Question
4) An electric motor is turned off, and its angular velocity decreases uniformly from 900rev/min to 400rev/min in 5.00 s a) Find the angular acceleration in rev/s^2 and the number of revolutions made by the motor in the 5 .00 s interval. b)How many more seconds are required for the motor to come to rest if the angular acceleration remains constant at the value calculated in part (a)? -1.67rev/s/s 54 rev, 3.98s more
Solution
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(235 Votes)
Evie
Professional · Tutor for 6 years
Answer
Here's the solution:**a) Angular Acceleration and Number of Revolutions**1. **Convert initial and final angular velocities to rev/s:** * Initial angular velocity (ω₀) = 900 rev/min * (1 min/60 s) = 15 rev/s * Final angular velocity (ω) = 400 rev/min * (1 min/60 s) = 6.67 rev/s2. **Calculate angular acceleration (α):** We can use the following equation: ω = ω₀ + αt * Rearranging for α: α = (ω - ω₀) / t * Plugging in the values: α = (6.67 rev/s - 15 rev/s) / 5.00 s = -1.67 rev/s²3. **Calculate the number of revolutions (θ):** We can use the following equation: θ = ω₀t + (1/2)αt² * Plugging in the values: θ = (15 rev/s)(5.00 s) + (1/2)(-1.67 rev/s²)(5.00 s)² * θ = 75 rev - 20.875 rev ≈ 54.1 rev (Rounding to two significant figures gives 54 rev)**b) Time to Come to Rest**1. **Set final angular velocity to zero:** ω = 0 rev/s2. **Use the same equation as in step 2 of part (a) to find the time (t):** * ω = ω₀ + αt * Rearranging for t: t = (ω - ω₀) / α * Since the motor is now starting from 400 rev/min (6.67 rev/s), this becomes our new ω₀. * Plugging in the values: t = (0 rev/s - 6.67 rev/s) / -1.67 rev/s² ≈ 3.99 s (Rounding to two significant figures gives 4.0 s)Therefore, it takes approximately 4.0 more seconds for the motor to come to rest. The provided answer of 3.98s is also correct within rounding error. I've shown the calculations with slightly different rounding choices to demonstrate how minor differences can arise.