Torque for sharpening pencils with different diameters

[patehi]

Hello guys,

if i sharpen 2 pencils with 2 different Diameters (thin and jumbo), which one will have a higher torque ( while sharpening), thus make it harder to sharpen?

Let's assume that the angle of the pencil tips are the same, both have same wood of same density, the push force is the same and the sharpening speed is the same.

Could someone explain which and why?

Thank you for your inputs!

 

[mjc123]

It seems self-evident that under the conditions you describe, the thicker pencil will (for that portion up to the diameter of the thin one) experience the same torque, plus whatever else is due to the portion outside that diameter, so the thick one will inevitably have the greater torque.

If we do the maths, then if we assume that the resisting force per unit length at the blade is constant along the sharpened surface, then the torque is proportional to R², where R is the pencil radius. If we assume that the force is proportional to the speed of the wood past the blade, and hence to the radius at that point (v = ωr), then the torque is proportional to R³. I leave the calculations as an exercise for you.

If by "harder to sharpen" you are actually referring to the force you need to apply to get the necessary torque, then F = T/R, then F is proportional to R or R² respectively. Either way, it's greater for the thick one.

 

[patehi]

Hi there mjc123,

thanks so much for your explanation! That really helped me further :)

 

[nasu]

The force you apply will produce a larger torque on the thick pencil too.
So it may be that the force will be the same unless the resistance force depends on the curvature of the pencil or on the diameter of the pencil. It will depend on what type of sharpener is used. If the sharpener is designed for the thick pencil the thin one won't touch the walls and the friction will be different. But if you are looking at a simple case when the resistance force applied to one end is the same in both cases, then the force you need to apply at the other end will be the same too.

 

[256bits]

Let's assume that the angle of the pencil tips are the same, both have same wood of same density, the push force is the same and the sharpening speed is the same.

It is not evident what you actually mean by push force or sharpening speed being the same for the pencils.
If the push force is that which you apply to the end of the pencil to push it into the tool, than an argument could be made that the torque is the same for both pencils, large and small, as friction in its rudimentary explanation does not depend upon area.
Sharpening speed could be interpreted as number of rotations of the tool or pencil per second, or as the number of pencils sharpened per second. The latter involves a greater torque for the larger pencil; the former not necessarily so.

 

[haruspex]

friction in its rudimentary explanation does not depend upon area.

but... the opposing force is not friction. See post #2.
Also, the longitudinal force on the pencil should not be that important. As long as there is sufficient to engage the blade, the angle of cut should be largely independent of it. If excessive force is applied along the pencil, there may be a frictional resistance that is unrelated to the actual cutting.

Sharpening speed could be interpreted as number of rotations of the tool or pencil per second, or as the number of pencils sharpened per second. The latter involves a greater torque for the larger pencil; the former not necessarily so.

I would have thought that in either case the faster speed requires greater power, not torque.

There are desk-mounted sharpeners with a handle to be turned instead of rotating the pencil. I assume that is not the context for the question.

 

[256bits]

but... the opposing force is not friction

Act similar to...
Since in this case since we want the mount of cutting action, we generally do not label the load as friction.
On the other hand, if we want the shaft in the hub to run with less torque, we label that as bearing friction, and attempt to eliminate any cutting action upon bearing surfaces.
An anecdote from years past was the Red River carts, where the dust would combine with axle grease and churn the inside of wooden axle and hub. Eliminate the grease, the dust did not stay within and the wheels hub and axle lasted much longer, but were squeaky.

General equation for cutting is F = K b h, where K is a factor to be determined, b is the length of the cut, h is the depth.
Depth would be a condition of the thrust force - ranging from not enough and the blade rides over the top, to too much and the blade cuts too deep breaking the tool, the workpiece, or stops the motor.

I would have thought that in either case the faster speed requires greater power, not torque.

with the same rpm, a greater torque requires greater HP.