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The propeller of an aircraft is spinning at 10 rev/s when the pilot shuts off the engine. The propeller reduces its angular velocity at a constant 2.0rad/s2 for a time period of 40 s. What is the rotation rate of the propeller in 40 s? Is this a reasonable situation?

(a) 30 rev/s
(b) 90 rev/s
(c) 50 rev/s
(d) 10 rev/s

User Mallory
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1 Answer

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Final answer:

Using the formula for angular deceleration, with the provided values, results in a negative angular velocity, which indicates the propeller has stopped and reversed direction, a situation that is not reasonable for an aircraft propeller. None of the given options are correct.

Step-by-step explanation:

The subject of the question is Physics, and it is at a high school level. To solve for the rotation rate of the propeller after 40 seconds, we can use the formula for angular deceleration, which is ω = ω0 - αt, where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular deceleration, and t is the time in seconds.

Given that the initial rotation rate is 10 rev/s, we first need to convert this rate to radians per second. Since 1 revolution is equivalent to 2π radians, we have ω0 = 10 rev/s × 2π rad/rev = 20π rad/s.

The angular deceleration is given as α = 2.0 rad/s2. Substituting these values into the equation gives us ω = 20π rad/s - (2.0 rad/s2 × 40 s) = 20π rad/s - 80 rad/s.

However, because this calculation results in a negative number for ω, it means that the propeller has completely stopped and even reversed direction, which is not a reasonable situation for an aircraft propeller.

Therefore, the rotation rate of the propeller after 40 seconds cannot be a positive number like the ones given in options (a) 30 rev/s, (b) 90 rev/s, or (c) 50 rev/s.

Option (d) 10 rev/s is also incorrect because the propeller decelerates. The mistake in this problem is either in the calculation or the premise; the aircraft propeller would never reverse direction under a constant deceleration condition.

User Derpirscher
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