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A Confusing Problem

  1. Mar 6, 2007 #1
    There is a truck with a rock stuck in the tire.
    The tire has a diameter of 36 inches.
    The truck is going 55 mph.
    There is a car behind it going 55 mph also.
    What is the minimum distance the car can be behind the truck so it won't get hit by the rock when it flies out of the tire?
    I did a problem like this before... but I never had to include the diameter of the tire.
    The answer has to be reported in inches.
    Can anyone please explain this problem to me?
    -o_0curiousraych:cry:
     
  2. jcsd
  3. Mar 6, 2007 #2

    HallsofIvy

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    Knowing the radius of the tire allows you to calculate the initial velocity of the rock. The tire has diameter 35 inches and so has circumference [itex]\pi d= 70\pi[/itex] which is about 220 inches.
    The car is going at 55 miles per hour which is 55 mi/hr(5280ft/mi)(12in/ft)= 3484800 in/hr= 3484800 in/hr(1/60 hr/min)(1/60 sec/min)= 968 in/sec. Since the tire is rolling over every inch of that, the tire must make 968/220= 4.4 revolutions per second. That means the rock is moving at 4.4 rev/sec(220 in/rev)= 968 in/sec. Now, imagine an object thrown up at angle [itex]\theta[/itex] with speed 968 in/sec. What will its range (distance until it hits the ground again) as a function of [itex]\theta[/itex]? What value of [itex]\theta[/itex] gives the maximum range and what is that maximum range? (The car had better be at least that far back to be certain it will not be hit by the rock.)

    I see I used the circumerence of the wheel, 220 in, twice in calculating the speed of the rock. It probably could have been done more simply!
     
  4. Mar 6, 2007 #3
    hmmm. obviously all the wheels are going 55mph and the rock flies backward at such...
    Edit: see above
     
    Last edited: Mar 6, 2007
  5. Mar 6, 2007 #4

    HallsofIvy

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    OMG. I was writing "NO" and pointing out that while the center of the wheel was moving forward at 55 mph, the points on the circumference were going at- then stopped to do the computation. While the center of the wheel (the axle and so the car) moves forward a distance of 220 cm (the circumference of the tire) a point on the circumference of the tire makes one rotation around the tire and so goes- 220 cm. Yes, the speed (but not the velocity) of a point on the circumference of the wheel is exactly the same as that of the car. That's why I divided and then multiplied by 220 above! Thanks.
     
  6. Mar 6, 2007 #5
    well while you were doing that, I was thinking about contact patch and how the size of the tire might influence the trajectory when the worst case approach was the obvious approach in terms of figuring a safe distance. (I should take heed having suffered 3 broken windshields here in Colo over the past 10 winters!)
     
  7. Mar 6, 2007 #6
    help

    I'm still not sure what you guys are talking about.
    please clarify. =[[[[:cry:
     
  8. Mar 6, 2007 #7
    you be in very good hands here, just some idle chatter about the problem.
    How are you coming along on this?
    So back on topic, did you figure out the angle which would throw the rock the farthest distance?
     
  9. Mar 6, 2007 #8
    yes. i already know that 45 degrees will fire the furthest. thats common sense. but what i'm still not getting is, why was i given the diameter if that isnt used in the range equation at all?
     
  10. Mar 6, 2007 #9
    Don't know. If someone knows, hopefully they will post. I think smaller tires would tend to propel a rock more vertically, as they are spinning faster and an object more likely to break loose higher up the rotation cycle. But Halls... said, figure the worst case, and I think thats good advice and arguable to the max if the TA marks you down for some oversight. (Trucks suck, they spit gravel like a cannon shot at about 5-10 degrees.)
     
  11. Mar 6, 2007 #10


    he said that the diameter was 35 in...then in his formula for pie times D....he has 70(pie)......i think this may be bad.....=\
     
  12. Mar 6, 2007 #11
    is their anyway....someone would redue the problem...using 35*pi...? so that way...we can figgure out how fast the tires movin? and then to do it in meters per second? that would be amazing
     
    Last edited: Mar 6, 2007
  13. Mar 6, 2007 #12
    that would be amazing wouldnt it
     
  14. Mar 6, 2007 #13
    Unless anyone has somrthing constructive to add to the soln, lets give it a rest and let the OP determine the flow from here.
     
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