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.)