While waiting to board a helicopter, you notice that the rotor's motion changed from 340 rev/min to 265 rev/min in one minute. Assuming that this acceleration remains constant, calculate how much longer (in seconds) it will take for the rotor to stop.
a = dv/dt = (265-340)/60 = -1.25 rev/sec
v = Vi + a t
0 = 265 - 1.25 t
t = 212 seconds
To calculate the time it will take for the rotor to stop, we need to determine the rate at which the rotor's speed is decreasing.
Given:
Initial speed (ω1) = 340 rev/min
Final speed (ω2) = 265 rev/min
The change in speed (Δω) = ω2 - ω1
= 265 rev/min - 340 rev/min
= -75 rev/min (negative sign indicates a decrease in speed)
We can calculate the rate of acceleration using the formula:
Acceleration (α) = Δω / Δt
Here, we know that Δt = 1 minute since the change in speed occurred over one minute.
Substituting the known values:
α = -75 rev/min / 1 min
= -75 rev/min^2
To determine the time it will take for the rotor to stop, we can use the formula for acceleration:
Final speed (ωf) = Initial speed (ωi) + α * t
Assuming the rotor stops completely, the final speed (ωf) will be 0 rev/min.
0 rev/min = 340 rev/min + (-75 rev/min^2) * t
Simplifying the equation:
-340 rev/min = -75 rev/min^2 * t
Dividing both sides by -75 rev/min^2:
4.53333 min^2 = t
Converting minutes to seconds:
t = 4.53333 min^2 * 60 s / min
≈ 272 seconds
Therefore, it will take approximately 272 seconds for the rotor to stop completely.
To calculate the time it will take for the rotor to stop, we need to use the concept of uniform acceleration. Here's how to do it step by step:
Step 1: Find the angular acceleration (α)
The change in angular velocity over time (Δω/Δt) is the angular acceleration (α). In this case, the change in angular velocity is given by:
Δω = ωf - ωi
where ωf is the final angular velocity (265 rev/min) and ωi is the initial angular velocity (340 rev/min).
Plugging in the values:
Δω = 265 rev/min - 340 rev/min
Δω = -75 rev/min (negative because the velocity is slowing down)
Next, we need to convert rev/min to rad/s. Since 1 revolution (rev) is equal to 2π radians, and 1 minute (min) is 60 seconds, we can set up the following conversion factors:
1 rev/min = (2π rad)/60 s
1 rad/s = 60/(2π) rev/min
Converting Δω to rad/s:
Δω = -75 rev/min * (2π rad)/60 s
Δω ≈ -7.85 rad/s
So, the angular acceleration (α) is approximately -7.85 rad/s.
Step 2: Find the time it takes for the rotor to stop (t)
As per the problem statement, the acceleration remains constant. Therefore, we can use the equation that relates angular acceleration, final angular velocity, initial angular velocity, and time:
Δω = αt
Plugging in the values:
-7.85 rad/s = -7.85 rad/s^2 * t
Solving for t:
t = (-7.85 rad/s) / (-7.85 rad/s^2)
t = 1 second
So, it will take 1 second for the rotor to stop.
However, the question asks for the amount of time in seconds beyond the initial one minute. To calculate that, we need to subtract the initial one minute from the time it takes for the rotor to stop:
Extra time = t - 1 second
Extra time = 1 second - 1 second
Extra time = 0 seconds
Therefore, the rotor will stop exactly 1 minute after the initial observation, and no additional time is required.