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Circular motion

  1. Aug 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle starts from rest at t=0s.
    It moves along a circular path of radius 18m and has an acceleration component along its path of 6.7m/s^2.
    What is the magnitude of the acceleration when t=2s

    2. Relevant equations
    [tex]\begin{array}{l}
    \left| a \right| = \sqrt {a_n ^2 + a_t ^2 } \\
    a_n = \frac{{v^2 }}{p} \\
    v = \frac{{2\pi r}}{T} \\
    \end{array}[/tex]

    3. The attempt at a solution
    The particle is acting in circular motion, hence:
    [tex]v = \frac{{2\pi r}}{T} = \frac{{2\pi \left( {18} \right)}}{2} = 18\pi \;ms^{ - 1} [/tex]

    The normal component of acceleration is then given by:
    [tex]a_n = \frac{{v^2 }}{p} = \frac{{\left( {18\pi } \right)^2 }}{{18}} = \frac{{18^2 .\pi ^2 }}{{18}} = 18\pi ^2 \;ms^{ - 2} [/tex]

    Then I can find the magnitude of the total acceleration:
    [tex]\left| a \right| = \sqrt {a_n ^2 + a_t ^2 } = \sqrt {\left[ {18\pi ^2 } \right]^2 + 6.7^2 } \approx 178ms^{ - 2} [/tex]

    Look OK? My first response would be that the figure seems quite high...
     
  2. jcsd
  3. Aug 1, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's an equation for uniform circular motion where T is the period of the motion. Not relevant here.

    Instead, use kinematics to calculate the tangential speed of the particle at the end of 2 seconds.
     
  4. Aug 1, 2008 #3
    OK thanks, I thought this might have been where I went wrong.

    How about?

    [tex]\begin{array}{l}
    a_t = \frac{{dv}}{{dt}} \Rightarrow dv = a_t .dt \\
    \int {dv} = \int {a_t .dt} \Rightarrow v = \int\limits_0^2 {a_t .dt} = \int\limits_0^2 {6.7.dt} = \left[ {6.7t} \right]_0^2 = 13.4\,ms^{ - 1} \\
    a_n = \frac{{v^2 }}{r} = \frac{{13.4^2 }}{{18}} = 9.98\;ms^{ - 2} \\
    \left| a \right| = \sqrt {a_t ^2 + a_n ^2 } = \sqrt {6.7^2 + 9.98^2 } = 12.0167\;ms^{ - 2} \\
    \end{array}[/tex]

    Cheers.
     
  5. Aug 1, 2008 #4

    Doc Al

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    Staff: Mentor

    Much better. :approve:
     
  6. Aug 1, 2008 #5
    ok thanks :)

    Just to confirm my understanding of this, the tangential acceleration is defined as the rate of change of velocity, in the 't' direction, yeah?
    There is no other way to calculate this than to use kinematics?
     
  7. Aug 1, 2008 #6

    Doc Al

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    Staff: Mentor

    Right.
    Not sure what you mean.
     
  8. Aug 1, 2008 #7
    Well I was just curious as to whether or not there was another way to get the same result.
     
  9. Aug 1, 2008 #8

    Doc Al

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    Staff: Mentor

    Not that I can see.
     
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