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Linear Acceleration of a turntable

  • Thread starter blue5t1053
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  • #1
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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.
 

Answers and Replies

  • #2
Doc Al
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[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]
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]
 
  • #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:
 
  • #4
23
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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]
[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?
 
  • #5
23
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Oh my, I forgot about radius, haha. I was too concerned with the conversion to meters from centimeters!
 
  • #6
Doc Al
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[tex]\frac{3.45575 \frac{rad}{sec}}{.36 m} = 9.59931\frac{m}{sec}[/tex]
Multiply (not divide) by the radius (not the diameter) to find the linear speed.

[tex](9.59931 \frac{m}{sec})^{2} \times .36 m = 33.1728 \frac{m}{sec^{2}}[/tex]
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.
 
  • #7
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[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!
 
  • #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:
 
  • #9
Doc Al
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[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!
Looks good. (But you would be wise to follow tiny-tim's advice about labeling your equations.)
 

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