A rotating merry-go-round has an angular velocity of 0.184 rad/s. Find the period of rotation of the merry-go-round.

Question

BlackTom AI · Accepted Answer

We start by identifying the relationship between angular velocity and period. The period T is the time for one full revolution, which relates to angular velocity by T = 2π / ω.
Option A: 3.91 s. If you plug ω = 0.184 rad/s into T = 2π/ω, you get about 34.1 s, not 3.91 s. This option misplaces the arithmetic and is too small by roughly a factor of 9. 
Option B: 13.6 s. This value is closer to what you’d get if ω were around 0.462 rad/s, which is not the given angular speed. It reflects a miscalculation in inverting or misreading units.
Option C: 5.43 s. This would correspond to an ω around 2π / 5.43 ≈ 1.158 rad/s, which is substantially larger than the provided 0.184 rad/s. The inconsistency shows a misunderstanding of how period scales with angular velocity.
Option D: 34.1 s. Using T = 2π / ω with ω = 0.184 rad/s gives T ≈ 6.2831853 / 0.184 ≈ 34.13 s, which rounds to 34.1 s. This aligns with the fundamental relation between period and angular speed and matches the expected magnitude for the given ω.
In summary, the key concept is that the period is inversely proportional to angular velocity, and precise calculation with 2π divided by 0.184 rad/s yields a result around 34.1 seconds, making this option the correct one.

A rotating merry-go-round has an angular velocity of 0.184 rad/s. Find the period of rotation of the merry-go-round.单项选择题

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