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A tire with a pebble

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A tire with radius 0.50m rolls in a straight line without slipping. The center moves with constant speed 6.7 m/s. A small pebble stuck in the tread of the tire is in contact with the road at time t=0. The total time is 14.1s. The horizontal position of the pebble at time t = 94.26m.
    Calculate the vertical position y of the pebble at time t.
    Calculate the horizontal velocity of the pebble at time t.
    Calculate the vertical velocity of the pebble at time t.


    2. Relevant equations
    the integral of velocity from t=0s to t=14.1s
    y=rsin(theta)


    3. The attempt at a solution
    I thought that since x=94.26m then the equation x=rcos(theta) could tell me what theta was.
     
  2. jcsd
  3. Jun 2, 2009 #2

    Hao

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    The position of the pebble is a superposition of pure rotational motion of the wheel, and pure translational motion of the wheel.

    You do not need to use integration for this question.

    Writing out the separate x and y components for the position of the pebble will help you find velocity.

    x=rcos(theta) only applies for a point on the wheel relative to the center of the wheel.
    In this case, 94.26m refers to a point on the wheel relative to its starting position at t = 0.
     
  4. Jun 2, 2009 #3
    Okay so how do find theta? Because you cannot use the x value i got since it refers to the position at t=0.
     
  5. Jun 2, 2009 #4

    Hao

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    Let us consider the rotational motion of the wheel alone.

    If we suppose that the center of the wheel is stationary, and the edge is rotating at a speed vr, we can find our angular velocity vr = ω r, where ω is the angular velocity, and r is the radius of the wheel.

    Since we know ω, we can find θ = ω t.

    As a result, for a stationary wheel, we can say that x = r sin(θ) = r sin(ω t).

    Our job is now to find vr.

    Since our wheel in the question is rolling without slipping, we know that the bottom of the wheel is always instantaneously at rest.

    Hence, vtranslational + vr = 0.

    Since vtranslational is given, we can find vr, and hence θ at anytime.

    As a result, we now know the position of the center of the wheel at any time, and the position of the pebble relative to the center of the wheel at any time. Adding the two gives us what we want.

    It helps to draw a vector diagram for positions of the pebble relative to the wheel and the center of the wheel.

    You can also draw a vector diagram of pure rotation and pure translation and add them together to see how you get pure rolling.
     
  6. Jun 2, 2009 #5
    okay so when I solve y=rsin(theta) that gives me the position of y, but in order to find due to the time I need to take the time given multiplied by the y?
     
  7. Jun 2, 2009 #6
    okay i got how to find the positions but how do i find the velocity of x and y?
     
  8. Jun 2, 2009 #7

    Hao

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    The definition of velocity in the x direction is

    [tex]\frac{\bigtriangleup x}{\bigtriangleup t} \rightarrow \frac{d x}{d t}[/tex]

    This means that we need to differentiate both x and y with respect to t to obtain velocity.

    Alternatively, we can resolve the velocity components for pure rotational motion to the x and y axis respectively, and add the velocity for pure translational motion.
     
  9. Jun 2, 2009 #8
    i am not quite sure what you mean. I know that inorder to find the y component I solved r-rcos(theta). Do I just differentiate that equation?
     
  10. Jun 2, 2009 #9

    Hao

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    Exactly - differentiating y with respect to time will give you the y component of velocity. Just remember to account for the fact that θ = ω t.

    Similarly, differentiating x with respect to time will give you the x component of velocity.
     
  11. Jun 2, 2009 #10
    I have tried to take the derivative and I keep getting the wrong answer. I get dr/dt - (dr/dt)cos(theta) + r*sin(theta)*(v/r). Then since I am given r, I assume that dr/dt = 0. Is this the correct way of thinking?
     
  12. Jun 2, 2009 #11

    Hao

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    r refers to the radius of the wheel, which is constant, so dr/dt = 0.

    The correct answer is therefore your last term vertical velocity = r*sin(theta)*(v/r).

    Note that you could have also gotten this answer by picturing the wheel undergoing pure rotation, and resolving the velocity to the y axis. You may or may not find it easier this way.
     
    Last edited: Jun 2, 2009
  13. Jun 2, 2009 #12
    thanks for the help
     
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