find the maximum possible error in the measurement of the centripetal force on an object of mass m travelling at velocity v in a circle of radius r if m=3.5 +or- 0.1 kg,v=20 +or- 1 m/s and r =12.5 +or - o.5 m
Asked by fathimaansari | 15th Jan, 2012, 05:04: PM
Centripetal force, F = mv2/r
?F/F = (?m/m) + (2 ?v/v) + (?r/r)
M= 3.5 +_0.1 kg
V= 20 +_1 m/s
R = 12.5+_ 0.5 m
?F/F = ( 0.1/3.5) +(2 x1/20) + (0.5/12.5)
= 0.028571 +0.1 + 0.04
=0.1685
maximum possible percentage error is:
(?F/F)x 100% =0.1685 x 100=16.8%
Centripetal force, F = mv2/r
?F/F = (?m/m) + (2 ?v/v) + (?r/r)
M= 3.5 +_0.1 kg
V= 20 +_1 m/s
R = 12.5+_ 0.5 m
?F/F = ( 0.1/3.5) +(2 x1/20) + (0.5/12.5)
= 0.028571 +0.1 + 0.04
=0.1685
maximum possible percentage error is:
(?F/F)x 100% =0.1685 x 100=16.8%
Answered by | 15th Jan, 2012, 06:17: PM
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