How fast (in rpm) must a centrifuge rotate if a particle 6.00 cm from the axis of rotation is to experience an acceleration of 125,000 g's?

Respuesta :

A particle at 6 cm from the axis of rotation has radius
r = 6 cm = 0.06 m
The expected acceleration is
a = 125,000 g's = 125000*(9.8 m/s²) = 1.225 x 10⁶ m/s²

Let ω =  the angular velocity.
The acceleration is centripetal, and it is given by
a = ω²r
Therefore
(ω rad/s)²*(0.06 m) = 1.225 x 10⁶ m/s²
ω² = 2.0417 x 10⁷
ω = 4518.5 rad/s
    = (4518.5 rad/s)*(1/2π rev/rad) 
    = 719.14 rev/s
ω = (719.14 rev/s)*(60 s/min)
    = 43148.5 rpm

Answer: 43149 rpm (nearest integer)

aksnkj

The speed of the particle in rpm is 43170.32 rpm.

Given:

The distance of particle (radius) from the axis is [tex]r=6\rm \: cm=0.06\; m[/tex].

The acceleration (centripetal) experienced by the particle is,

[tex]a=125000\rm \; g's\\a=125000\times 9.81 \rm \; m/s^2\\a=1226250\rm \; m/s^2[/tex]

Let ω be the angular velocity of the particle and N be the speed in rpm.

So, ω will be  [tex]\omega=\dfrac{2\pi N}{60}[/tex]

Now, the acceleration of the particle can be written as,

[tex]a=\omega^2r\\a=\left(\dfrac{2\pi N}{60}\right)^2\times 0.06[/tex]

So, the speed of the particle in rpm will be,

[tex]a=\left(\dfrac{2\pi N}{60}\right)^2\times 0.06\\12262550=0.01095N^2\times 0.06\\N=43170.32\rm \; rpm[/tex]

Therefore, the speed of the particle in rpm is 43170.32 rpm.

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https://brainly.com/question/8825608?referrer=searchResults