Linear Acceleration of a turntable

[blue5t1053]

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:
T = \frac{2 \pi r}{v}; v = \frac{2 \pi r}{T}

a_{r} = \frac{v^{2}}{r}

\frac{33 rev}{min} \times \frac{1 min}{60 sec} = \frac{.55 rev}{sec}

\frac{2 \pi .36 m}{\frac{.55 rev}{sec}} = 4.1126 \frac{m}{sec}

\frac{(4.1126\frac{m}{sec})^{2}}{.36 m} = 46.9826 \frac{m}{sec^{2}}

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

 

[Doc Al]

\frac{33 rev}{min} \times \frac{1 min}{60 sec} = \frac{.55 rev}{sec}

\frac{2 \pi .36 m}{\frac{.55 rev}{sec}} = 4.1126 \frac{m}{sec}

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

You can also use a different formula for centripetal acceleration:

a_{r} = \frac{v^{2}}{r} = \omega^2 r

 

[tiny-tim]

Hi blue!

Definitely use Doc Al's formula:

a_{r} = \frac{v^{2}}{r} = \omega^2 r

And even more important:

r = diameter/2 ​

 

[blue5t1053]

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

You can also use a different formula for centripetal acceleration:

a_{r} = \frac{v^{2}}{r} = \omega^2 r

2 \times \pi \times \frac{.55 rev}{sec} = 3.45575 \frac{rad}{sec}

\frac{3.45575 \frac{rad}{sec}}{.36 m} = 9.59931\frac{m}{sec}

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

Is this correct?

 

[blue5t1053]

Oh my, I forgot about radius, haha. I was too concerned with the conversion to meters from centimeters!

 

[Doc Al]

\frac{3.45575 \frac{rad}{sec}}{.36 m} = 9.59931\frac{m}{sec}

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

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

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.

 

[blue5t1053]

2 \times \pi \times \frac{.55 rev}{sec} = 3.45575 \frac{rad}{sec}

3.45575 \frac{rad}{sec} \times .18 m = .622035 \frac{m}{sec}

\frac{(.622035 \frac{m}{sec})^{2}}{.18 m} = 2.1496 \frac{m}{sec^{2}}

Thank you!

 

[tiny-tim]

… don't save words …

Hi blue!

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

Your first line should begin \omega\,=.

Your second line should begin \omega^2r\,=.

But because you're trying to save words, you're getting completely mixed up, and even in this post you've only got \omega r.

 

[Doc Al]

2 \times \pi \times \frac{.55 rev}{sec} = 3.45575 \frac{rad}{sec}

3.45575 \frac{rad}{sec} \times .18 m = .622035 \frac{m}{sec}

\frac{(.622035 \frac{m}{sec})^{2}}{.18 m} = 2.1496 \frac{m}{sec^{2}}

Thank you!

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