Linear Acceleration of a turntable

In summary, the linear acceleration of a point on the rim of a 36 cm diameter turntable which is turning at 33 rev/min is .55 rev/sec.
  • #1
blue5t1053
23
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.
 
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  • #2
blue5t1053 said:
[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
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
Doc Al said:
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
Oh my, I forgot about radius, haha. I was too concerned with the conversion to meters from centimeters!
 
  • #6
blue5t1053 said:
[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
[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
… 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
blue5t1053 said:
[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.)
 

What is linear acceleration of a turntable?

Linear acceleration of a turntable refers to the rate of change of the speed of the turntable in a straight line. It measures how quickly the turntable is increasing or decreasing its speed in a linear direction.

How is linear acceleration calculated for a turntable?

Linear acceleration of a turntable can be calculated by dividing the change in speed by the time taken for the change to occur. This can be represented mathematically as a = (vf - vi) / t, where a is the linear acceleration, vf is the final speed, vi is the initial speed, and t is the time.

What factors can affect the linear acceleration of a turntable?

The linear acceleration of a turntable can be affected by several factors, such as the mass of the turntable, the force applied to it, and the friction between the turntable and its surface. Additionally, the shape and size of the turntable can also impact its linear acceleration.

Why is understanding linear acceleration important for turntable design?

Understanding linear acceleration is crucial for turntable design because it helps in determining the appropriate size, shape, and materials for the turntable to achieve the desired speed and smoothness of rotation. It also helps in ensuring the safety and stability of the turntable during operation.

How does linear acceleration affect the sound quality of a turntable?

The linear acceleration of a turntable can have a significant impact on the sound quality produced. A higher linear acceleration can result in a faster and smoother rotation, reducing any potential audio distortions. Additionally, a consistent linear acceleration can also contribute to maintaining accurate and consistent playback speed, resulting in better sound quality.

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