You have a helium balloon in the center of your car, on the ceiling (where it naturally sits). You're going 60mph and you slam on the brakes. What happens to the balloon?

It pops when you slam into the back of the guy in front of you, who you didn't see in time because you were fiddling with the balloon.

:rofl:

Hmm... Gonna look like an idiot, and give this one a shot.

I think it would move forward. It has 60mph worth of momentum, and will have to lose this just like everything else in the car.

Think about it this way - gravity pulls a helium balloon down, yet it goes up. Why does this happen?

^buoyancy.

but submarines/ships have tons of buoyancy, and still have a buttload of momentum.

balloon moves forward.

I think the baloon would go backward. It is lighter than air. It would float on the air that is still moving forward while the car is stopping. Think about the little glass things with different colored liquids in them.

^Ah... good thought. The bulk of air in the car would be forced to the front, thus increasing it's density. This would force the much lighter helium to "rise" to a less dense area, which would be further back. I hate it when I miss something as obvious as that. :)

it'll roll towards the front of the car, like everything else. Inertia.
Unelss it's proped against something or it's oddly shaped.

Helium is less dence/lighter than air. That's why the baloon floats.

It also depends on how much your front end sinks when you hit the breaks.
If it creates an angle in the car it won't roll forward too far, and once the speed has gone down it'll roll back.

I think the baloon would go backward. It is lighter than air. It would float on the air that is still moving forward while the car is stopping. Think about the little glass things with different colored liquids in them.

Correct!
Vectorwolf too.

That's not what happened when my mom slammed on the breaks.
All of the balloons went forward. :D:

Every time I've driven with helium ballons in the car they move forward when you accelerate and backward when you brake.
Air is denser than Helium and has more momentum.
Try it; bet almost anyone here can afford a helium balloon.

Remains there i think, law of relativity. I could be totally wrong.

Everyone------try it and report your experience. This aint rocket surgery!

This is a much harder problem... than one might first think...

its true that when braking air density does increase in the front of the car, but all the air does not go to the front of the car... there is still air in the back of the car... as proven by non suffocating rear seat passengers...

pressure from the higher density of air in the front of the car causes a pressure gradient that decreases towards the back of the car...

in order to figure out which way the balloon moves.. you'd have to calculate the pressure gradient... and then figure out how much force the balloon puts on in the direction of the front of the car due to an opposite force slowing down the balloon from moving all the way to the front of the car...

anyway... its much too complicated of a calculation... and i don't even have the volume of air inside a car cabin...


its much easier to just go try it...

Simple - the balloon is traveling at 60 mph with the car. The car slows down, the balloon will still be traveling at 60 mph until friction or an immovable object in the car slows it as well. In this case the balloon is already contacting the roof of the car, and it's a round object. If it doesn't have a string on it, it will roll like a ball forward along the roof of the car until it contacts the windshield or other more vertical surface where it will stop. If it has a string and weight to keep it from rolling, then the friction of the roof against the surface of the balloon will "brake" it while it slows from 60 mph.

If it were tethered in the car but not touching anything, then it would swing forward relative to its position in the car, as it's tether point would be slowing but the balloon would not be.

If it were somehow freefloating in a car or other moving object (like a large trailer) with just enough helium in it to make it neutrally buoyant in the air, then it would float forward until the drag of air on the surface of the balloon brought it to a stop.

This is all due to the fact that the balloon, moving at 60 mph, has momentum. See below.

Its buoyancy relative to the air has jack to do with the equation - it just means that instead of rolling around on the floor like a tennis ball would be (or if the tennisball would be on a string suspended from the roof, then it would swing forward as you brake), it's floating in the air. But the mass of the latex and the helium is still moving at 60 mph at that movement has to be arrested somehow. The friction of the air, or the car's ceiling or impact with the car's interior will do that nicely.

About the balloon's momentum. It has some. That's why it will move forward. Now that all said, it will not be a very "violent" thing as the balloon has very little momentum. The physics equation for momentum is "momentum = mass*velocity" and the unit of measure of momentum is mass*distance/time. One might argue that the balloon is "weightless" because it floats - and that is essentially true - because weight is not a measure of mass, but is instead a measure of relative gravitational pull. The buoyancy of the helium in the balloon "outweighs" the gravitational pull on the mass of the balloon, therefore it floats upward relative to other objects. But the balloon still has mass. The latex in the balloon has a mass of a few grams, and the helium in the balloon also has a mass of a few grams. If it has mass and it is in motion, it has momentum. That momentum is what causes the balloon to move forward in the car.

Easy one.

Do the damn experiment for yourself!! Please report your findings.

every time i try this i keep forgetting the sunroof is open .so mine goes out the sunroof

Here's another question for y'all.

A RX8 of mass M_L drives around a circular track of radius R and mass M_T. The track forms the edge of an otherwise massless wheel which is free to rotate without friction about a verticle axis. The RX8 starts at rest and is accelerated to a final speed v relative to the track. What is the final speed v_f of the RX8 relative to the floor that the track is suspended above?

If somebody get's this I'll post another one, that has a counter-intuitive answer. :)

0 ;)

Nope :p:

Are we to disregard all gyroscopic (sp?) effects of clutch wheels, drive shaft, wheels, axles, etc.? or do we assume these all cancel each other out? BTW, I have no idea (read: too lazy to do the math), it's been years since my statics and dynamics classes. :)

The latex in the balloon has a mass of a few grams, and the helium in the balloon also has a mass of a few grams. If it has mass and it is in motion, it has momentum. That momentum is what causes the balloon to move forward in the car.

Easy one.

The mass of the air + latex for the given volume of the balloon is less than any equivalent volume of air in the car. That is, it is less dense, even when you include the latex. When the car decellerates, all of the air in the car has the same forward speed. It all has momentum. The parcel contained in the volume of the balloon has the least momentum of any parcel of air in the car. It will be displaced by other parcels of air which have a greater momentum. - it will be displaced towards the rear of the car. Easy one :).

I always drive with my windows down. So I guess I can't do the experiment. Man I'm lazy.

Here's another question for y'all.

A RX8 of mass M_L drives around a circular track of radius R and mass M_T. The track forms the edge of an otherwise massless wheel which is free to rotate without friction about a verticle axis. The RX8 starts at rest and is accelerated to a final speed v relative to the track. What is the final speed v_f of the RX8 relative to the floor that the track is suspended above?

If somebody get's this I'll post another one, that has a counter-intuitive answer. :)

The final "ground" speed would be the same as the top speed on a flat road, or until redlined in 6th gear, whichever came first; depends on the relative masses.

For example, if the car and wheel had the same mass, they would have the same velocity, in opposite directions. By the time the speedometer reads 148, the car's "ground" speed is only 74. The track's speed is 74 in the opposite direction. The car is "seeing" only 74mph worth of wind resistance. That means it can continue to accelerate, whereas normally this would be the aerodynamic limit. By the time it reached 9000 rpm in 6th gear , the speedo would be reading 180 mph (?) but the car would only be experiencing 90 mph worth of air resistance. The ground speed would thus be redline-limited. I am assuming no air resistance for the track itself.

If the track was much lighter than the car, the ground velocity of the track would be much higher than the car's. Thus, redline would come at a lower "ground speed". If the track was much heavier than the car, the car would have a higher ground speed than the track's. Given a heavy enough track, the car's ground speed would be limited by wind drag rather than redline.

Are we to disregard all gyroscopic (sp?) effects of clutch wheels, drive shaft, wheels, axles, etc.? or do we assume these all cancel each other out? BTW, I have no idea (read: too lazy to do the math), it's been years since my statics and dynamics classes. :)

Yeah, just answer in terms of v, M_L, and M_T. I won't make it that hard, but the next problem is a gyroscope problem, although one that's pretty simple as gyroscopes go.

The final "ground" speed would be the same as the top speed on a flat road, or until redlined in 6th gear, whichever came first; depends on the relative masses.

For example, if the car and wheel had the same mass, they would have the same velocity, in opposite directions. By the time the speedometer reads 148, the car's "ground" speed is only 74. The track's speed is 74 in the opposite direction. The car is "seeing" only 74mph worth of wind resistance. That means it can continue to accelerate, whereas normally this would be the aerodynamic limit. By the time it reached 9000 rpm in 6th gear , the speedo would be reading 180 mph (?) but the car would only be experiencing 90 mph worth of air resistance. The ground speed would thus be redline-limited. I am assuming no air resistance for the track itself.

If the track was much lighter than the car, the ground velocity of the track would be much higher than the car's. Thus, redline would come at a lower "ground speed". If the track was much heavier than the car, the car would have a higher ground speed than the track's. Given a heavy enough track, the car's ground speed would be limited by wind drag rather than redline.

I dont' think wind resistance matters in this one, because it is simply assumed the car reaches a final speed v to be left in variable form. Don't use numbers in place of variables like assuming redline in 6th gear. Just assume it reaches speed v and answer using a quantitative expression using the abstract variables given. :)

I dont' think wind resistance matters in this one, because it is simply assumed the car reaches a final speed v to be left in variable form. Don't use numbers in place of variables like assuming redline in 6th gear. Just assume it reaches speed v and answer using a quantitative expression using the abstract variables given. :)

If we disregard friction, drag and mechanical limits, Vf approaches the speed of light. Vf --> C.

If it was being continually accelerated, yes it would approach c, but it is not continually being accelerated. It is accelerated to a final speed v, so therefore if it is a final speed, then the speed is constant (ie acceleration=0). v can be 40, 100, or 150mph, or any reasonable value for the car to be going, so long as the effect from General Relativity is negligible. Treat v like an independant variable with the domain being any speed attainable by an RX8. Drag doesn't matter. The final speed is constant: it is in equilibrium. Whether this is because it reached a terminal speed from drag or whether there was a driver (whose mass is included in the mass of the car) who decided to stop accelerating at 30mph is irrelevant.

So you could say that the car is continually radiating heat and losing energy and experiencing drag, but then the engine is supplying enough power to have the net force affecting the linear speed sum to zero. Likewise you could disregard drag and say no forces are affecting the linear speed of the car: it doesn't matter so long as you use the car was accelerated to a final speed, and the new assumption that this equation is only being applied over possible values for a car, so the effect of General Relativity doesn't matter.

An example (but wrong) answer could look something like:

v_f=v/(1-(M_T/M_L)^(1/2)) :)

Here's another question for y'all.

A RX8 of mass M_L drives around a circular track of radius R and mass M_T. The track forms the edge of an otherwise massless wheel which is free to rotate without friction about a verticle axis. The RX8 starts at rest and is accelerated to a final speed v relative to the track. What is the final speed v_f of the RX8 relative to the floor that the track is suspended above?
Hmmmm... It's been 10 years since I did any problems like this, so I'm rusty. But I'll try to stumble through the logic...

The force of the RX8's wheels causes the RX8 to move in one direction along the track while the wheel spins in the other direction. The ratio of the two speeds depends on moment of inertia of the wheel and of the car. For a thin circular plate, moment of inertia is:


I = 0.5 * m * (R ^ 0.5)


Assuming the track is thin, its moment of inertia is:


It = 0.5 * M_T * (R ^ 0.5)


If the car was fixed and only the track moved, the forces involved would be:

Tw = torque at track axis caused by force of car's wheels
Fw = Force of wheels
R = track radius
Tw = Fw * R


If the only thing causing the wheel to spin is the force from the car, then:


At = angular acceleration of the track
Tw = I * At


combining:


Tw = Fw * R = I * At
At = Fw * R / I


However, some of the force from the car's wheels is used to accelerate the car and keep it going. If the car were on a track that it couldn't move:


Ac = acceleration of car
Fw = M_L * Ac


And here's the part where I scratch my head. I have two equations for two different situations.... can we combine them by substituting Fw? I'm not sure, though this is what it'd look like:


At = M_L * Ac * R / I
At = M_L * Ac * R / (0.5 * M_T * (R ^ 0.5))


And for kicks the ratio of how quickly the two accelerate, while simplifying the equation (again, I'm not sure how valid this is):


At/Ac = 2 * M_L / M_T * (R ^ 0.5)


If that was valid (though I don't think it is), I suppose one could then determine what the angular velocity is in terms of the car's velocity, which would be known depending on how long the car actually accelerated before the system reached steady state. However, this reasoning ignores the moment of inertia of the car itself spinning around the track's axis. If we were to account for that, then maybe we need to treat this as a conservation of energy or angular momentum problem.

If the car's center of mass is radius R away from the center, I think (but I'm not positive) its moment of inertia is:


Ic = M_L * (R ^ 2)


But I gotta get back to work, so I can't spend time working through summing up the moments. I suspect I've taken the wrong approach anyway and have made things more complicated than they need to be (back in the day, I used to find this stuff intuitive and easy... now I can't remember any of it!)

This is the part where I used to hope for the professor's mercy (and partial credit) when I ran out of time :mdrmed:

Well, you have part of it right. It is esentially an angular momentum problem, with frames of reference thrown in there too.

You are right about the moment of inertia being M_L*R^2 for the track.

Angular momentum is conserved because there are no external torques being applied. The total angular momentum L_z is equal to zero since the system starts from rest. You can solve for the angular velocity using angular momentum, and then turn that into a linear speed. Then you can relate it to the other speed given in the problem.

Ahhh... For all the logic that seems to hold true in this case, the illogical is true. Given a flat interior roofline, the balloon will likely move backward, not forward. Weird, eh?
We actually tested this a couple weeks ago in my wife's MDX without actually intending to test it.

We went to a local restaurant for dinner, and the kids got balloons as a parting gift. On our way home, then had just let the balloons free roam in the car. They basically pretty much stayed hovering around the second row of seats. Suddenly a complete dillhole pulled out in front of us and I had to slam on the brakes to avoid crinkling up the front end. Immeidatly upon doing this, the balloons scooted to the back of the vehicle. Why? Because as the nose dipped, the roofline tilted forward making the section of roof at the back of the car higher than that in the front. I'm assuming that beause the mass of the balloon was small enough the energy it had as a passenger in the vehicle wasn't enough to overcome the energy pushing it up against the roofline of the car.

I found it pretty amusing actually, though my wife wasn't as impressed.

What's all this then? Friction? Preposterous!

Probably what has most people confused about the balloon question is that they don't realize that the air they breathe has mass, and when accellerated in the car, that mass has momentum. Thus, when you stomp on the brake, the mass of the air has a tendency to move forward. Since the air is under pressure though (14.6 psi at sea level) the pressure gradient changes very little when coming to a stop. However, the lower mass of the balloon will be displaced backwards by the momentum of the air. Imagine a soda bottle full of water with an air buble in it. If you hold the bottle on its side, the air bubble will float roughly centered at the top. if you move it forward at constant velocity, the air bubble will stay pretty close to centered after the initial acceleration (this is assuming that you keep it perfectly level). Upon coming to a quick stop though, the air bubble will move backwards. The lower mass of the air has less momentum than the higher mass of the water per unit volume, and thus the air bubble is displaced to the back or bottom of the bottle. The air in your car will act similarly to the water in this bottle experiment, and the balloon will act similarly to the air bubble.

^^^ That's an excellent analogy actually.

If it was being continually accelerated, yes it would approach c, but it is not continually being accelerated. It is accelerated to a final speed v, so therefore if it is a final speed, then the speed is constant (ie acceleration=0). v can be 40, 100, or 150mph, or any reasonable value for the car to be going, so long as the effect from General Relativity is negligible.

Ok, I misinterpreted the original problem! Doh.

How about Vf = V * (M_L/(M_L+M_T))

You are extremely close, but not quite there.

You are extremely close, but not quite there.


Oops

Vf = V * (M_T/(M_L+M_T))

Yep, there you go :)

Vf = V * (M_T/(M_L+M_T))
How about working through the solution for us old timers who don't remember dynamics that well? (and for those that aren't familiar with conservation of angular momentum)

The balloon will move forward IF and only IF it is in contact with the roof of the car. being in contact with the car would make it... a part of the car so inertia and momentum will play an affect.

BUT, if the helium baloon is suspended in mid air inside the car.. meaning the balloon itself is free from contact anywhere else on the car, except for the string, THAN the balloon will move backwards.

The balloon will move forward IF and only IF it is in contact with the roof of the car. being in contact with the car would make it... a part of the car so inertia and momentum will play an affect.

BUT, if the helium baloon is suspended in mid air inside the car.. meaning the balloon itself is free from contact anywhere else on the car, except for the string, THAN the balloon will move backwards.
People have brought up friction with the car's ceiling more than once... I'm a little surprised.

Contact with the car's ceiling will make the balloon roll a bit, but the air moving forward has a much more significant effect: it'll push the balloon backward whether or not the balloon is touching the ceiling (since the balloon is lighter than air).

It'll only cost ya $.50 for a balloon at a card store to prove it to yourself in all situations. Then give it to some little kid to play with.

How about working through the solution for us old timers who don't remember dynamics that well? (and for those that aren't familiar with conservation of angular momentum)

Total angular momentum combines the track and the car, and equals 0. That is L_Tot= -I_T*w_T+M_L*R*v_f. Angular momentum is conserved, so L_Tot=0.

M_T*R^2*w_T=M_L*R*v_f

so w_T*R=(M_L/M_T)*v_f

I solved for w_T*R because linear velocity in terms of w is v=wr.

The track has a velocity towards the car, so v=v_f + w_T*r
v=v_f + (M_L/M_T)*v_f

Solving for v_f you get

v_f=v/(1+(M_L/M_T))

i gotta retake dynamics....