Answer to Question #121857 in Mechanical Engineering for Michael

Question

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%

Answer 1

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 = mrw²

= mr (2*3.14 N/60)²

= 0.011(mrN²)

N² = (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/rm³)

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

= -4.767 square root (600/(0.12³*0.025)

= -17765.56

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

= -4.767 square root (F/mr³)

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

= -4.767 square root (600/(0.12*0.025³)

= -85274.7

calculate the uncertainty for angular velocity

W_N = square root ( [1.2*3.3553²] +[0.0005*-17765.56]² +[2*10^-5 *-85274.7]²)

= +- 10 rpm

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