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Consider a particle that is moving along the perimeter of a circle at a uniform
rate (figure X). The following equations apply in this circular motion:
T = 2π / ω
x = r cos (θ) = r cos(ωt)
y = r sin (θ) = r sin(ωt)
V = r . ω
θ = S / r
r : Radius (m)
T : Time to complete one revolution (s)
θ : Angle (radials)
ω : Angular velocity (rad / s)
S : Arc length (m)
The circular motion is described by defining a vector r
from the origin to the particle.
The position of the particle is
then defined by r
and its angle θ. The angular velocity
ω equals 2π / T. The arc length S on the perimeter of the circle is r . θ (for one
complete revolution θ = 2π, so we get S = r . θ = 2πr, which is the perimeter of
a circle). The particle’s tangential speed follows from V = dS/dt. Substituting
S = r . θ gives V = r . d(θ)/dt = r . ω.
A rotor of a helicopter is turning at 500 rpm. The blade length is 5 metres. What
is the tip speed in m/s?
V = r . ω.
ω = 2π . 500 / 60 = 100π/6
V = 5 . 100π / 6 = 262 m/s.
Again, consider a particle that is moving along the perimeter of a circle at a
uniform rate (figure X). We now introduce a force Fcp that is directed towards the centre of the circular motion. We call this force a centripetal force. When the
particle has mass m and a tangential speed V, then this force will accelerate the
particle by changing the direction of travel, such that it will travel in a circular
motion with radius r. The following equation describes this situation:
Fcp = m . v2
or with ω = V / r
Fcp = m . ω2
The formula above calculates the centripetal force that is needed to sustain a circular
motion of a particle. Note that this force is not doing any work. This can be easily
understood by realizing that the force vector and the direction of travel are orthogonal.
The same result follows from looking at the kinetic energy. This energy stays the same, because the speed doesn't change, only the direction of travel.
Note that the centripetal force is not the same as the centrifugal force! The centrifugal
force only makes sense when looking at a rotating reference frame
which leads to the introduction of the coriolus and centrifugal forces.
We will again consider the example of the rotor that is turning at an
rpm of 500. We will now make the assumption that the blade does not have any mass. Instead, all the mass
is concentrated at one point at the end of the blade. This point of
mass equals 5 kg. How great is the centripetal force needed to allow this mass to
make a circular motion with a radius of 5 metres?
Fcp = m . v2
/ r = 5 . 2622
/ 5 = 68,644 N. Note that this force is almost equivalent to the weight of a rock of 6,800 kg. The rotorhead and blades must
be able to cope with these strains.