Inertia for a special yo-yo help

[ahello888a]

Inertia for a special yo-yo help!

Homework Statement

A yo-yo is made of two solid cylindrical disks, each of mass M and diameter D, , joined by a (concentric) thin solid cylindrical hub of mass m and diameter d. Use conservation of energy to calculate the linear speed of the yo-yo just before it reaches the end of its long string length L, if it is released from rest.

Homework Equations

KE for translational: 0.5*m*v^2
KE for rotational: 0.5*I*w^2
PE: mgh

for this problem: potential = KE(trans) + KE(rot)

The Attempt at a Solution

My main concern is how to treat the inertias for the 3 cylindrical objects (2 identical big cylindrical disks and 1 small cylindrical hub). At first I tried to simply combine the inertias for all 3 into one KE equation (i.e. I = all 3 cylindrical objects...KE(rot) = .5*I*w^2.) but when I tried to convert the w to v/r I couldn't decide which r to use (disk or hub). So then I tried to separate the KE(rot) into 2 (1 for hub, 1 for 2 disks). this way allowed me to have different r for the w=v/r substitution, but my answers came out wrong. How then should I do this problem?

 

[Dick]

Use the radius of the hub to compute w. You can certainly add the moments of inertia. They are all rotating at the same rate. Beyond that it's hard to say why you are getting wrong answers until you tell us the answers you are getting.

 

[ahello888a]

is there any reason to use that over the radius of the cylindrical disks?

 

[Dick]

I used my technical knowledge of yoyo's. They consist of two big disks with a smaller hub in between. The string wraps around the hub.

 

[ahello888a]

Heres what I get after simplifying the Kf=Ki

v is velocity, d is for the disks, h is for the hub, L is for height/string length

v$$^{2}$$ = $$\frac{2m_{tot}gL}{m_{tot}+M_{d}\frac{R_{d}}{R_{h}}^{2}+0.5M_{h}}$$

I plugged in the numbers that my book version that the question provides and I still don't get the right answer. If you want to try,
mass of disk = 0.050kg
mass of hub = 0.005kg
diameter of disk = 0.075m
diameter of hub = 0.010m

answer comes out to be: 0.84m/s

I didn't bother to convert the diameter to radius since it just cancels out in my equation.

 

[ahello888a]

...so i guess its just something that you'd have to know about yo-yos

 

[Dick]

Heres what I get after simplifying the Kf=Ki

v is velocity, d is for the disks, h is for the hub, L is for height/string length

v$$^{2}$$ = $$\frac{2m_{tot}gL}{m_{tot}+M_{d}\frac{R_{d}}{R_{h}}^{2}+0.5M_{h}}$$

I plugged in the numbers that my book version that the question provides and I still don't get the right answer. If you want to try,
mass of disk = 0.050kg
mass of hub = 0.005kg
diameter of disk = 0.075m
diameter of hub = 0.010m

answer comes out to be: 0.84m/s

I didn't bother to convert the diameter to radius since it just cancels out in my equation.

That looks correct to me. What's L?

 

[ahello888a]

L is the length of the string which I assume to basically be the height since the string goes that length down.

 

[Dick]

L is the length of the string which I assume to basically be the height since the string goes that length down.

I meant, what's the numerical value you put in for L?

 

[ahello888a]

oh I am sorry. I put in 1m.

 

[Dick]

oh I am sorry. I put in 1m.

I get the same formula you get, but when I put your numbers in I don't get 0.84m/s. I get something pretty close, though. Can you check your arithmetic??

 

[ahello888a]

im getting .613

0.84 is what the book says is the answer.

 

[Dick]

The value I calculated was sqrt(4*9.8/(2+7.5^2+1/2)). I divided numerator and denominator by the mass of one of the large disks. That's 0.817. Not 0.84. So I guess I don't know.

 

[ahello888a]

okay i just assume the book has rounding mistakes...thanks very much for your abundance of help though!