A laboratory centrifuge on earth makes n rpm (rev/min) and produces an acceleration of 7.00 g at its outer end.
What is the acceleration (in g's, i.e., acceleration divided by g) at a point halfway out to the end?
Duplicate post; answered elsewhere.
To find the acceleration at a point halfway out to the end of the centrifuge, we can use the concept of centripetal acceleration.
Centripetal acceleration is given by the formula:
a = rω²
where a is the acceleration, r is the radius, and ω is the angular velocity.
In this case, we know the acceleration at the outer end of the centrifuge is 7.00 g. Let's assume the radius of the centrifuge is denoted by R.
To find ω, we need to convert rpm (revolutions per minute) to angular velocity in radians per second. Since one revolution is equal to 2π radians, we can use the following conversion:
ω = (2π × n) / (60)
where n is the rpm of the centrifuge.
Now, let's calculate ω:
ω = (2π × n) / (60)
Next, let's calculate the radius halfway out to the end of the centrifuge. Since the halfway point is at a distance of R/2 from the center, the radius will be:
r = R/2
Now, we can substitute the values we have into the centripetal acceleration formula:
a = rω²
a = (R/2) × [(2π × n) / (60)]²
Finally, to find the acceleration in g's (acceleration divided by the acceleration due to gravity, g), we divide the centripetal acceleration by g:
acceleration in g's = a / g
Please provide the values of 'n' (rpm) and 'R' (radius) to calculate the acceleration at a point halfway out to the end of the centrifuge.