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Step-by-Step Solution
Step 1: Identify the Forces
The coin on the rotating disc experiences a centripetal force that keeps it moving in a circle. This centripetal force is provided by friction.
Step 2: Express Centripetal Force in Terms of Angular Velocity
Centripetal force on the coin is given by:
$ F_{\text{centripetal}} = m \omega^2 r $
where
$m$ is the mass of the coin
$\omega$ is the angular velocity of the disc
$r$ is the distance of the coin from the center
Step 3: Relate Friction to Centripetal Force
Since friction provides the centripetal force, we have:
$ m \omega^2 r = \mu \, m g $
where
$\mu$ is the coefficient of friction
$g$ is the acceleration due to gravity
Step 4: Find the Angular Velocity
The disc rotates at 3.5 revolutions per second (rev/s). One revolution is $2\pi$ radians, hence
$ \omega = 3.5 \times 2\pi \,\text{rad/s} $
Step 5: Substitute the Known Values
The radius $r$ is given as 1.25 cm = $1.25 \times 10^{-2}$ m, and $g = 10 \,\text{m/s}^2$.
From the equation
$ m \omega^2 r = \mu \, m g $,
we get
$ \omega^2 r = \mu \, g $
$ \mu = \frac{\omega^2 r}{g} $
Substituting:
$ \omega = 3.5 \times 2\pi \,\text{rad/s}, \quad r = 1.25 \times 10^{-2} \,\text{m}, \quad g = 10 \,\text{m/s}^2. $
$ \mu = \frac{(3.5 \times 2\pi)^2 \times (1.25 \times 10^{-2})}{10}
= 0.60
$
Step 6: Conclude the Coefficient of Friction
Hence, the coefficient of friction between the coin and the disc is 0.6.
