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Spinning blade - Effort and Force

by Centrifugal
Tags: blade, effort, force, spinning
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Centrifugal
#1
Jun18-14, 05:09 PM
P: 3
How do I calculate the force exerted on the fulcrum (center) from a spinning blade and the effort required to spin it.

In a practical example:

A spinning lawnmower blade. If the blade of length x and mass y was exchanged with a heavier longer blade how much more effort would be required to spin that blade.
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crador
#2
Jun18-14, 06:37 PM
P: 25
I think this has to do with the air resistance (and grass resistance if we are talking about a lawnmower). Otherwise you do not need any torque (force) to keep it spinning, since we have conservation of angular momentum. To get it spinning in the first place you need some power.

With regards to the air resistance I have no idea and leave the topic for others.
Centrifugal
#3
Jun18-14, 08:11 PM
P: 3
To develop the practical example further, or rather, the situation that got me thinking about it...

A lawnmower has a 1450w motor and has a 33cm blade. The next model up has a 35mm blade and a 1500w motor.

Both models have the same chassis. Only the blades and motor are different.

Therefore, is there an increase in the effort require from the motor to keep a slightly larger heavier blade spinning.

Would the increase in length and mass of the blade require the extra 50w in motor power, or is this just model/range marketing policy by the manufacturer.

Taking the "grass resistance" into account, as the motor would be required to drive the blade in light and heavy grass conditions, one would assume that there must be ample power to drive a slightly larger blade.

crador
#4
Jun18-14, 08:47 PM
P: 25
Spinning blade - Effort and Force

Quote Quote by Centrifugal View Post
Taking the "grass resistance" into account, as the motor would be required to drive the blade in light and heavy grass conditions, one would assume that there must be ample power to drive a slightly larger blade.
I think you've hit it on the head here. The motor definitely has enough power to spin a larger blade. A larger blade covers a larger area though (area going up as the square of the radius, so even 2cm makes a difference). This larger area means you're cutting a bit more grass, and ripping apart all that cellulose is probably what requires the extra power.

If you really want to get into it you could look up the rest of the manufacturers data, and by using the average speed of the mower cook up a graph of average turf cut per second vs engine power.
Centrifugal
#5
Jun19-14, 09:08 PM
P: 3
Looking at the specs of the mowers I notice that all are rated to spin at 4300rpm.

The three mowers all have the same chassis and only differ in the following:

30cm blade 1400w motor

33cm blade 1450w motor

35cm blade 1500w motor.

As all three are rated to run at 4300rpm, (assuming this is an accurate measurement) could there be a relationship between the motor power and blade length in order to maintain the rpm of 4300?
Khashishi
#6
Jun20-14, 02:19 AM
P: 887
To be honest, physics isn't too good at handling these practical problems because there're just so many unknowns. But, we can try.
The resistance will be higher when actively cutting grass. But the resistance is pretty low when not cutting grass, so there might be some kind of rate limiting device in the motor to keep it from spinning too fast. I don't know much about lawn mowers, so this is just an assumption.

Let's assume that when the mower is pushed forward with constant velocity into uncut grass. Then the grass resistance is proportional to the wingspan of the blades. In other words, the power needed should be proportional to the diameter of the blade. Your data shows that this is incorrect. Maybe the higher speed of the edge of the larger blade or the higher moment of inertia makes it better at cutting, so the power needed grows smaller than proportionally to the wingspan.
CWatters
#7
Jun20-14, 04:23 AM
P: 3,143
What Khashish said but...

Then the grass resistance is proportional to the wingspan of the blades. In other words, the power needed should be proportional to the diameter of the blade. Your data shows that this is incorrect
It is roughly proportional to the diameter (or at least more so than to the area). However the makers data is not intended to be used for this purpose. The maker may well specify a 1500W motor but that doesn't mean it will draw 1500W. You would have to measure actual consumption.

It's not unknown for a manufacture to fit the same motor to all models to reduce costs and just change the spec. A 1500W rated motor will happily run at 1400W and may live longer as a result. Most types of electric motors won't over speed if they are lightly loaded.


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