Relative error of radius when derived from diameter

[smoze]

Hi guys

Have lurked the forums for a while but this is my first contribution. I am in first year physics at uni and I was having a discussion with my tutor regarding the relative error for an experiment (circular motion / moment of inertia). (I tried searching for an answer to this with no luck)

The relative error for the radius of a wheel needs to be calculated as part of the experiment.

The radius was calculated by measuring the diameter of the wheel with verniers and dividing by 2. The smallest division on the verniers is 0.05 mm and for arguments sake the diameter of one of the wheel is 20 mm (there are actually several wheels that need measuring).

I understand relative error to be half the smallest division on the measuring instrument divided by the measured total, units irrelevant provided they are both the same. In the above instance this equates to an error rate of 0.025/20 = 0.125% for the diameter.

Now for the radius my tutor is telling me that you keep the 0.025 (half smallest division on verniers) and divide by the radius which doubles your error to 0.025/10 = 0.25%.

I believe this to be incorrect and that relative error should be the same as for the diameter. My argument for this is that the radius is defined as exactly half of the diameter, and as such exactly half of any absolute error in the diameter should be attributed to the radius. I don't know whether this relative error is best explained by halving the smallest division when you halve the measurement or by leaving the error for the diameter untouched. The other possibility is that I am indeed wrong.

Looking forward to your comments, thanks.

 

[sophiecentaur]

It all hangs upon what would be introducing the error. If the wheel were eccentric yet circular, the radius would vary a lot but the diameter measurement error could be small and wouldn't reveal the radius error. This could be very relevant when angular momentum is involved and would be a totally hidden error.

If the wheel is balanced and the only error is introduced by your caliper measurement then the same %error would apply to radius and diameter, imo.

 

[Naty1]

"I understand relative error to be half the smallest division on the measuring instrument divided by the measured total, units irrelevant provided they are both the same."

I'd say 'ok for direct measurements'...as I think you imply...
but calculated lengths I think would be different...see below.

My argument for this is that the radius is defined as exactly half of the diameter, and as such exactly half of any absolute error in the diameter should be attributed to the radius. I don't know whether this relative error is best explained by halving the smallest division when you halve the measurement or by leaving the error for the diameter untouched.

Suppose you measured the radius and had to decide on the diameter error??
Any measurement you make can be +/- .025, right??

...if the radius is 10mm +/- .025, and you double it that's one thing...; if you directly measure the diameter, that's different...error is still +/- .025.

2[10 + .025] = 20.05; 2[10 - .025] = 19.95 etc...

[I guess this comes out as Sophie explained for a calculated size.]

[attribution of error can be really tricky because as Sophie details in the prior post; different things may contribute and you may never even realize some...for example, what makes you think the vernier is accurate?? Did you calibrate it somehow and what is the calibration error?? Are you using it at the same tmperature at which it was calibrated??...just food for thought...]

 

[Chestermiller]

Hi guys

Have lurked the forums for a while but this is my first contribution. I am in first year physics at uni and I was having a discussion with my tutor regarding the relative error for an experiment (circular motion / moment of inertia). (I tried searching for an answer to this with no luck)

The relative error for the radius of a wheel needs to be calculated as part of the experiment.

The radius was calculated by measuring the diameter of the wheel with verniers and dividing by 2. The smallest division on the verniers is 0.05 mm and for arguments sake the diameter of one of the wheel is 20 mm (there are actually several wheels that need measuring).

I understand relative error to be half the smallest division on the measuring instrument divided by the measured total, units irrelevant provided they are both the same. In the above instance this equates to an error rate of 0.025/20 = 0.125% for the diameter.

Now for the radius my tutor is telling me that you keep the 0.025 (half smallest division on verniers) and divide by the radius which doubles your error to 0.025/10 = 0.25%.

I believe this to be incorrect and that relative error should be the same as for the diameter. My argument for this is that the radius is defined as exactly half of the diameter, and as such exactly half of any absolute error in the diameter should be attributed to the radius. I don't know whether this relative error is best explained by halving the smallest division when you halve the measurement or by leaving the error for the diameter untouched. The other possibility is that I am indeed wrong.

Looking forward to your comments, thanks.

I agree with your assessment. (2r ± δ ) /2 = r ± δ/2

 

[smoze]

Thank you all for your replies