Find radius of curvature and rate of change of speed at times t = 1 & 2 seconds...

1. Sep 3, 2015

bkw2694

1. The problem statement, all variables and given/known data

A football player releases a ball at 35° with initial velocity of 80 ft/sec. Determine the radius of curvature of the trajectory at times t = 1 sec and t = 2 sec, where t = 0 is the time of release from the quarterback's hand. For each case, compute the time rate of change of the speed.

Book answers: at t = 1: p = 142.2 ft, at = -6.58 ft/sec^2
at t = 2: p = 149.7 ft, at = 8.75 ft/sec^2

2. Relevant equations
an = v^2/ρ
at = dv/dt = acceleration in tangential direction
v = ρ(dΘ/dt)
a = dv/dt

3. The attempt at a solution

I'm really confused about how to tackle a problem like this that involves specific times. In class and in all the homework up to this point we haven't had an example of how to approach these types of problems, but I'm guessing there's a derivative or integral that I'm not seeing or understanding that's killing me, but the closest thing I could come up with is:

magnitude of acceleration = g = 32.2 ft/sec^2
v = ρ(dΘ/dt), which I split into: v dt = ρ dΘ.
I then made an integral: ∫(from 0 to t) 80 dt = ρ ∫(from 0 to 35*π/180) dΘ.
This gave me 80t = (.6109)ρ, when t = 1 gives me ρ = 131 ft.
I then tried solving for an with v^2/ρ, which gave me an = 48.87, which can't be possible as it makes magnitude a = g false...

2. Sep 3, 2015

SteamKing

Staff Emeritus
You are assuming that the football follows a circular arc after it leaves the player's hand. This may not be a valid assumption.

BTW, there is a mathematical definition of radius of curvature, which you should know:

I would evaluate the football's trajectory just like any other projectile motion problem, and apply the definition of radius of curvature as given in the linked article.

3. Sep 3, 2015

TSny

I would determine an first, and then use an = v2/ρ to get the radius of curvature.

4. Sep 3, 2015

bkw2694

So if it isn't following a circular arc, this means that ρ is changing, correct? Or is it that Θ is changing and causing ρ to change?

Also, I see the radius of curvature equation is: [1+(dy/dx)^2]^(3/2) / [(d^2y/dx^2)]. But I don't understand how to make that apply to this particular question since I'm not given any equation for y(x) or x(y).

5. Sep 3, 2015

SteamKing

Staff Emeritus
Like most projectiles thrown in a gravity field (ignoring air resistance), the ball's trajectory is going to be a parabola, rather than a circle. This is why I suggested that the path of the ball be analyzed like projectiles in general, without making unnecessary assumptions about the trajectory.

In short, the R.O.C. of the ball's path will be changing w.r.t. time and will not be a constant like it would if its path were in fact circular.

Pay attention to the formula for the radius of curvature where x and y are functions of time, IOW, x = x(t) and y = y(t).

By analyzing the trajectory of the football, you should be able to determine the coordinates of the ball's trajectory at the times requested, along with any time derivatives which are required to calculate the R.O.C.

6. Sep 3, 2015

bkw2694

Yeah, sorry, I just now realized what you meant and used the normal projectile equations to find ρ. By using the correct formula, I ended up getting ρ at t=1 is 142.19 ft.

So now the only problem I'm having is I can't seem to work out the correct at. Am I wrong in assuming that the magnitude of acceleration = g? Or does that change as well? Because I tried solving for an with 80^2/142.19 = 45.01, then substituting that into a = sqrt(an^2 + at^2) and it's giving me a way off answer.

7. Sep 3, 2015

TSny

What angle does the velocity vector make to the horizontal at t = 1 s?
What angle does the acceleration vector make to the normal direction at t = 1 s?
Yes, the magnitude of the acceleration vector must be g.

Last edited: Sep 3, 2015
8. Sep 3, 2015

SteamKing

Staff Emeritus
Now that you have calculated the R.O.C. of the ball's path at a particular time, you can assume for a brief instant that its path approximates that of a circle with a radius = R.O.C. In that case, knowing the tangential velocity and the R.O.C., you should be able to use the circular motion equations and calculate the tangential acceleration of the ball.

9. Sep 4, 2015

bkw2694

Okay, so I can see that the angle will be getting smaller, but I'm not sure what equation to use to figure this one out. Is this just a projectile equation I'm overlooking or is there calculus involved here?

Since I knew the answer from the book I worked backwards and found an, then solved to see what the velocity should be, which is 66.96 ft/sec.

10. Sep 4, 2015

haruspex

In case it is not clear from the other posts, you are asked for the rate of change of speed, whereas g is the magnitude of the rate of change of velocity. For the rate of change of speed you need to restrict to the tangential acceleration. How might you derive that from g?

11. Sep 4, 2015

bkw2694

Okay, so I used trigonometry to find at = gcos(Θ). And I suppose I can use the same thing to find an = -gsin(Θ). I tried plugging those into a = sqrt(an^2 + at^2), but I think that causes the Θ to disappear due to sin^2 + cos^2 = 1.

12. Sep 4, 2015

bkw2694

^whoops, I mixed those two up, at = -gsin(Θ); an = gcos(Θ)

13. Sep 4, 2015

TSny

It's basically just a projectile motion problem with the additional formula an = v2/ρ. You know the magnitude of the acceleration. Therefore, you can find the normal component if you know the angle between the acceleration vector and the normal direction. How is the normal direction related to the velocity direction? How can you get the direction of the velocity vector?

14. Sep 4, 2015

Staff: Mentor

If x = x(t) and y = y(t), show using the above equation for the radius of curvature that:

$$R=\frac{\left[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2\right]^{\frac{3}{2}}}{\left[\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{dy}{dt}\frac{d^2x}{dt^2}\right]}$$

Chet

15. Sep 4, 2015

bkw2694

I must be solving for v (when t = 4) incorrectly. a_n = v^2/p, but when I try doing v = 80cos(theta), then plug gcos(theta) = (80cos(theta))^2/142.2 I end up with theta = 44.3 degrees, but I know the correct answer should be more like 11 degrees.

16. Sep 4, 2015

TSny

Did you mean t = 1?
I don't follow what you are doing here. In the equation an = v2/ρ, v is the speed of the particle: v = (vx2+ vy2)1/2.
So you can get v at t = 1 s if you first determine vx and vy at t = 1 s using standard projectile motion equations.

Last edited: Sep 4, 2015
17. Sep 4, 2015

bkw2694

Sorry, I one of my other HW problems had t = 4 so I screwed that up.

Okay, I figured out my problem. Aside from using the wrong time, I was also using the wrong v. I was using v = 80cosθ, but v was obviously 80 ft/sec and provided as the initial velocity. The incorrect v value with cosθ was giving me some crazy answers and really throwing me off.

Huge thanks to everyone in this thread for bearing with me on this one. Now that I see how easy this problem actually was I'm really sorry for taking up everyone's time, but I really appreciated it!