I'm sorry this thread was even started. The fact that it has stirred up so much controversy makes it look like the oppositng point of view has more merit than it does.
I'm sorry this thread was even started. The fact that it has stirred up so much controversy makes it look like the oppositng point of view has more merit than it does.
marijuologist wrote:
I'm sorry this thread was even started. The fact that it has stirred up so much controversy makes it look like the oppositng point of view has more merit than it does.
Teach the controversy!!!
Isn't that interesting, Marijuoligist!
Here is an interesting running physics question that I've been thinking about:
Why do you run slower times on the track in windy conditions? Since you are running in a circle, shouldn't the wind cancel itself out and have no negative effects on performance?
I have my own explanation for this, but I was just wondering what other people will come up with to answer this question.
Question for Jim Fiore wrote:
I'm a bit rusty on my physics 101 I must admit. Could someone provide me the answers (step by step) to the following questions.
1) I have an object on a frictionless incline (30 degree slope) that rests 3 meters high. If I let the object go, what will be the horizontal speed of the object when it gets to the end of the incline?
2) Same object on a frictionless incline, same height (3 m), 45 degree angle.
I see. With all the great physicists here, including JimFiore, not one of you could solve this basic physics problem for me?
Just conserve energy:
.5*m*Vtotal^2 = m*g*h
Vtotal = sqrt(2*g*h)
Vtotal will be the same for both problems because h = 3m in both instances.
1) Vhoriz = Vtotal*cos(30)
2) Vhoriz = Vtotal*cos(45)
I answered it twice but both times the messages got removed.
if it's so basic, then why did you have to ask it?
Question for Jim Fiore wrote:
I see. With all the great physicists here, including JimFiore, not one of you could solve this basic physics problem for me?
Many people could. No one wanted to waste their time doing it, because although objects on inclined planes (including runners) tend to accelerate down the inclined plane, this doesn't address the far more prevalent case: motion on a flat plane.
I'd also like to know how the Pose Method deals with runners going uphill--a valid concern, wouldn't you say? When you have a ball on an inclined plane, like you indicated before, will gravity somehow push the ball UP the inclined plane?
No. It won't. You have to do work on it.
roger penrose wrote:
Just conserve energy:
.5*m*Vtotal^2 = m*g*h
You're wrong already. Mass does not affect velocity. Wouldn't the instantaneous V be something like the f'(x) = 2*9.8m/s*h for a free falling object.
Please don't conserve energy. It's been awhile, someone who really knows the answer walk me through the problem?
He's right.
If you look closely he eliminated mass.
Vx = |Vf| cosine(theta)
Vx = [sqrt (2gh)]cosine(theta)
You can use energy to solve for final speed. That's what energy is used for. But to find the x component it is necessary to multiply by cosine(theta)
The other method is to use a free body diagram/ kinematics.
Tilting the X axis along the inclined plane of theta degrees, starting from rest, assuming no friction, no wind resistance then...
Sum Fx = ma
mgsin(theta)=ma
a= gsin(theta)
Vf^2= Vi^2 + 2ax
Vf^2= 0 + 2[gsin(theta)][h/sin(theta)]
Vf^2= 2gh
|Vf| =sqrt(2gh)
Vx = |Vf|cos(theta)
Vx = [sqrt (2gh)]cosine(theta)
Its the same difference.
Only in this particular case energy is a quicker solution. In projectile motion energy is only good to tell you the speed. But you must use forces and kinematics to find direction. Remember the V in energy is speed NOT velocity. To get velocity we need a direction. In this case of the incline, you already know direction. So energy is a good choice for 1D motion or a case like this incline where you know the direction. Then the horizontal component is simply |V|cosine(theta).
esmoke
One Clarification:
The original kinematics solution tilted the x-axis along the plane. This is the standard method to solve such a physics problem. The |Vf| is the final velocity, down the plane at theta degrees above the horizontal.
To find the Vx along the ground it is necessary to multiply by cosine(theta). So the final coordinate system is tilted back to make the x component parallel to the ground. This is also called the projected velocity along the ground (or the dot product).
esmoke
THAT'S WHAT I'M TALKIN BOUT!
He used to do physics
For girls on a message board
But gravity always wins.
This thread wears him out, it wears him out
It wears him out, it wears him out.
roger penrose wrote:
Here is an interesting running physics question that I've been thinking about:
Why do you run slower times on the track in windy conditions? Since you are running in a circle, shouldn't the wind cancel itself out and have no negative effects on performance?
I have my own explanation for this, but I was just wondering what other people will come up with to answer this question.
anybody got any ideas?
Similarly, shouldn't the same also be true of hilly courses where there is no net elevation gain or loss?
I think the answer is that while on paper the positives and negatives may look the same, their actual effect is that the negatives take a greater toll in hindering you than the benefits do in making up for them.
roger penrose wrote:
Why do you run slower times on the track in windy conditions? Since you are running in a circle, shouldn't the wind cancel itself out and have no negative effects on performance?
I told myself I wasn't going to post on this thread anymore because it had gotten so goofy. I will offer some observations on this one for roger though because it has nothing to do with gravity and because it's something I had a conversation with jtupper about last year.
The simple answer is a question: "Who said the effect of wind on a runner has to be a linear function?" (or more correctly, a linear function with bilateral symmetry. The analysis technique you're referring to is sometimes called "superposition" and it requires the system under study to be linear and bilateral). Wind load doesn't have to be (and isn't) linear. jtupper has a nice graph in RDF showing the effects of wind on a runner (head and tail winds). The graph is clearly non-linear and not bilateral. At one point I did an estimation using jack's data to come up with a simple ballpark value and the curve is around the 1.5 power (that is, the resistance increases as the 1.5 power of velocity of the wind). This is not precisely true at very small wind values. What is obvious from this curve is that you never get out of a tailwind what you put into the headwind, in much the same way as you never get out of a downhill what it took to scale the uphill.
dunes runner wrote:
Do Not Agree wrote:Gravity didn't change here. Of course a several hundred pound horse runs faster than a 130 pound runner.
In fact a 180 pound sprinter runs faster than a 130 runner, too.
and a 200 pound running back from USC is also faster than your hypothetical 130 pound runner.
This is why a hippopotamus runs faster than a gazelle.
In fact a 180 pound sprinter is faster at the 10k than Kenenisa Bekele.
And a 350 pound football player is faster than Paul Terget at the marathon.
Now you are adding facts. Before we were talking about raw speed. Now the only way you can make your point is by making the comparison over a longer distance. Way to distract the conversation from the original point. Go back to flipping burgers, George.
That's exactly what I was thinking JimFiore. It takes more energy to overcome a 10 mph wind for 10 minutes than it would take to overcome a 5 mph wind for 20 minutes if the resistance function follows the curve for a power greater than 1.
I'm just curious. What do you do for a living, Jim?
College professor. I teach in an electrical engineering technology program. Mostly circuit theory, semiconductor devices, and computer programming. "Superposition" is a well known analysis technique for a certain subset of circuits. The misapplication of it can lead to some crazy results.