The coin revolves with the record in the case when the force of friction is enough to provide the necessary centripetal force . If this force is not sufficient to provide centripetal force the coin slips on the record Now the frictional force is mu R where R is the normal reaction and Rc = mg
Hence force of friction = mu mg and centripetal force required is `(m upsilon^(2))/(r)` or `m r omega^(2)`
`mu omega` are same for both the coins and we have different valuse of r for the two coins
So to prevent slipping i e causing coins to rotate `mu g ge m r omega^(2)` or `mu g ge r omega^(2)`
` r = 4 cm = (4)/(100) m`
`n = 33 (1)/(3) rev // min . = (100)/(3 xx 60) rev.// sec`
`omega = 2 pi n = 2 pi xx (100)/(180) = 3 .49 s^(-1) `
`:. r omega^(2) = (4)/(100) xx (3. 49 )^(2) = 0.4 ms^(-2) and mu g = 0. 15 xx 10 = 1.5 ms^(2) `
As `mu g ge omega^(2)`, therefore this coin will revovle with the record
`r = 14 cm = (14)/(100) m , omega = 3 .49 s^(-1)`
`r omega^(2) = (14)/(100) xx (3.49)^(2) = 1. 705 ms^(-2)` and `mu g = 1.5 ms^(-2)`
Here `mu g ge r omega^(2)` is not satisfied so this coin will not revolve with the record .
Note that we have noting to do with the radius of the record (=15cm) .
