A coin is placed on a rotating disc whose angular velocity is 1 rev/s. If the coefficient of static friction between the coin and the disc is 0.6 what is the maximum distance from the center of the disc that the coin can be placed without it slipping off?

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BlackTom AI · Accepted Answer

We start by identifying the forces on the coin in the rotating frame. The coin experiences a centripetal requirement of m ω^2 r to stay in circular motion, which must be supplied by friction between the coin and the disc. The maximum static friction force is F_f,max = μs N, and since the disc is horizontal, the normal force N equals mg, so F_f,max = μs mg.

Option by option reasoning:
- 0.35 m: If the coin were at r = 0.35 m, the required centripetal force would be m ω^2 r = m (2π)^2 (0.35). This value would exceed μs mg = 0.6 × m × 9.8, so friction would be insufficient to provide the necessary centripetal acceleration. Therefore, this position would cause slipping.
- 0.05 m: At r = 0.05 m, the required centripetal force m ω^2 r is smaller, since it scales with r. It could be supported by friction, but we must compare against the maximum friction. If r is within the actual maximum, slipping would not occur here; however, this option does not represent the maximum allowable distance, so while it might be safe, it is not the extreme limit.
- 0.15 m: This value is obtained by setting μs g = ω^2 r, i.e., r = μs g / ω^2, which ensures the frictional force exactly provides the needed centripetal force at the threshold of slipping. Substituting μs = 0.6, g = 9.8 m/s^2, ω = 2π rad/s gives r ≈ (0.6 × 9.8) / (2π)^2 ≈ 5.88 / 39.48 ≈ 0.149 m, which rounds to 0.15 m. This is the maximum safe radius.
- None of these: Since we found a consistent maximum radius around 0.15 m, this option would be incorrect because a true maximum exists.
- 0.5 m: This far exceeds the calculated maximum and would require a centripetal force much greater than μs mg, so the coin would slip before reaching this distance.

In summary, the maximum distance from the center at which the coin can be placed without slipping is obtained by equating the available static friction to the needed centripetal force, yielding r_max = μs g / ω^2, which evaluates to about 0.15 m. The option that matches this maximum is 0.15 m.

2025 Fall PHYS 2205 TR 11:00 - 12:15 _ 89122 2205 Final Exam

A coin is placed on a rotating disc whose angular velocity is 1 rev/s. If the coefficient of static friction between the coin and the disc is 0.6 what is the maximum distance from the center of the disc that the coin can be placed without it slipping off?

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