Video: Conservation of Angular Momentum during a Projectile Motion

An ice-skater is preparing for a jump in which he will rotate while in the air. When he is on the ground, with his arms extended, his moment of inertia is 2.2 kg⋅m², and he is spinning at 0.31 rev/s. He launches himself into the air at a speed of 12.6 m/s at an angle of 45° above the horizontal. At the moment that he leaves the ground, he contracts his arms, negligible time, and changes his moment of inertia to 0.62 kg⋅m². How many revolutions can he complete while airborne?

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Video Transcript

An ice skater is preparing for a jump in which he will rotate while in the air. When he is on the ground with his arms extended, his moment of inertia is 2.2 kilograms meter squared, and he is spinning at 0.31 revolutions per second. He launches himself into the air at a speed of 12.6 meters per second at an angle of 45 degrees above the horizontal. At the moment that he leaves the ground, he contracts his arms, taking negligible time, and changes his moment of inertia to 0.62 kilograms meter squared. How many revolutions can he complete while airborne?

We’ll call this number of revolutions capital 𝑁. And we’ll start by drawing a sketch of the situation. When the ice skater jumps, he has an initial moment of inertia we’ve called 𝐼 sub 𝑖 and an initial angular speed we’ve called 𝜔 sub 𝑖, both given to us in the problem statement. He launches himself at an angle we’ve called 𝜃 above the horizontal, with an initial take-off speed we’ve called 𝑣.

Just as he launches off the ice, he pulls his arms into his chest and achieves a final moment of inertia we’ve called 𝐼 sub 𝑓, also given to us. While the skater is in the air, he’s rotating at an angular speed we’ve called 𝜔 sub 𝑓 and, after rotating at this speed while he’s airborne, lands back on the ice. Over that time and spinning at that rate, he completes some number of revolutions we’ve called capital 𝑁. And that’s what we wanna solve for.

We’ll begin by solving for 𝜔 sub 𝑓, the skater’s angular speed while he’s in the air. And to do that, we’ll rely on the principle of the conservation of angular momentum. In a closed system that experiences no net torques, we can say that the system’s angular momentum is equal at any one time to the angular momentum of the system at any other time. That is, the system’s initial angular momentum, wherever we define initial to be, is equal to its final angular momentum.

In our case, we define the initial moment to be a time before the skater has left the ice and the final time to be after the skater has left the ice and changed his moment of inertia. So, we can write that 𝐿 sub 𝑖 is equal to 𝐿 sub 𝑓. And recalling that the magnitude of angular momentum is equal to moment of inertia multiplied by an object’s angular speed, we can expand this statement to say that 𝐼 sub 𝑖 times 𝜔 sub 𝑖 is equal to 𝐼 sub 𝑓 times 𝜔 sub 𝑓.

In all this, it’s 𝜔 sub 𝑓 we want to solve for. We see it’s equal to the ratio of the initial moment of inertia to the final moment of inertia multiplied by the initial angular speed. We’re given all these values in the problem statement, so we can plug in and solve for 𝜔 sub 𝑓. Plugging these values in and entering them on our calculator, we find 𝜔 sub 𝑓 is 1.1 revolutions per second. That’s the angular speed of the skater when he’s in the air.

Knowing that, if we knew how long the skater was in the air, then we could figure out how many revolutions he performed and solve for capital 𝑁. While the skater is airborne, he’s essentially a projectile, under the rules of projectile motion. If we call the total horizontal range that the skater travels capital 𝑅, there’s a relationship we can recall which helps us solve for 𝑅 based on the initial launch speed 𝑣 and the angle 𝜃.

Sometimes called the range equation, this relationship says that the horizontal range of an object which is launched and lands at the same elevation is equal to the square of its initial speed, 𝑣 sub zero, times the sin of two times its launch angle 𝜃, all divided by the acceleration due to gravity 𝑔. If we let 𝑔 be exactly 9.8 meters per second squared, then we can say that the range our ice skater travels while in the air is equal to 𝑣 squared sin two 𝜃 over 𝑔, or 12.6 meters per second squared times the sin of two times 45 degrees, or the sin of 90 degrees, all divided by 9.8 meters per second squared.

We can simplify this last equation a bit because we know that the sin of 90 degrees is equal to one. Before we calculate this result though, let’s consider how it fits into the bigger picture of solving for 𝑁. 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. This range value 𝑅 that we’re calculating is the distance the skater travels in the horizontal direction.

That means that if we were to apply this relationship to our scenario, we would use the ice skater’s horizontal speed 𝑣 times the cos of 𝜃 and set that equal to the horizontal distance traveled 𝑅 divided by the time the skater is in the air 𝑡. Really, it’s not 𝑅 we wanna solve for, but 𝑡, because if we have 𝑡 and we multiply it by 𝜔 sub 𝑓, we’ll get a total number of revolutions the skater has traveled. In other words, we’ll have solved for capital 𝑁.

We can rearrange and say that 𝑡 is equal to 𝑅 divided by 𝑣 times the cos of 𝜃. And now rather than calculating the range 𝑅, we can calculate the time 𝑡 by dividing the range 𝑅 expression by 𝑣 times the cos of 𝜃. So, now we’re solving for 𝑡, which is equal to 𝑣 squared over 𝑔 times 𝑣 times the cos of 45 degrees. And we see that a factor of the speed 𝑣 cancels out from numerator and denominator. Entering this simplified fraction on our calculator, we find that 𝑡 is approximately 1.818 seconds. That’s the amount of time that the skater is in the air.

We’re now ready to solve for capital 𝑁, which is 𝑡 times 𝜔𝑓, or 1.818 seconds times 1.1 revolutions per second. To two significant figures, this is equal to 2.0 revolutions. That’s how many turns the skater would make while airborne.

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