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Question
The coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the center of the coin is 2 cm away from the center of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim? (use g = π2m/s2)
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Solution
Given: μs = 0.5
rD = 8 cm = 8 × 10−2 m
rC = 2 cm = 2 × 10−2 m
Fmin = ?
mrω2 = μN
mrω2 = μ mg ...[N = mg]
∴ ω2 = `(mu g)/r`
⇒ 4 π2 f2 = `(0.5 xx pi^2)/(2 xx 10^-2)` ...[ω = 2 π f ⇒ ω2 = 22 π2 f2 ⇒ ω2 = 4 π2 f2]
⇒ `4 cancel(pi^2)f^2 = (0.5 xx cancel(pi^2))/(2 xx 10^-2)`
⇒ f2 = `(0.5 xx 10^2)/(4 xx 2)`
⇒ f2 = `50/8`
⇒ f2 = `25/4`
⇒ f = `5/2 = 2.5` rev/s
`f prop 1/sqrtr, f_n prop 1/sqrt(r_n)`
`f_n/f = sqrt(r/r_n)`
= `sqrt((2 xx 10^-2)/(8 xx 10^-2))`
= `sqrt(1/4)`
fn = `1/2 f`
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