Question #121862

A mechanical speed control system works on the basis of centrifugal force, which is related to angular velocity through the formula. F = mrω2 The following values are measured to determine ω: r = 20 ± 0.02 mm, m = 100 ± 0.5 g, and F = 500 ± 0.1% N. Find the rotational speed in rpm and its uncertainty.

Expert's answer

value of mass, *m*+*w*_{m}* *is 100 +- 0.5 g that is 0.1 +- 0.0005 kg

value of radius , r +- w_{r }is 20 +- 0.02 mm that is 0.02 +- 2*10-5 m

value of the force F +- w_{f} is 500+- 0.1%N that is 500 +- 0.5 N

Calculate the expression for rotational speed *N*

*F = mrw*^{2}

= mr (2*3.14 N/60)^{2}

= 0.011(mrN^{2})

N^{2} = (90.9F/mr)

N = SQUARE ROOT (90.9F/mr)

substitute 500 N for F 0.1kg for m and 0.02m for r in N

N = SQUARE ROOT ( (90.9*500)/(0.1*0.02))

= 4767 rpm

Therefore the value of rotational speed is 4767 rpm

Calculate the expression for angular velocity

w = square root (F/mr)

Calculate the partial derivative of angular velocity with respect to F

dw/dF = d/dF (square root (F/mr))

= 1/2 square root (1/Fmr)

substitute 500N for F 0.1kg for m and 0.02 m for f

dw/dF = 1/2 square root (1/(500*0.1*0.02)

= 0.5

Calculate the partial derivative of angular velocity with respect to m

dw/dm = d/dm (square root (F/mr))

= -1/2 square root (F/rm^{3})

substitute 500 N for F 0.1kg for m and 0.02m for r

= -1/2 square root (500/(0.1^{3}*0.02)

= -2500

Also calculate the partial derivative of angular velocity with respect to r

= -1/2 square root (F/mr^{3})

substitute 500 N for F 0.1kg for m and 0.02 m for r

= -1/2 square root (500/(0.12*0.02^{3})

= -12500

calculate the uncertainty for angular velocity

W_{N} = square root ( [0.5*0.5^{2}] +[0.0005*-2500]^{2} +[2*10^{-5} *-12500]^{2})

= +- 1.3 rpm

Hence the value of uncertainty for the angular velocity +- 1.3 rpm

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