How Is the Position Function H(t) for a Yo-Yo Derived in Calculus?

[farnworth]

H(t) = t^3-6t^2+5t+30 this is a yo yo 30 inches above ground at t =0, at 4 secs it is 18 inches above ground. Please tell me how these figures are derived; t^3,6t^2, 5t; I realize the 30 is initial position. I am 81 but very curious. Thank you.

 

[Vanadium 50]

Where did you get this equation? We can't tell you where it comes from until you tell us where you found it.

 

[Delta2]

Hard to answer this without knowing any context of where did you find this equation. But I can try
It is well known fact that a body that is thrown with initial velocity $v_0$ upwards and from initial height $y_0$ will have an equation of motion, where y is the height from the ground at time t as follows:
$$y=y_0+v_0t-\frac{1}{2}gt^2$$
where g is the gravitational acceleration $g=10m/s^2$
So I believe this explains (almost) the $-6t^2$ and $5t$ terms (for $v_0=5$ and $-\frac{1}{2}g=-5$ and $y_0=30$). For the $t^3$ term I believe the explanation lies within the rotational dynamics of a yo-yo. As the yo-yo goes upward, some of its rotational kinetic energy is converted to gravitational potential energy and this might explain the presence of the $t^3$ term. But I haven't studied myself a lot the rotational dynamics of the yoyo so I can't tell you exactly how we get this term.

 

[farnworth]

Thank you Delta2 I think I am almost there, I found this problem in 'Calculus for Dummies' page181. I really should have used a metric based problem. I really appreciate your kind reply. Regards.

 

[Vanadium 50]

I found this problem in 'Calculus for Dummies' page181.

Then it doesn't necessarily represent an actual physical system. It's just there to practice the mechanics of calculus.