I was confused by another equation again. Would appreciate it if anyone sheds light for me!

Consider the following question:

A teacher demonstrates a giant pirouette (legs fully extended, body straight) on the uneven bars for a young student. She swings 2 times around the bar in 5 seconds. The length of her body from the tops of her hands to the tips of her toes is 6 feet.

Now it's the young gymnast's turn. She is 2 feet shorter and, after a little practice, she swings her feet at the same average speed as her teacher. How many pirouettes can the student do during 5 seconds?

The answer to this question was The young gymnast does exactly one more pirouette than her teacher in 5 seconds.

Essentially, the explanation was roughly like so:

The teacher and the young gymnast are rotating around the bar, which is the axis of rotation in this problem. Since we were told that their feet move at the same speed, we will try to relate this fact to the lengths of their bodies to obtain the young gymnast's rate of rotation. Then, we can calculate how many rotations the young gymnast completes in 5 seconds.

We start by relating the speed of the teacher's feet, $$v_t$$ to her rotation rate. $$f_t$$ In the last pane, we focused on the relationship of speed to the rotation rate and radius:

$$v_t = 2 \pi r_t f_t$$

Because she holds onto the bar with her hands, we can assert that the radius of rotation of her feet is equal to the length of her fully extended body, 6 feet. We also know her rotation rate is 0.4 rotations per second (because she does 2 pirouettes in 5 seconds).

With the same relationship, the gymnast's feet can be related to her unknown rotation rate, : where is the radius of rotation of the gymnast's feet, 4 feet. The speed of her feet are also $$v_t$$ the same as the teacher's feet.

We can set our two expressions for $$v_t$$ equal and cancel out common factor $$2 \pi$$: The only unknown in this equation is the gymnast's rotation rate, $$f_g$$, which we can solve for:

$$f_g = \frac {r_t f_t}{r_g} = 0.6$$

and we conclude that the gymnast's rotation rate is 0.6 rotations per second. That's 1.5 times larger than the teacher's, so we expect the number of pirouettes to increase by the same factor. Indeed, the number of rotations performed by the gymnast at a rate of 0.6 rotations per second during 5 seconds is 3, exactly one more than the teacher.

So my question is this: how can $$\frac {r_t f_t} {r_g}$$ be used to calculate the second gymnast's (the student's) rotation rate $$f_g$$? As far as I know, what we're doing here is to multiply 0.6 with 6 feet and get it divided by 4 feet (the student's height). Nothing else explains to me this relationship.

Thanks in advance.


So using formulas for the rotation around an axis to get the linear velocity of the feets of the teacher $(v_t)$, and of the gymnast $(v_g)$, you get:

$v_t=2\pi R_t f_t$, and also $v_g=2\pi R_g f_g$. Where R is the radius of the motion, i.e the length of each, and where f is the frequency, the inverse of the period.

From the statement of the problem you get that $v_t=v_g$, then: $$2\pi R_t f_t=2\pi R_g f_g\\R_t f_t=R_g f_g\\f_g=\frac {R_t f_t}{R_g}$$

You have $R_t=6$, $f_t=\frac 2 5=0.4$, and you have $R_g=4$, so you get the result $f_g=0.6$ directly.

I think this is a pretty straightfoward calculation, so if this answer is not what you are looking for, it would help if you could explain in more detail what you mean by "How can $\frac {r_t f_t} {r_g}$ be used to calculate the gymnasts rotation rate?"

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  • $\begingroup$ Hi! Sorry for the late reply. Thanks for answering! As for the explanation, I meant "How can $$ \frac {r_t f_t}{r_g} $$be used to calculate the gymnasts rotation rate?" by, how can we actually calculate the second gymnast's rotation rate using that equation? I understand how the equation works and everything, but just how can $$0.6 * 6 / 4 $$ be used to calculate the second gymnast's rotation rate (0.4). I'm trying to understand the concept here. I'm not sure if you understand what I meant (I'm just pure bad at explaining things) but that's all I can think of right now. Thank you in advance! $\endgroup$ – Jeffrey10th Oct 13 '18 at 5:23

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