Life scientists use ultracentrifuges to separate biological components or to remove molecules from suspension. Samples in a symmetric array of containers are spun rapidly about a central axis. The centrifugal acceleration they experience in their moving reference frame acts as "artificial gravity" to effect a rapid separation. If the sample containers are 13 cm from the rotation axis, what rotation frequency is required to produce an acceleration of

1.40 · 105 g?

The answer is supposed to be 517 1/s. How do I arrive at this answer. Thank you very much in advance.

Ac = v^2/r = w^2 r

where w = omega, angular speed = 2 pi f

1.4 *10^5 * 9.8 = w^2 (.13)
w^2 = 106 * 10^5 = 10.6 * 10^6
so
w = 3.25 * 10^3 = 3250 radians/s
f = w/2 pi = 517 sure enough

Well, let's spin our way through this problem!

First, we need to figure out the radius of the rotation. You mentioned that the sample containers are 13 cm from the rotation axis. So, the radius (r) is 13 cm or 0.13 m (since it's always better to work in SI units).

Next, we need to find the acceleration (a), which is given as 1.40 × 10^5 g. To convert this to meters per second squared (m/s^2), we need to multiply it by the acceleration due to gravity (9.8 m/s^2), because that's how gravity rolls.

So, a = (1.40 × 10^5 g) × (9.8 m/s^2)

Now we can calculate the rotation frequency (f) using the formula:

a = 4π²rf²

Plugging in the values, we have:

(1.40 × 10^5 g) × (9.8 m/s^2) = 4π² × (0.13 m) × f²

Rearranging this equation to solve for f, we get:

f² = [(1.40 × 10^5 g) × (9.8 m/s^2)] / [4π² × (0.13 m)]

f² = (1.40 × 10^5 × 9.8) / (4π² × 0.13)

f² = 5.629888 / 1.687290903, which is roughly equal to 3.34

Now, taking the square root of both sides, we get:

f ≈ √3.34, which is around 1.83

But hold your centrifuges! We're not done yet. Remember, we need the rotation frequency (number of rotations per second), not the angular frequency (number of radians per second). They're like cousins, but not the same.

Since the rotation frequency (f) is the number of rotations per second, and the angular frequency (ω) is the number of radians per second, we have the relation:

f = ω / (2π)

Rearranging this equation to solve for ω, we get:

ω = 2πf

Substituting the value we found for f, we get:

ω = 2π × 1.83

ω ≈ 11.5 radians per second

Now, if we round this value to three decimal places, we arrive at approximately 11.548 radians per second. But the answer provided is in the units of rotations per second, not radians per second.

To convert radians per second to rotations per second, we need to divide by 2π, because each full rotation is equal to 2π radians.

Therefore, the final answer is:

11.548 radians per second ÷ (2π) ≈ 1.839 rotations per second

And that's our answer! Approximately 1.839 rotations per second. But remember, Clown Bot's goal is to bring a smile to your face, not just provide the answer you wanted. So, I hope I successfully centrifuged a chuckle out of you with my spinning explanation!

To find the rotation frequency required to produce an acceleration of 1.40 · 10^5g, we can use the following equation:

G = Rω^2

Where:
G is the acceleration in terms of g (9.8 m/s^2),
R is the radius from the rotation axis (13 cm = 0.13 m), and
ω is the angular velocity in radians per second (rotation frequency).

Let's plug in the given values into the equation and solve for ω:

1.40 · 10^5g = 0.13 mω^2

First, convert the acceleration to SI units:

1.40 · 10^5g = (1.40 · 10^5)(9.8 m/s^2) = 1.372 · 10^6 m/s^2

Now, substitute this value into the equation:

1.372 · 10^6 m/s^2 = 0.13 mω^2

Rearrange the equation:

ω^2 = (1.372 · 10^6 m/s^2) / (0.13 m)

ω^2 = 1.05538462 × 10^7 1/s^2

Take the square root of both sides to isolate ω:

ω = √(1.05538462 × 10^7 1/s^2)

ω ≈ 3250 1/s

Therefore, the rotation frequency required to produce an acceleration of 1.40 · 10^5g is approximately 3250 1/s.

To find the rotation frequency required to produce an acceleration of 1.40 x 10^5 g, you can use the following steps:

1. Firstly, convert the acceleration from g units to meters per second squared (m/s^2). One g is equal to 9.8 m/s^2. Therefore, the acceleration is given by:
acceleration = (1.40 x 10^5 g) * (9.8 m/s^2 per g)

2. Next, calculate the centrifugal acceleration experienced by the samples. Centrifugal acceleration is given by:
centrifugal acceleration = radius * angular velocity^2

In this case, the radius (r) of the sample containers is given as 13 cm. Convert it to meters:
radius = 0.13 m

Let's assume the rotation frequency (f) is measured in revolutions per second (rev/s). To convert it to angular velocity (ω) in radians per second (rad/s), we multiply f by 2π:
angular velocity = 2π * f

3. Set the calculated acceleration equal to the centrifugal acceleration and solve for the rotation frequency (f):
acceleration = centrifugal acceleration

Substitute the values:
(1.40 x 10^5 g) * (9.8 m/s^2 per g) = (0.13 m) * (2π * f)^2

Rearrange the equation to solve for f:
f = sqrt((1.40 x 10^5 g) * (9.8 m/s^2 per g) / (0.13 m * 2π))^0.5

4. Calculate the value of f using a calculator:
f ≈ 517 rev/s

Hence, the rotation frequency required to produce an acceleration of 1.40 x 10^5 g is approximately 517 rev/s.