A centrifuge rotor has a moment of inertia of 3.85x10^-2 kg per m squared. How much energy is required to bring it from rest to 8490 rpm?
To find the energy required to bring the centrifuge rotor from rest to a certain speed, we can use the principle of conservation of energy. The initial kinetic energy of the rotor at rest is zero, and the final kinetic energy at a given speed can be calculated using the moment of inertia.
The formula for rotational kinetic energy is:
E = (1/2) * I * ω^2
Where:
E is the energy
I is the moment of inertia
ω is the angular velocity (in radians per second)
First, we need to convert the given speed from rpm (revolutions per minute) to radians per second. A full revolution is equal to 2π radians.
Given:
Moment of inertia (I) = 3.85x10^-2 kg/m^2
Speed (ω) = 8490 rpm
1. Convert rpm to radians per second:
ω = (8490 rpm) * (2π radians / 1 minute) * (1 minute / 60 seconds)
Simplifying the conversion:
ω = (8490 rpm) * (2π / 1) * (1 / 60)
ω = 2π * 8490 / 60
ω ≈ 888.5 radians per second
2. Substitute the values into the formula:
E = (1/2) * (3.85x10^-2 kg/m^2) * (888.5 radians per second)^2
Calculate the squared value:
E = (1/2) * (3.85x10^-2 kg/m^2) * (888.5 radians per second) * (888.5 radians per second)
Evaluate the expression:
E ≈ 1.61 joules
Therefore, approximately 1.61 joules of energy is required to bring the centrifuge rotor from rest to 8490 rpm.