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A centrifuge, r = 0.06 cm, revolves at 24 000 rpm. If its tensile strength is 120000 N,...
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You should remember the formula for centripetal force such that:
F = m*centripetal acceleration
You need to evaluate the centripetal acceleration such that:
centripetal acceleration `= omega^2*r`
You should evaluate velocity omega such that:
`omega = 2pi*N/60`
Substituting 24 000 for N yields:
`omega = 2pi*24000/60`
`omega = 2pi*400 = 800pi`
Substituting 800pi for omega in centripetal acceleration yields:
centripetal acceleration = `640000*pi^2*0.06`
Hence, substituting `640000*pi^2*0.06` for centripetal acceleration in F = m*centripetal acceleration yields:
`120 000 = m*640 000*pi^2*0.06`
`12 = m*64*pi^2*0.06 =gt 3 = m*16pi^2*0.06`
`m = 1/(16pi^2*0.02)`
`m = 0.316 Kg`
Hence, evaluating the mass that the tensile can hold yields `m ~~ 0.316 = 0.32 Kg.`
Posted by sciencesolve on June 29, 2012 at 5:30 PM (Answer #1)
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