Tonight's mythbuster

[Brian Fein]

yeah, that's what I meant when I said the helium ball would be in the air longer.

[BeefStewert]

Beef, I'm not sure I'm getting what you're saying about decelerating quicker.  All objects fall to earth at the same rate, thanks to gravity (and Mr. Newton).

I'm talking about deceleration in the x, not the y.  Gravity is a downward force, but the ball is traveling in both the x and y.  The air resistance (and ground) are the only things appling an opposite force in the x.  And ultimately we are talking about which ball will fly furtur in the x, not overall flight path.

[bsfins]

I try to give a little love to the the Local U,and it turns into a 3 page debate...

A stupid question,I don't think it really applies here,but wouldn't both the Gravity,and the wind resistance be constant,or the same for both balls?

Both balls are inflated to 13 PSI
The Wieghts of the balls are different,but they don't change during flight....

So Newtons law that inertia,applies?

Or have you guy's already explained this,and I just missed it...

[BeefStewert]

I try to give a little love to the the Local U,and it turns into a 3 page debate...

A stupid question,I don't think it really applies here,but wouldn't both the Gravity,and the wind resistance be constant,or the same for both balls?

Both balls are inflated to 13 PSI
The Wieghts of the balls are different,but they don't change during flight....

So Newtons law that inertia,applies?

Or have you guy's already explained this,and I just missed it...

Yes both gravity and wind resistance are the same forces.  But the effect of these forces changes with mass - which is changing with the addition of helium in to the ball.  So, if you consider F=ma if F is constant with wind resistance and m is smaller, then a (deceleration) has to be larger.  In the case of gravity and a downward force, F=mg.  Since g is constant and m is smaller, the downward force is smaller.  The problem with the downward force in a non-vacuum system is that there is a maximum (terminal) velocity that the ball can reach.  Once this is reached, it doesn't matter what the gravitational force is on the ball, it won't fall any faster - making air resistance in the x the only differing opposing force.

[bsfins]

Thanks,beef

I kind of assumed that they used the sledge hammer,and showed that both balls leave with the same velocity,and the wieghts don't change in flight,it would revert to inertia (being the rest is the same.)

I didn't think about the terminal velocity.

Damn,over Ten years away from doing physics,It all rushes back in one day...My head hurts....

[Brian Fein]

The flight in the x depends upon the flight int he y.  The flight in the x can continue forever if there's no force in the y (mg).  So the time in flight can only be found by simplifying the problem to firing a football straight up in the air and see when it comes down. 

Imagine the example of if you fired a bowling ball and a beach ball out of a cannon at the same time, same shape, same field.  Which would go farther?  I'd think the beach ball would, since its mass is less and it requires less impulse to change its inertia.

[BeefStewert]

The flight in the x depends upon the flight in the y.  The flight in the x can continue forever if there's no force in the y (mg).  So the time in flight can only be found by simplifying the problem to firing a football straight up in the air and see when it comes down. 

Imagine the example of if you fired a bowling ball and a beach ball out of a cannon at the same time, same shape, same field.  Which would go farther?  I'd think the beach ball would, since its mass is less and it requires less impulse to change its inertia.

The flight in the x only depends on the y since there is a fixed limit due to the ground.  So you want to simplify the problem by calculating the hang time with a straight up kick and then using average velocity to determine distance?  The problem with that is that the average velocity changes due to wind resistance when you kick with a trajectory.  You logic only works in a vacuum.  But I've already said that in a vacuum I believe the lighter object will go furthur.  To make the lighter object go less of a distance you can't be assuming a vacuum. 

Funny you use the bowling ball and beach ball example.  Because I think that in the real world, the beach ball doesn't go furthur.  I've kicked a soccer ball and a beach ball and I can't kick the beach ball nearly as far.  Maybe you are more special than me when it comes to this. 

[BeefStewert]

and it requires less impulse to change its inertia.

Brian this is why the air resistance makes such a difference with a ligher object. 

[Phishfan]

Imagine the example of if you fired a bowling ball and a beach ball out of a cannon at the same time, same shape, same field.  Which would go farther?  I'd think the beach ball would, since its mass is less and it requires less impulse to change its inertia.

Incorrect. As previously stated, air drag affects a lighter object more than a heavy object.

[Brian Fein]

I'm not disputing the existance of air resistance.

Never have

I also never stated anything about average velocity, nor simplifying the problem to a striaght up kick.  Its hard to explain what I'm trying to communicate over a message board.

At the same time, imagine a beach ball at a football game, being batted around in the crowd.  Imagine someone serves it up, volleyball style, hits the ball and it goes flying.

Now change it to a bowling ball.  Does it go flying?  Or does it move 2 inches and fall to the ground?

Maybe a bowling ball is too extreme, but what's the function of distance vs mass?

[Sunstroke]

Incorrect. As previously stated, air drag affects a lighter object more than a heavy object.

Not doubting the theory, but in practice, I can't see this as accurate. If I have a cannonball weighing 10 lbs in one catapult and a beach ball of the same size and shape, but only weighing in at 4 ounces next to it in an identical catapult. Both catapults fire with the same force and at the same angle. The mass of the bowling ball will affect its inertia significantly more than the lesser mass of the beach ball will affect its inertia.

That's it...I'm running away from this topic before I feel even geekier.




modified to add: While I was typing this, Brian posted that example above...the same example exactly that I used with Fausto at about the exact time Brian was typing. Interesting...

[BeefStewert]

You said time of flight could be simplified to a straight up kick.  I guess I didn't see how that was relative except to assume that time of flight is the same for both types of kicks - at which point you can jut multiply by average v to get total distance traveled (in both x and y).  Anyway, that doesn't matter if thats not what you were saying.

As for serving a bowling ball.... The bowling ball won't fly as far because the initial force from the hand is not sufficient to project it.  The beach ball obviously is.  So air resistance doesn't even become a factor.  You need to compare two object that can be projected with the kicking force to allow air resistance to enter the equation.  If you compare a volleyball to a beach ball, which can you serve furthur?

[Brian Fein]

OK, I looked it up - you guys are peaking my curiosity.

drag force is independant of mass.  Drag force is defined as:

F= ( Cd A p v^2 ) / 2

where
Cd = drag coefficient
A = frontal cross sectional area
p (rho) = density of the fluid its traveling thru (in this case, air)
v = velocity of the object

So, please tell me how the drag force differs for different mass objects.

If the drag force is the same on both, how does it affect one differently than the other?

The only thing I can see is the resultant force in the y (mg - Fd) of the falling object is different for each cause each has different weight, thus different net accelleration towards the ground.  But for the helium one, its mg - Fd - Fb, so there should be a small difference.

Am I not thinking of this right?

[BeefStewert]

Maybe a bowling ball is too extreme, but what's the function of distance vs mass?

There are 2 functions here.  First the impact force and the resulting acceleration on the object due to that force.  The second is the air resistance force and the resulting deceleration once the object is in motion.  Obviously if the first force is not sufficient to incude significant motion, the second factor never even enters the problem.

[BeefStewert]

OK, I looked it up - you guys are peaking my curiosity.

drag force is independant of mass.  Drag force is defined as:

F= ( Cd A p v^2 ) / 2

where
Cd = drag coefficient
A = frontal cross sectional area
p (rho) = density of the fluid its traveling thru (in this case, air)
v = velocity of the object

So, please tell me how the drag force differs for different mass objects.

If the drag force is the same on both, how does it affect one differently than the other?

The only thing I can see is the resultant force in the y (mg - Fd) of the falling object is different for each cause each has different weight, thus different net accelleration towards the ground.  But for the helium one, its mg - Fd - Fb, so there should be a small difference.

Am I not thinking of this right?

Force doesn't change.  deceleration changes - apply it to F=ma.  If m is bigger, a is smaller.

I've always stated that air restistance is the same for the same size balls.