Physics Analysis: Coin on Rotating Disc
The problem asks for the coefficient of friction ($\mu_s$) required to keep a coin stationary on a horizontally rotating disc. The coin experiences a centripetal force due to the rotation, which must be balanced by the static friction force between the coin and the disc.
Physics Principles and Formulas
For the coin to remain at rest, the static friction force ($f_s$) must provide the necessary centripetal force ($F_c$).
- Centripetal Force: $ F_c = m a_c $, where $m$ is the mass of the coin and $a_c$ is the centripetal acceleration.
- Centripetal Acceleration: $ a_c = \omega^2 r $, where $\omega$ is the angular velocity and $r$ is the radius.
- Angular Velocity: $ \omega = 2 \pi f $, where $f$ is the frequency of rotation.
- Maximum Static Friction: $ f_{s,max} = \mu_s N $, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.
- Normal Force on a horizontal surface: $ N = mg $, where $g$ is the acceleration due to gravity.
The condition for the coin to remain at rest is $F_c \le f_{s,max}$. For the minimum required coefficient of friction, we consider the case where $F_c = f_{s,max}$.
$ m a_c = \mu_s N $ $ m (\omega^2 r) = \mu_s (mg) $ $ \mu_s = \frac{\omega^2 r}{g} $Calculation Steps
- Calculate the angular velocity ($\omega$): Given $f = 3.5$ rev/s. $ \omega = 2 \pi f = 2 \pi (3.5 \text{ s}^{-1}) = 7 \pi \text{ rad/s} $.
- Substitute values into the formula for $\mu_s$: Given $r = 1.25$ cm $= 0.0125$ m and $g = 10$ m/s². $ \mu_s = \frac{(7 \pi \text{ rad/s})^2 \times (0.0125 \text{ m})}{10 \text{ m/s}^2} $ $ \mu_s = \frac{49 \pi^2 \times 0.0125}{10} $
- Approximate and calculate $\mu_s$: Using the approximation $ \pi^2 \approx 10 $: $ \mu_s \approx \frac{49 \times 10 \times 0.0125}{10} $ $ \mu_s \approx 49 \times 0.0125 $ $ \mu_s \approx 0.6125 $
- Determine the answer: The calculated coefficient of friction is approximately $0.6125$. This value is closest to the option $0.6$.
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