A disc revovles with a speed of `33 (1)/(3) rev//min` and has a radius of 15 cm Two coins are palaced at 4 cm and 14 cm away from the center of the record If the coefficient of friction between the coins and the record is 0.5 which of the coins will revolve with the road ?

Explanation

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) .