Re: Centripetal Force & g forces

Bradster wrote in message
aec2641d-e0d7-488f-92e9-7c9e88670bb8@xxxxxxxxxxxxxxxxxxxxxxxxxxxx

Hi,
I trying to calculate the forces experinced by a sensor placed on a
drill bit. The main forces i'm looking at are the Centripetal and
Tangentail forces that the sensors undergo.

Drill Bit Radius = 5mm
Sensor Mass = 1gram = 0.001kg
Speed = 7000rpm => v = 2pi*7000/60 * 0.005 = 3.66ms-1

a = v^2/r = 3.66*3.66/0.005 = 2686.73ms-2 => 274g's
f = ma = 0.001 * 2686.73 = 2.6867N

I'm looking at the tangential component as the drill starts up from
rest. and using the following formula.
f = ma = m*(v1-v0)/(t0-t1)

Total G Force experencied = (ac^2+at^2)^0.5/9.81 < tan-1(-at/ac)
I obtained these formula's from textbooks.
I am probably more interested in the g forces that the sensor will
experience as these are often qouted in component data sheets.

I'm just assuming you divide accel by gravity to get corresponding g
forces.

I need to know if what i'm doing is correct?
Should i be using different forumla's?

I've seen the link on g forces which seems to be in line with what i'm
doing, but i just ain't sure

I'd pick a mathematical model for the drill spin up and
work from that. An exponential curve would probably
suffice:

w(t) = W*(1 - e^(t/tau))
where:
W = Final angular velocity = 7000*2*pi/60sec
tau = Spin-up time constant. Probably about 1 sec.

An exponential curve "settles" in approximately 5 time
constants.

Then you can write the centrifugal acceleration at any
time as:

Ac(t) = r*(w(t))^2
where:
r = Radius of rotation

For the tangential acceleration we can first find the
angular acceleration by taking the derivative of the
angular velocity:

alhpa(t) = d w(t)/dt = (W/tau)*e^(-t/tau)

and then write the tangential acceleration as:

At(t) = r*alpha(t) = (r*W/tau)*e^(-t/tau)

Now you can determine the total acceleration for
any time after the drill starts turning by the
usual vector methods.


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