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Rolls-royce trent 1000 is a high-bypass turbofan engine and is one of the two engine options for the boeing 787 dreamliner. The 112" fan spins at over 2700 pm (rotations per minute). If the fan reaches from 1500 pm to 2200 pm in 7s, what is the average angular acceleration during this period (in rad/s²)? Express your answer correct to two decimal places. Take n = 3.14159.

User Guymid
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Final answer:

The average angular acceleration of the Rolls-Royce Trent 1000 fan during the specified time interval is approximately 628.32 rad/s². This calculation is derived from the change in angular velocity, which is 700 pm, over 7 seconds. The conversion to rad/s² involves multiplying by 2π, resulting in the final answer.

Step-by-step explanation:

To find the average angular acceleration, we first need to calculate the change in angular velocity. From the given information, the fan goes from 1500 pm to 2200 pm in 7 seconds. The change in angular velocity (δω) can be found by subtracting the final angular velocity (2200 pm) from the initial angular velocity (1500 pm): δω = 2200 pm - 1500 pm = 700 pm. To find the average angular acceleration (α), we can use the formula: α = δω / δt. Where δω is the change in angular velocity and δt is the change in time. Plugging in the values, we get α = (700 pm) / (7s) = 100 pm/s. Since we need to express the answer in rad/s², we need to convert the units. One rotation (rev) is equivalent to 2π radians. Therefore, we can convert pm/s to rad/s by multiplying by 2π: α = (100 pm/s) * (2π rad/rev) = 200π rad/s² ≈ 628.32 rad/s².

User Tej Shah
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