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Mechanics problem - Serway/Jewett 4.61

  1. Oct 27, 2009 #1
    A person standing at the top of a hemispherical rock of
    radius R kicks a ball (initially at rest on the top of the rock)
    to give it horizontal velocity vi. (a) What
    must be its minimum initial speed if the ball is never to hit
    the rock after it is kicked?





    I tried solving this but am getting the wrong answer. Please tell me what I'm doing wrong. I first took the vertical distance the ball travels and used R= 0 x t + 1/2 g t2. So I got time t of travel as sq. root (2R/g). Then I took this travel time and used it for the equation of motion in the horizontal direction. So in the horizontal, R < vi x t + 1/2 a t2. Since a is 0, taking time from the above equation, I get vi > (Rg/2)1/2.

    But the solution says the answer is (Rg)1/2. What am I doing wrong?
     
  2. jcsd
  3. Oct 27, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You are assuming that the ball hits the ground exactly at the edge of the hemisphere, for one thing. Is there any reason to think that such a trajectory would not hit the rock at some point?

    Instead, consider the forces acting on the ball and apply Newton's 2nd law. Hint: Make use of the fact that the rock is spherical.
     
  4. Oct 27, 2009 #3
    Well, I assumed that in the time it takes the ball to hit the ground, i.e., time of the vertical path, it will have to travel a distance > R in the horizontal direction for it to go beyond the sphere's horizontal extent, so I adjust vi accordingly. If the ball hits the ground at a distance > R in the horizontal, isn't it true that it won't have touched the sphere at any point in its path?

    I'm not sure I get your hint, all I can think is there is an impulse applied to the ball when I kick it but that's at the start of its path and it disappears immediately after, following which the only force is gravity in the vertical direction. So there is acceleration in the vertical but not horizontal in which case the equations would be what I used. The rock being spherical gives me distance in the vertical and horizontal that I need the ball to travel, beyond that I can't think of anything.

    What am I missing?
     
  5. Oct 27, 2009 #4

    Doc Al

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    Staff: Mentor

    You cannot assume that. The path of the ball (if it doesn't hit the rock) will be parabolic, not circular. (Assume some speed V that just gets the ball to x=R when it hits the ground. Write the vertical position as a function of x. Compare that to the height of the semicircle as a function of x.)


    Forget the impulse. As you say, it just provides the initial speed. Don't forget the normal force of the rock on the ball. You want the minimum speed such that the normal force is zero.

    Apply Newton's 2nd law. Hint: For the ball to just skim the surface at the top of the spherical rock, what kind of motion must it execute?
     
  6. Oct 27, 2009 #5
    The path of the ball must be a parabola, so even if the range is greater than R, you're saying its possible that it might touch the sphere on its way down? I'm trying to visualize it and am I right in thinking that the only place it could happen is near the top of the sphere? If thats possible then my conception of geometry was at fault. I think I get how to do it now, I'll have to make sure that the vertical position of the ball is greater than the respective y co-ordinate of the sphere throughout its motion. Where y would be (R2- x2)1/2.

    But I still don't get why you're mentioning the normal force, once the ball is off the surface in that initial instant then the normal force won't act anymore. How can the normal force help me with the velocity?
     
  7. Oct 27, 2009 #6

    Doc Al

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    The normal force is the key. The physical criteria for losing contact with the surface is that the normal force be zero.

    What forces act on the ball? What kind of motion must the ball execute to barely maintain contact with the spherical surface of the rock? Apply Newton's 2nd law.

    If the speed is too low, what happens to the ball? Start by analyzing the forces in that case.
     
  8. Oct 29, 2009 #7
    I think you've finally pushed me to the right way of solving this. Here's what I've stubmled to, please correct me where I'm wrong.

    If the velocity is too small it will simply move along the sphere with a centripetal force directed towards the centre. This force will be mvi2/R which should counter the weight of the ball mg. So if the ball moves along the surface a certain horizontal distance x then at that point mvi2/R = mg cos (whatever angle is between the weight and the radial line from the centre to the ball). So, if I want the ball not to touch the surface of the sphere then the centripetal force must be greater than this weight component, or the velocity should be greater. Therefore at the top of the sphere where weight mg = normal reaction mvi2/R (am I right in calling this the normal reaction?), the vi must be just high enough to unbalance this equation which will give me vi = [tex]\sqrt{}gR[/tex].

    Is the logic valid?
     
  9. Oct 29, 2009 #8

    Doc Al

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    Staff: Mentor

    OK. You've got the answer and your reasoning is almost right. Let's tune it up.
    Good!
    Careful. The centripetal force doesn't counter the weight--the weight is what provides the centripetal force. "Centripetal force" is the net radial force on a object that is centripetally accelerating. In this case, the only forces on the ball are weight and the normal force. Thus, at the top of the sphere:
    ΣF = mac
    mg - N = mv2/r

    The faster the ball moves, the more centripetal force is required. While the weight of the ball is obviously fixed, the normal force is not--the normal force decreases as the speed increases. When the ball moves fast enough that it loses contact with the sphere, that's when the normal force = 0. So, at that point, the speed satisfies this equation:
    mg = mv2/r

    (And you can easily show that if the speed it great enough to not hit the rock at the top, it will certainly be great enough to miss all other points of the rock as well. As you go down the rock, the available centripetal force--the radial component of the weight--decreases while the speed increases.)

    Make sense?
     
  10. Oct 29, 2009 #9
    Completely! I was looking at the thing from a much narrower perspective, I have a much better view on whats going on now. You rock, Doc!
     
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