Question #121857

A mechanical speed control system works on the basis of centrifugal force, which is related to angular velocity through the formula F = mrw2. The following values are measured to determine ω: r = 25 ± 0.02 mm. m = 120 ± 0.5 g. and F = 600 ± 0.2% N. Find the rotational speed in rpm and its uncertainty. All measured values have a confidence level of 95%

Expert's answer

value of mass, *m*+*w*_{m }is 120 +- 0.5 g that is 0.12 +- 0.0005 kg

value of radius , r +- w_{r} is 25 +- 0.02 mm that is 0.025 +- 2*10^{-5} m

value of the force F +- w_{f} is 600+- 0.2%N that is 600 +- 1.2 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)

Calculate the value of the rotational speed N

N = 9.534 SQUARE ROOT (F/mr)

substitute 600 N for F 0.12kg for m and 0.025m for r in N

N = 9.534 SQUARE ROOT ( 600/(0.12*0.025))

= 4263.8 rpm

Therefore the value of rotational speed is 4263.8 rpm

Calculate the partial derivative of rotational speed with respect to F

dN/dF = 9.534 d/dF (square root (F/mr))

= 4.767 square root (1/Fmr)

substitute 600N for F 0.12kg for m and 0.025 m for f

dN/dF = 4.767 "squareroot" (1/(600*0.12*0.025)

= 3.553

Calculate the partial derivative of angular velocity with respect to m

dN/dm = 9.534 d/dm (square root (F/mr))

= -4.767 square root (F/rm^{3})

substitute 600 N for F 0.12kg for m and 0.025m for r

= -4.767 square root (600/(0.12^{3}*0.025)

= -17765.56

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

= -4.767 square root (F/mr^{3})

substitute 600 N for F 0.12kg for m and 0.025m for r

= -4.767 square root (600/(0.12*0.025^{3})

= -85274.7

calculate the uncertainty for angular velocity

W_{N} = square root ( [1.2*3.3553^{2}] +[0.0005*-17765.56]^{2} +[2*10^{-5} *-85274.7]^{2})

= +- 10 rpm

Hence the value of uncertainty for the angular velocty +- 10 rpm

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