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Linear Acceleration

  1. Mar 23, 2008 #1
    Problem:
    What is the linear acceleration of a point on the rim of a 36 cm diameter turntable which is turning at 33 rev/min?

    My Work:
    [tex]T = \frac{2 \pi r}{v}; v = \frac{2 \pi r}{T}[/tex]

    [tex]a_{r} = \frac{v^{2}}{r}[/tex]

    [tex]\frac{33 rev}{min} \times \frac{1 min}{60 sec} = \frac{.55 rev}{sec}[/tex]

    [tex]\frac{2 \pi .36 m}{\frac{.55 rev}{sec}} = 4.1126 \frac{m}{sec}[/tex]

    [tex]\frac{(4.1126\frac{m}{sec})^{2}}{.36 m} = 46.9826 \frac{m}{sec^{2}}[/tex]

    Is my work correct? The final answer is suppose to be in [tex]\frac{m}{sec^{2}}[/tex]. The reason I ask is that the multiple choice answer that is nearest to my calculated answer is [tex]46.3 \frac{m}{sec^{2}}[/tex]. My professor admits that the exactness can be off a few times for the possible choices.
     
  2. jcsd
  3. Mar 23, 2008 #2

    Doc Al

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    Staff: Mentor

    To convert the angular speed from rev/s to radians/s, multiply by [itex]2\pi[/itex]. To go from angular speed in radians/sec to linear speed, use [itex]v = \omega r[/itex].

    You can also use a different formula for centripetal acceleration:

    [tex]a_{r} = \frac{v^{2}}{r} = \omega^2 r[/tex]
     
  4. Mar 23, 2008 #3

    tiny-tim

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    Hi blue! :smile:

    Definitely use Doc Al's formula:

    [tex]a_{r} = \frac{v^{2}}{r} = \omega^2 r[/tex]

    And even more important:
    :redface: r = diameter/2 :redface:
     
  5. Mar 23, 2008 #4
    [tex]2 \times \pi \times \frac{.55 rev}{sec} = 3.45575 \frac{rad}{sec}[/tex]

    [tex]\frac{3.45575 \frac{rad}{sec}}{.36 m} = 9.59931\frac{m}{sec}[/tex]

    [tex](9.59931 \frac{m}{sec})^{2} \times .36 m = 33.1728 \frac{m}{sec^{2}}[/tex]

    Is this correct?
     
  6. Mar 23, 2008 #5
    Oh my, I forgot about radius, haha. I was too concerned with the conversion to meters from centimeters!
     
  7. Mar 23, 2008 #6

    Doc Al

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    Multiply (not divide) by the radius (not the diameter) to find the linear speed.

    When using linear speed to find the radial acceleration, divide by the radius. If you use my alternate formula and angular speed (in rad/s, not m/s), then you can multiply by the radius. Don't mix up the two versions of the formula for radial acceleration.
     
  8. Mar 23, 2008 #7
    [tex]2 \times \pi \times \frac{.55 rev}{sec} = 3.45575 \frac{rad}{sec}[/tex]

    [tex]3.45575 \frac{rad}{sec} \times .18 m = .622035 \frac{m}{sec}[/tex]

    [tex]\frac{(.622035 \frac{m}{sec})^{2}}{.18 m} = 2.1496 \frac{m}{sec^{2}}[/tex]

    Thank you!
     
  9. Mar 23, 2008 #8

    tiny-tim

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    … don't save words …

    Hi blue! :smile:

    General tip: put [itex]\omega[/itex] (or whatever) at the beginning of your lines, so you remember what each line is.

    Your first line should begin [itex]\omega\,=[/itex].

    Your second line should begin [itex]\omega^2r\,=[/itex].

    But because you're trying to save words, you're getting completely mixed up, and even in this post you've only got [itex]\omega r[/itex]. :cry:
     
  10. Mar 23, 2008 #9

    Doc Al

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    Looks good. (But you would be wise to follow tiny-tim's advice about labeling your equations.)
     
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