sport. 9.81m/s is about 22 MPH. A bowler could easily release a bowling ball in that range. However, 29.43m/s is about 66 MPH. At that speed, the bowling pins would splinter. The bowling ball probably would too if it hit a solid object.

I would do the math to show just how fast a single pin would be moving if it absorbed 100% of the energy from a bowling ball in a perfect horizontal collision (impossible in reality since theNEOone seems to care about reality) at 29.43m/s, but I am tired. A pin is 1.64 kg max if someone feels like doing the math. (1/2 * 1.64 kg * x = 2165.31)

 

[Born2bwire]

Originally posted by: silverpig
There is so much bloody fail in this thread. It's like, a botched abortion worth of fail. On a failboat. The SS abortaboat.

OP: You can pretty much ignore every post in this thread. There are a few that are pretty close, but you'll have to sift through a mountain full of pure crap to get to it.

If you want something pretty good, Fenixgoon's post just above this one is pretty correct, although the nitpick about F != m*a is a little much (even though he's correct that it's F = dp/dt = mdv/dt + vdm/dt).

Speg pretty much has it right. It isn't about the force of gravity, the energy involved, or just the velocity at impact. It is about the time it takes for the ball to come to a complete stop from a given velocity. You can take a ball that is moving at 1000 m/s or another ball that has a kinetic energy of 1 MJ but the real factor in what the OP is asking about is the amount of time that ball takes to decelerate upon impact. You can work this out as an impulse and get an equivalent constant force that is applied during the impact or treat it like a momentum problem, same diff.

Maybe we should put the bowling ball on a treadmill...

Originally posted by: Evadman

Originally posted by: theNEOone
Think about the scenario of when a ball reaches terminal velocity....assume that terminal velocity is achieved after 60 seconds, and that a 60 second drop translates into 5,000 feet. Your explanation implies that a ball dropped from 10,000 feet would have a more violent impact because it's in the air longer, despite the fact that the both balls (5,000 ft. 10,000 ft.) will have the same velocity at impact.

Uh, no. Unless you are taking aerodynamics, or a class way more advanced than it looks like, aerodynamic drag will not be included. However I am making it as easy as possible to understand, since you are trying to wrap your head around f=ma.

Hell, the formula is for a constant acceleration force. Where in the equation is anything about an opposing force (aerodynamic drag), where you need to net the 2 out? Especially considering the aerodynamic drag force is anything but constant. At low speeds the drag is almost nothing, while a falling human body at about 53 m/s will experience a (roughly) 9.81m/s^2 aerodynamic force that directly counters gravity. That means that there will be no further acceleration, and the velocity remains constant.

Long story short, since it's bedtime. 481.1805 newtons or 962.361 newtons are how much "energy" is put into the bowling ball by the acceleration of gravity over a 1 or 2 second drop. However a newton is a measure of force, while a joule is a measure of energy, and they can not be directly converted. to get joules, you need to take the end result of the force, and compute the energy (joules) now in the object as kenetic energy.

I accelerated the bowling ball using 481.1805 or 962.361 newtons. But now that the bowling ball is traveling at 9.81m/s or 29.43m/s the ball must lose that velocity (to 0 in this example), and the ball does that by transferring the kinetic energy into many different kinds of energy at impact (sound, heat, etc). That equation is KE=1/2mv^2. So:

1/2 * 5kg * 9.81m/s^2 = 240.59 joules.
1/2 * 5kg * 29.43m/s^2 = 2165.31 joules.

Now you have units that you can wrap your head around. Even if you don't know what a joule is, you can see that allowing a bowling ball to fall for 2 seconds results in an impact that must dissipate more than 9 times the energy than allowing a bowling ball to fall for 1 second.

Or, conver it into imperial units, and relate it to the bowling sport. 9.81m/s is about 22 MPH. A bowler could easily release a bowling ball in that range. However, 29.43m/s is about 66 MPH. At that speed, the bowling pins would splinter. The bowling ball probably would too if it hit a solid object.

I would do the math to show just how fast a single pin would be moving if it absorbed 100% of the energy from a bowling ball in a perfect horizontal collision (impossible in reality since theNEOone seems to care about reality) at 29.43m/s, but I am tired. A pin is 1.64 kg max if someone feels like doing the math. (1/2 * 1.64 kg * x = 2165.31)

Your units are wrong though, you are not calculating the force but the impulse, Newton-seconds, which is the same thing as the change in momentum. So you are not calculating how much "energy" you are putting in but the change in momentum.

Wait... you're taking the square of the acceleration? That isn't force either, your units are then kg-m^2/s^3, not kg-m/s^2. You could get energy from 0.5*m*a^2*t^2 assuming you make the correct adjustments for constants. You want to do m*v and use momentum conservation for collisions.

 

[Fenixgoon]

Originally posted by: silverpig
There is so much bloody fail in this thread. It's like, a botched abortion worth of fail. On a failboat. The SS abortaboat.

OP: You can pretty much ignore every post in this thread. There are a few that are pretty close, but you'll have to sift through a mountain full of pure crap to get to it.

If you want something pretty good, Fenixgoon's post just above this one is pretty correct, although the nitpick about F != m*a is a little much (even though he's correct that it's F = dp/dt = mdv/dt + vdm/dt).

hey, that's why rockets work

 

[silverpig]

Originally posted by: Fenixgoon

Originally posted by: silverpig
There is so much bloody fail in this thread. It's like, a botched abortion worth of fail. On a failboat. The SS abortaboat.

OP: You can pretty much ignore every post in this thread. There are a few that are pretty close, but you'll have to sift through a mountain full of pure crap to get to it.

If you want something pretty good, Fenixgoon's post just above this one is pretty correct, although the nitpick about F != m*a is a little much (even though he's correct that it's F = dp/dt = mdv/dt + vdm/dt).

hey, that's why rockets work

Rocket bowling ball? Now I'm interested