Calculation of Centrifugal Force

[svansvan]

Hello

I am making a vibration project which requires me to mount eccentric weights on both sides of an electric motor. I must submit the "generated force(in Kg)" by this motor.

The RPM of motor = 1000

Total weight on both sides of the motor = 16.8Kg (8.4Kg on both sides)

The radius(distance from the center of the shaft to the end of the weight) = 103 mm

What will be creation of Centrifugal force when this motor is run at 1000 RPM ??

Dhiraj

[Robittybob1]

 

Hello

I am making a vibration project which requires me to mount eccentric weights on both sides of an electric motor. I must submit the "generated force(in Kg)" by this motor.

The RPM of motor = 1000

Total weight on both sides of the motor = 16.8Kg (8.4Kg on both sides)

The radius(distance from the center of the shaft to the end of the weight) = 103 mm

What will be creation of Centrifugal force when this motor is run at 1000 RPM ??

Dhiraj

 

The set-up seems extraordinary, either that or it a very heavy high powered motor for the vibration (centrifugal force) is going to be a frighteningly large quantity.

What formulas are you going to use?

[studiot]

Look here

 

http://www.precisionmicrodrives.com/application-notes-technical-guides/application-bulletins/ab-004-understanding-erm-characteristics-for-vibration-applications

[Enthalpy]

Normal materials and standard hardware withstand it, but you better balance the rotor carefully before running the motor.

[Robittybob1]

Normal materials and standard hardware withstand it, but you better balance the rotor carefully before running the motor.

I thought the whole purpose was to create an out of balance situation to make vibration. Would you still get vibration if it was well balanced?

[Enthalpy]

O yes. Maybe "both sides" is "both ends"? If the masses ar imperfectly balanced, I don't wee why one should use two instead of a single lighter one, and if they are, one gets no vibration.

With unbalanced masses, the motor's bearing will work under very unsound conditions. I would be worth checking if they can survive it.

[Robittybob1]

O yes. Maybe "both sides" is "both ends"? If the masses ar imperfectly balanced, I don't wee why one should use two instead of a single lighter one, and if they are, one gets no vibration.

With unbalanced masses, the motor's bearing will work under very unsound conditions. I would be worth checking if they can survive it.

That is why I was thinking for the masses talked of the motor, bearings, shafts and the like would have to be heavy duty.

Edited November 6, 2014 by Robittybob1

[Robittybob1]

That is why I was thinking for the masses talked of the motor, bearings, shafts and the like would have to be heavy duty.

It might make a good paint shaker except for its frequency.

[Robittybob1]

Every time I try and find out how to calculate centrifugal force they kept telling me it is a fictitious force.

[studiot]

 

Every time I try and find out how to calculate centrifugal force they kept telling me it is a fictitious force.

 

 

It is, but it is a very useful fictious force as it transforms a problem from one of dynamics to one of static equilibrium.

 

It is known as D'Alambert's Principle.

 

This link is simpler than Wiki for a first reading

 

http://www.britannica.com/EBchecked/topic/150132/dAlemberts-principle

[John Cuthber]

The force can be calculated as m v^2/r no matter whether it's real or not.

[Robittybob1]

The force can be calculated as m v^2/r no matter whether it's real or not.

Then you add into that an expression of the tangential speed v = 2*Pi()*r/T, T being the period.

(edited)

Edited November 9, 2014 by Robittybob1

[Strange]

Shouldn't that be [math]\displaystyle {v = \frac{2 \pi r}{T}}[/math]?

Edited November 9, 2014 by Strange

[Robittybob1]

Shouldn't that be [math]\displaystyle {v = \frac{2 \pi r}{T}}[/math]?

How do you even begin to write formulas like that? "[math]\displaystyle {v = \frac{2 \pi r}{T}}[/math]"

[Strange]

You mean the math/Latex stuff?

There is a guide here: http://www.scienceforums.net/topic/3751-quick-latex-tutorial/

And an interactive editor here: http://www.codecogs.com/latex/eqneditor.php

[Robittybob1]

You mean the math/Latex stuff?

There is a guide here: http://www.scienceforums.net/topic/3751-quick-latex-tutorial/

And an interactive editor here: http://www.codecogs.com/latex/eqneditor.php

Thanks

Could the force be in the order of

18975.9594 Kgs?

That seems a way higher than I had expected. 19 tonnes down and 19 tons up 17 times a second, something will break before the hour is up!

Edited November 10, 2014 by Robittybob1

[Robittybob1]

Thanks

Could the force be in the order of

18975.9594 Kgs?

That seems a way higher than I had expected. 19 tonnes down and 19 tons up 17 times a second, something will break before the hour is up!

I have now been told I was using an incorrect formula

 

I started with Force = m*V^2/r

and then introduced velocity as the circumference/period =2*Pi*r/T but when I squared that I didn't square the bottom line as well.

(2*Pi*r/T)^2 = 4*pi^2*r^2/T^2

Using that for V^2

Force = m*4*pi^2*r^2/(T^2*r)

the r top and bottom cancel.

 

Force = m*4*pi^2*r/(T^2)

 

So I think I need to divide the answer by the period for I was using T rather than T^2

 

Thanks for correcting me.