How fast would a 750 lb. car have to be moving to drive a loop-de-loop and be held to the track by centrifugal force? Is there an equation that figures in weight and speed that could be used to figure this out?
I don’t have my physics 101 book in front of me but you probably don’t need to know the mass of the car. Do need to know the radius of the loop and if speed is constant through the maneuver. Are you on earth? Then you just need to make sure that the centrifugal force is enough to cancel out the approx. 9.8m/s[SUP]2[/SUP] acceleration of gravity.
the question that jumps to mind is: why do you want to know this, evil caneival?
Don’t forget to include the weight of the person driving the car in the equation, or you could end up the recipient of a Darwin Award.
I would imagine you’d need a large radius, so that the bumpers of the car don’t hit the ramp on too steep an incline, meaning you’d probably have to be going pretty damn fast.
The weight doesn’t matter. The formula for centripital acceleration is a=v[sup]2[/sup]/r where v is velocity and r is the radius. You want a to be equal to or greater than the acceleration due to gravity.
If your car is going 30 meters per second, a radius of ~90 meters should suffice, for example.
hijack-
“How fast would a 750 lb. car have to be moving”
A 750lb car would be one tiny car. My motorcycle wet weighed more than that.
end of hijack
Eric
For those mathematically challenged (approximated)
30 m/sec = 90 ft/sec = 5400 ft/min = ~1 mile/min = ~60 miles/hour
So apparently doing 75 mph would be more than enough, accounting for approximation errors.
it would probably be difficult to maintain a speed of 30 meters/sec (64mph) around a loop with a radius of 90 meters. but who knows, hey try it!
bryanmcc has a good point. You need to be going fast enough so that you are still going 30 m/s = 60 mph at the top of the loop, after you’ve converted some of your kinetic energy into potential energy.
The reduction in Kinetic Energy must be equal to the increase in Potential Energy. So,
[ul]
[li]mv[sub]1[/sub][sup]2[/sup]/[sub]2[/sub] - mv[sub]2[/sub][sup]2[/sup]/[sub]2[/sub] = mgh[/li]
or
[li]v[sub]1[/sub][sup]2[/sup] - v[sub]2[/sub][sup]2[/sup] = 2gh[/li][/ul]
Let’s plug in 9.8 ms[sup]-2[/sup] for g, 90*2 (2r) for h, and since we want the minimum speed to be about 60 mph, 30 ms[sup]-1[/sup] for v[sub]2[/sub].
[ul]
[li]v[sub]1[/sub][sup]2[/sup] - 900 m[sup]2[/sup]s[sup]-2[/sup] = 3528 m[sup]2[/sup]s[sup]-2[/sup][/li]
[li]v[sub]1[/sub][sup]2[/sup] = 4428 m[sup]2[/sup]s[sup]-2[/sup][/li]
[li]v[sub]1[/sub] = 66 ms[sup]-1[/sup][/li][/ul]
You’d have to start at over twice the final speed (~130 mph) assuming your car was coasting around the loop. With any kind of power over and above that lost by friction you could reduce your initial speed.
Like a loop with a constantly changing radius. Ever notice the shape of the loop on looping roller coasters? They’re not circular. They’re shaped like an inverted teardrop, with the radius gradually tightening towards the top.
I seem to remember the engineers doing this so the riders wouldn’t pull excessive Gs upon entering the loop but would still have enough speed to complete the loop.
But I’m no engineer. 
If you can make a car that weighs 750lbs with an engine that can make it go fast enough to go thru a loop you can probably just add wings and fly
Answer: as fast as possible.
Look, I don’t know know about you guys, but I’m not going to be driving through that loop at one mile over the hypothetical speed just to test what would happen.
Well, from what I can tell, it would be possible to try it, with a car, a test track, and a remote control. The problem one would run into would be making the track such that it wouldn’t fall apart.
Although, to be honest, I would be willing to try this, with the proper team on standby, and a lucrative Fox contract… hehehehe
Thanks for the answers, y’all.
Regarding the weight of the car, I haven’t a clue as to how much the average car weighs. I do recall hearing commercials for half-ton pickup trucks, though, and I figured a regular car would be a little lighter. Guess I was wrong.
130 mph, eh? Hmmm, so I won’t have to strap those Air Force surplus rockets to my Toyota Tercel after all… 
maybe 750 kg would be a better guess
-b
Cars generally weigh between 2000 and 3000 lbs. My current car (300ZX) weighs 3200, and is one of the heftier ones. My previous car (Integra) weighed around 2300 i think.
Quarter ton, half ton, and full ton trucks generally refers to the load they can carry, AFAIK. Could be wrong, but I do know they weigh a lot more than a quarter, half or full ton.
Another problem is that if you are close to the hypothetical minimum required speed at the top of the loop, this vehicle would be experiencing a net force upward, but just barely enough to keep it on the track. But because it is moving at 60+ MPH, wouldn’t the flow of air over the car tend to lift it off the track?
Suppose the vehicle and driver are approaching the top of the circle. Their speed and circular motion cause them to experience, say, a force 0.01 G away from the surface of the earth, because they are just barely compensating for gravity. From the driver’s inverted perspective, it as as if he is driving a car that weighs 1% of the car’s weight on the ground. When I try to imagine driving my car on the Beltway, with the car normal size but weighing only thirty pounds, I think it would probably go airborne and flutter through the air like a leaf.
But maybe if I put one of those big-ass spoilers on it…
You people are practically wrong completly! You don’t want centripidal acceleration, you want centripidal force!
If your car weighs 750lbs (very light) then that is it’s force due to gravity (yes, weight is a force.)
Now, the formula for centripidal force is:
F=((m)*(v^2))/R.
m is the mass of your car (NOT WEIGHT, MASS, to obtain mass, divide weight by force due to gravity (32.2 m/s/s), to help you out, the mass of your car is ~ 23.3.)
Now, you also need to know the radius of youe loop to compute the equation. But F in the equation must equal your car’s force due to gravity (750 lbs).
Let’s take the seemingly agreed upon speed of 130 mph. Plugging that into our equaiton we get a radius of 525 feet.
That is one LARGE loop. To use a more reasonable loop size (90 feet) we have a speed of 54 miles per hour. Going 130 miles per hour in a loop of 90 feet would pull excessive G’s, (a force of 4375 pounds, yeah, WAY too much, remember, the force increases exponentially with speed (or veloicity, rather.)) I don’t remember my lbs to G’s equaiton, but I guaruntee that you would probably die at that speed in that sized loop.
Hope this helps you out.
Thanks, Einstein. Did you actually read the thread before you spewed forth the Truth[sup]tm[/sup]?
Very clever, except for misspelling centripetal. Except that F=ma, so having either acceleration or force will work. Ain’t math wonderful?
Pick some units and stick with them. g = 9.8 ms[sup]-2[/sup] = about 32 ft s[sup]-2[/sup]. The units make it an acceleration, not ‘force due to gravity’. Since you started with a faulty value for g, your answer is going to be off.
So far so good.
Again, some reading comprehension would be useful. Bobort proposed a loop of 90 m radius (about 180 feet) in order to have the car only going about 60 mph. It was pointed out that in order to be going 60 mph at the top of the loop, it would have to be going faster upon entering the loop, unless it had an arbitrarily powerful engine. The 130 mph figure was the speed at which a car must enter a 90 m radius loop in order to be going 60 mph at the top of the loop. Get it?
Where to start? First, there is no lbs to G’s equation!!! G’s is a measure of acceleration, lbs is a measure of force. The closest thing you would have was to compare the force (your figure of 4375 lbs) to the weight of the car (750 lbs) and you’d get 5.8 Gs. However, your math is so wrong that I am compelled to start over.
Bobort’s post was completely correct. In order to not fall off the top of the loop, the car must be travelling fast enough so that the centripetal acceleration is equal to or greater than the acceleration due to gravity. Thus, we want a > 9.8 ms[sup]-2[/sup], and a = v[sup]2[/sup]/r. So, v[sup]2[/sup]/r > 9.8 ms[sup]-2[/sup].
For a loop of radius r, you want to be going at least sqrt(r*9.8ms[sup]-2[/sup]) at the top of the loop. For your 90 foot radius loop (27.4 meters), the speed is 16.4 ms[sup]-1[/sup]. Conversion to mph is 2.24 mph/ms[sup]-1[/sup], so the speed is only 36.7 mph, not your 54 mph figure.
Furthermore, you again have to take into consideration how fast you must go at the bottom in order to be going 36.7 mph at the top of the loop. From my post:
v[sub]1[/sub][sup]2[/sup] - v[sub]2[/sub][sup]2[/sup] = 2gh
Speed at the top (v[sub]2[/sub]) is 16.4 ms[sup]-1[/sup], so in this case (g = 9.8ms[sup]-2[/sup], h = 2 * 90 ft = 54.8m) we get:
v[sub]1[/sub][sup]2[/sup] - 268.96 m[sup]2[/sup]s[sup]-2[/sup] = 1074.08 m[sup]2[/sup]s[sup]-2[/sup]
Or,
v[sub]1[/sub][sup]2[/sup] = 1343.04 m[sup]2[/sup]s[sup]-2[/sup]
v[sub]1[/sub] = 36.6 ms[sup]-1[/sup] = 82.1 mph.
The car must be going 82.1 mph at the bottom of the loop to be going fast enough at the top to ‘stick’ to the track. Ignoring friction and power added by the engine.
The acceleration undergone by the car upon entering the loop is a=v[sup]2[/sup]/r, or (36.6 ms[sup]-1[/sup])[sup]2[/sup]/27.4m, or 48.9 ms[sup]-2[/sup]. This is about 5 Gs, and when added to the one G due to gravity would give a total downforce on the car of 6 times the car’s total weight.
Only in as much as it inspired me to go through the math again. Thanks for your help.