# A centrifuge, r = 0.06 cm, revolves at 24 000 rpm. If its tensile strength is 120000 N, what is the maximum massthat it can hold? Answer: 0.32 kg

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### 1 Answer

### User Comments

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.` **

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