Video: Time, Velocity, and Speed

A helicopter blade spins at exactly 140 revolutions per minute. Its tip is 5.00 m from the center of rotation. Calculate the average speed of the blade tip in the helicopter’s frame of reference. Give your answer in meters per second. What is its average velocity over one revolution? Give your answer in meters per second.

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Video Transcript

A helicopter blade spins at exactly 140 revolutions per minute. Its tip is 5.00 meters from the center of rotation. Calculate the average speed of the blade tip in the helicopter’s frame of reference. Give your answer in meters per second. What is its average velocity over one revolution? Give your answer in meters per second.

We’re told in this statement that the helicopter blade spins at exactly 140 revolutions per minute. We can call that number 𝜔. We’re also told that the distance from the axis of the blade’s rotation to its tip is 5.00 meters. We’ll call that distance lower case 𝑟. In part one, we want to solve for the average speed of the tip of the blade in the helicopter’s frame of reference. We’ll call that 𝑠 sub avg. And in part two, what we want to solve for its average velocity over one revolution. We will call this 𝑣 sub avg.

To begin on our solution, let’s draw an overhead view of this rotating helicopter blade. Looking down from an aerial view on this rotating helicopter rotor, we see that it’s rotating with an angular speed 𝜔 of 140 rpm and that each blade on the rotor has a length 𝑟 of 5.00 meters. If we mark the tip of one of the blades of the rotor and follow its path as it moves around a revolution, we want to solve for the average speed of that tip of the blade as well as its average velocity.

Starting with average speed, we can recall that the average speed of an object is equal to the distance it travels divided by the time it takes to travel that distance. In our instance, we can say that the distance we’re interested in is one complete revolution. That is, the distance that the tip of the blade would travel is two times 𝜋 times the radius of the rotor. We’re given that radius in our problem statement. So, all this left us to solve for is the time it takes for the blade to make one complete rotation.

If the rotor makes 140 revolutions every minute, then that tells us that every 60 seconds it rotates 140 times. This tells us that Δ𝑡, the time it would take the rotor to go through one complete rotation, is the inverse of this fraction. In other words, in sixty one hundred and fortieths of a second, the rotor goes through a complete rotation, or every three-sevenths of a second.

Now that we’ve solved for Δ𝑡 and we know 𝑟, we’re ready to plug in and solve for our average speed. When we enter this expression on our calculator, we find, to three significant figures, that 𝑠 sub avg is 73.3 meters per second. That’s the average speed of the tip of the helicopter blade.

Now, we move on to solving for the average velocity of the tip of the blade over one rotation, which we’ll find is not the same as average speed. Recall that average velocity is equal to displacement over time, where displacement is the difference in position from the start to the end of an object’s motion. With that understanding, we start to see what the average velocity of a rotor blade would be over one revolution.

When the single rotation begins, the position of our rotor blade is the same as its position when that rotation ends. This means that our displacement, which is the difference between those two positions, is equal to zero because those positions are the same. This means that our average velocity over one complete rotation is zero meters per second. Considered over that time span, the tip of the blade’s displacement is zero and so so is its average velocity.

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