A coin is placed on a disk that is rotating at 2rev/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

Question restatement: A coin sits on a disk rotating at 2 revolutions per second. With a static coefficient of friction μs = 0.6, what is the maximum distance r from the center the coin can be placed without slipping, given the rotation rate?

Option analysis:
- Option 0.15 m: To assess this, recall the condition for no slipping is that the required centripetal acceleration a_c = ω^2 r must be provided by friction, with f_max = μs N = μs m g. This yields m ω^2 r ≤ μs m g, or r ≤ μs g / ω^2. If we plug ω = 2 rev/s → ω = 4π rad/s, r_max = μs g / ω^2 = 0.6 × 9.8 / (4π)^2 ≈ 5.88 / 157.9 ≈ 0.037 m. Since 0.15 m is much larger than the maximum, slipping would occur before reaching 0.15 m. This option is inconsistent with the physics.

- Option 0.5 m: Using the same derivation as above, the maximum radius is about 0.037 m. 0.5 m is far larger than allowed, so at such a radius the coin would certainly slip. This is incorrect.

- Option "Need the mass of the coin": In the inequality m ω^2 r ≤ μs m g, the mass m cancels out, so the mass is not needed to determine the maximum radius. This statement is misleading; mass is not required for the calculation, so this option is not applicable.

- Option 0.35 m: As with 0.15 m and 0.5 m, this exceeds the computed maximum radius (~0.037 m). Therefore, this choice cannot be correct because the coin would slip at a much smaller radius.

- Option 0.03 m: This is the closest to the calculated maximum radius of about 0.037 m. Since the true maximum lies near 0.037 m, a value of 0.03 m is consistent with being within the non-slipping range, though slightly conservative. This option aligns with the physics when rounded to two significant figures.

Overall, each incorrect option either overestimates the allowable radius given the friction and rotation rate, or misstates the need for mass. The correct boundary comes from r ≤ μs g / ω^2 with ω = 4π rad/s, yielding a result around 3.7 × 10^-2 m, so values near 0.03 m fit the result better than the larger radii.

2025 Fall PHYS 2205 TR 11:00 - 12:15 _ 89122 PHYS 2205 Exam 2- Requires Respondus LockDown Browser

A coin is placed on a disk that is rotating at 2rev/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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