Calculate yoyo friction energy loss

[dislect]

Hi everyone!
Im having a project and i need to estimate the return hight of a falling yoyo
The yoyo is built of two 120cm radius and 2cm long cylinders connected by another one with a radius of 2.5cm and 30cm long. It is droped from the height of 30m with a rope that's 20m long.
How can i estimate how high it will bounce back (in the first time)? I tried looking for friction coefficients online and i found mew=0.25 for nylon which is close enough, but how can i estimate the energy loss due to rope friction?


thanks for your help, i know its not an east one and I am tight in time (need it by tommorow and just now remembered this forum). I set on it with a couple of friends and couldn't figure out a way to calc it

 

[Quinzio]

Friction of the rope will be nearly impossible to calculate. The way the rope wraps around the inner cylinder will influence all.
I think the air friction of the surface of the yo-yo will not be negligible. You could start with that.

 

[dislect]

thanks for your comment
ive already done the drag force calculation but it turns out pretty negligible, the friction of the rope has much more influence

is there any way but experimentel to estimate the energy loss due to the rope friction?
Assume the rope is 1cm thick and wraps ten times before creating another 'loop', meaning it wraps ten times around the cylinder axis and then around the rope itself fore another 18-17 times

 

[cassio]

I am not entirely sure but will the nylon friction not be negligible as the rope just wraps back onto the cylinder and word done is force * distance. No real work done by the friction of the rope. Maybe a loss in the stretching of the nylon and in air resistance.

 

[cassio]

Sorry for double posting but how negligible was air resistance.

 

[DickL]

Is the rope fixed to the yo-yo's axle, or is it looped around the axle as trick yo-yos are strung. If the former, I don't think the rope's friction really comes into play, unless you need to consider that the winding may not be ideal, that is it pinches against itself. Taking this as a somewhat elementary dynamics problem, it is a combined potential vs. kinetic energy problem, with rotational acceleration and energy being part of the analysis.

Now, if this is a real experiment I can see other problems. If the rope is fixed to the yo-yo, there is a change the rope will break when the yo-yo hits the bottom of its drop. The change in the vertical velocity is pretty sudden if the rope doesn't much stretch or can't slip on the axle (nylon is a good choice for the rope as it has good stretch properties). As the yo-yo drops and then climbs back, lateral balance will be affected as the rope unwinds to 1 side or the other (there is a reason real yo-yo's have short axles).

 

[cjl]

I don't think friction is where the energy loss comes from. I think the energy loss is because when it turns around at the bottom, the linear kinetic energy is not recovered. So, to find out how high it will go making some simplistic assumptions, you could assume that all of the linear kinetic energy it has just before turning around is dissipated in the turnaround, and that the new kinetic energy to begin the climb back comes from the rotational energy of the yoyo.