A test tube in a centrifuge is pivoted so that it swings out horizontally as the machine builds up speed. If the bottom of the tube is 160.0 mm from the central spin axis, and if the machine hits 40500 rev/min, what would be the centripetal force exerted on a giant amoeba of mass 8.00E-9 kg at the bottom of the tube?
F = m r omega^2
= 8e-9 * .160 * omega^2
You'll need to convert rev/min into rad/sec for omega
To find the centripetal force exerted on the giant amoeba, we can use the formula:
F = m * ω^2 * r
Where:
F is the centripetal force
m is the mass of the amoeba
ω is the angular velocity of the centrifuge
r is the distance from the central spin axis to the bottom of the test tube
First, let's convert the angular velocity from revolutions per minute (rev/min) to radians per second (rad/s). We know that:
1 revolution = 2π radians
So, to convert rev/min to rad/s, we can use the conversion factor:
ω = (2π * ω) / 60
where ω is the angular velocity in rev/min. Plugging in the given value:
ω = (2π * 40500) / 60
≈ 4265.37 rad/s
Next, let's convert the distance from millimeters to meters:
r = 160.0 mm = 160.0 / 1000 = 0.16 m
Now we can substitute the values into the formula:
F = (8.00E-9 kg) * (4265.37 rad/s)^2 * (0.16 m)
≈ 0.0468 N
Therefore, the centripetal force exerted on the giant amoeba at the bottom of the tube would be approximately 0.0468 Newtons.