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Radius of Curvature Question

  • Thread starter Khamul
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  • #1
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Homework Statement


A golfer hits a golf ball from point A with an initial velocity of 50 m/s at an angle of 25° with the horizontal. Determine the radius of curvature of the trajectory described by the ball (a) at point A, (b) at the highest point of the trajectory.


Homework Equations


p = V2/an
Vy = V*sin([tex]\theta[/tex])
Vx = V*cos([tex]\theta[/tex])

The Attempt at a Solution


Hello again you wonderful people! Firstly, I would like to say that I only need help with part (a). Part (b) was quite easy as I found that the highest point of trajectory is when the velocities' x component was at its' lowest, and the normal acceleration component acting on it is gravity. (The answer was 209 m!)

My question however, more or less pertains to solving for part a's radius of curvature. My book says that it is 281 m. Yet...I am unsure how they found this. I thought of using kinematics to find some additional information, but there is too much lacking. We are given no time, final velocity, or even starting/ending distances.

Could someone help me out with this problem? I feel like i'm overlooking something silly...again....
 

Answers and Replies

  • #2
gneill
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At point A the ball has just been struck. You know its speed, and you know its direction (so you have a vector to describe it). What is the acceleration acting?
 
  • #3
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At point A the ball has just been struck. You know its speed, and you know its direction (so you have a vector to describe it). What is the acceleration acting?
That sadly, is what i'm having difficulty deducing. I don't have time or distance, i'm not sure what the acceleration is. The initial velocity is 50 m/s, with a 25 degree angle with the horizontal..I do not have a final velocity to go with this, nor an area of distance traveled, nor a time component.

Edit: I attempted to use gravity as the acting acceleration; as I did with part B. It didn't work..or am I doing something wrong?
 
  • #4
gneill
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You're only interested in what is going on at the instant in time that the ball is at point A. You don't need distance traveled nor time elapsed; there's no time or distance associated with an "instant". Final velocity will happen elsewhere and elsewhen.

What things go into making up a velocity vector?
 
  • #5
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An instantaneous velocity vector is the quotient of the position vector over a change in time t, no? I still do not believe I am following what you are advising I do.. I am missing a part of the picture, I know, but you are asking the very same question that is making me have pause with this. The highest point of trajectory was when the only acceleration was gravity, with the velocity being the Vx tangent to the line of trajectory.

There is no distance nor time associated with this instant, alright. But both comprise velocity, and then there's my problem of again, since I have this wonderful lack of all this knowledge (if I do not need it, I am unable to ascertain why.) I find myself unable to think of what the acceleration also might be.
 
Last edited:
  • #6
tiny-tim
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Hi Khamul! :wink:

The acceleration at A has two perpendicular components: radial and tangential …

you know the total acceleration, and you know the formula for radial acceleration, so … ? :smile:
 
  • #7
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Hello, Tiny! Thank you for getting back to me so quickly! :smile:

Well, I was able to finally through much head-mashing to figure out the answer, but i'm not exactly sure WHY it's the answer.

To obtain what the answer was, the equation was manipulated as before, however it was the acceleration, 9.81, that was multiplied by cos(25) (Got the answer of 281m!). Since this is at point A, where the ball was initially hit from in the first place, why did we do that? Sorry if this seems like a silly question, but in my head, the fact alone that gravity is a straight downward acceleration, wouldn't that be enough to make it the normal at any instant point for the golf ball?

Edit: I know that's a bit of a tricky thing to explain (I think, at least...?) but is it correct to assume this is the way it is, because the velocity is at a given angle, and we are only interested in the normal (which is perpendicular), the cos(25) is sort of a correction vector to compensate for the angle?
 
Last edited:
  • #8
tiny-tim
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Hello Khamul! :smile:
To obtain what the answer was, the equation was manipulated as before, however it was the acceleration, 9.81, that was multiplied by cos(25) (Edit: got the answer of 281m!). Since this is at point A, where the ball was initially hit from in the first place, why did we do that? Sorry if this seems like a silly question, but in my head, the fact alone that gravity is a straight downward acceleration, wouldn't that be enough to make it the normal at any instant point for the golf ball?
No, as I said, the acceleration has a radial and a tangential component.

You know the formula for radial is v2/r, but you don't know the tangential formula, but you do know the whole thing must be -gj

so the component of -gj in the radial direction must be v2/r (which is what you did :wink:)

goodnight! :zzz:​
 
  • #9
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Alright, I think i'm getting the hang of it a bit more. (Visualizing trigonometry has never been my strong point, as you may have noticed.) Thanks for the help, Tiny-Tim! Sleep well you sweet prince! :biggrin:
 
  • #10
tiny-tim
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oooh, it's morning! :zzz: :approve:

I wonder what's happened overnight? :rolleyes:
Sleep well you sweet prince! :biggrin:
hmm … I'm not sure about that …

the last time I heard it, the prince in question was about to sleep for ever

"goodnight sweet prince, and flights of angels to sing thee to thy rest" :redface:
 

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