A coin placed on a disc rotates with speed of $33 \frac{1}{3}\;rev/min$ provided that the coin is not more than $10 \; cm$ from the axis. Calculate the coefficient of static friction between the coin and the disc.

Given,

Frequency of rotation $(f) = 33 \frac{1}{3}\;rev/min = \frac{100}{3\;*\;60}\;rev/sec = 0.56\;rev/sec$

Radius $(r) = 10\;cm = 0.1\;m$

Coefficient of static friction $(\mu_s) = \;?$

When the coin just sustain on the disc, the frictional force is equal to the centripetal force,

i.e $F_s = \frac{m\;V^2}{r}$

or, $\mu_s\;R = m\; \omega^2\;r$          [∵ $F_s = \mu\;R$]

or, $\mu_s = \frac{m\;\omega^2\;r}{m\;g} = \frac{(2\;\pi\;f)^2\;r}{g}$        [∵ $R = m\;g$]

or, $\mu_s = \frac{4\;\pi^2\;(0.56)^2\;0.1}{10} = 0.123$

Coefficient of static friction $(\mu_s) = \;0.123$

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A certain string breaks when a weight of $25\;N$ acts on it. A mass of $500\;gm$ is attached to one end of the string of $1\;m$ long and is rotated in a horizontal in a horizontal circle. Find the greatest number of revolutions per minute which can be made without breaking the string?

Given,

 

Maximum string $(T_{max}) = 25 \; N$

Mass of the object $(m) = 500 \; gm = 0.5 \; kg$

Radius of circle $(r) = 1\;m$

Frequency $(f) = \; ?$

We have,

$T_{max} = $$\frac{mv^2}{r}$$ = m\;\omega^2\;r$ = Centripetal Force

or, $T_{max} = m\;(2\;\pi\;f)^2\;r$

$25 = 0.5 * 4\pi^2 * f^2 * 1$

$f^2 = $$\frac{25}{0.5\; * \;4\; *\; \pi^2 \;*\; 1}$$ = 1.266$ rev/sec

$f = 1.125\;$ rev/sec

⇒ $f = 1.125 * 60\;rpm = 67.5\;rpm$

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