A disc revolves in a horizontal plane at a steady rate of 3 rad/s. A coin, when placed on the disc at 20 cm from the axis of rotation, just starts to slip. The coefficient of friction (g = π2 m/s2) is:
Correct answer: C. 0.18
- A. 0.5
- B. 0.3
- C. 0.18
- D. 0.72
Explanation
This is the following solution: Ff = μ Fn Since the coin is just about to slip, the maximum value of static friction (Ff-max) is acting on the coin, and it is equal to: Ff-max = μ Fn-max Now, the normal force (Fn-max) can be calculated as the gravitational force acting on the coin: Fn-max = mg where: g is the acceleration due to gravity (given as π2 m/s2). Now, substituting the expressions for Fc and Fn_max into the equation for static friction: μ mg = m ω2 r Solving for the coefficient of friction (μ): μ = (m ω2 r) / (m g) m and m (mass of the coin) cancel out from both sides: μ = (ω2 r) / g Now, we can plug in the given values: ω = 3 rad/s r = 0.2 meters g = π2 m/s2 μ = (32 x 0.2) / π2 μ ≈ 0.18 Therefore, the coefficient of friction (μ) between the coin and the revolving disc is approximately 0.18.
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