Question

A .0250 kg sock spins in a dryer of radius .225 m once every .304 seconds. How much centripetal force acts on the sock?

Answer 1

`24.025"N"`

Explanation:

`F_("centripital")=(mv^2)/r`
`m = 0.250 kg`
`r = 0.225 m`

`v = ?`
`v = ("distance")/("time")`

Here, distance `= 2pir`

`=> v= ((2pi0.225)m)/((0.304)s) = 4.65m/s`

Now, substituting,
`F_("centripital")=(0.25kg(4.65m/s)^2)/(0.225m)`

With a little help from Google,😁

`F_("centripital")= 24.025N`

Answer 2

The force is found from F =`mv^2/r` or more simply in this case F =`momega^2r`

Explanation:

You calculate `omega` from `2pi.f`

Should work out ok.