Dynamics of rotational motion: the yo-yo

[adrian116]

the question is about that:
Ayo-yo is made from two uniform disks, each with mass m and radius R,
connected by a light axle of radius b. A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

the question from me is that , what is the use of radius b of axle?
the question did not give me the mass of axle, and i can't find the
moment of inertia of the axle...
can anyone tell me?

 

[Galileo]

I think you can ignore the mass/moment of inertia of the axle compared to those of the uniform disks since they gave information other than that it's light.

The radius b is important, since that will relate the falling velocity of the yo-yo to the rotational velocity of the disks.

 

[adrian116]

how the b be so important?

 

[Galileo]

Did you read my post? "The radius b is important, since that will relate the falling velocity of the yo-yo to the rotational velocity of the disks"
Now you have to find this relation, assuming the yo-yo unwinds without slipping.
If the yo-yo falls a distance $2\pi b$, how many revolutions will the disks have made?

 

[adrian116]

javascript:;/2PI?

 

[adrian116]

2*pi*b/2*pi?

 

[adrian116]

i am sorry , i am making the mistake

 

[adrian116]

the axle and the disks are connect together?
is the disks also make 2*PI*b revolutions?

 

[Galileo]

No. Imagine the disks are turning. How much rope is unwound per disk revolution?

EDIT: Yeah, the disks and axle are rigidly connected ofcourse, so they have the same angular velocity.

 

[adrian116]

2*PI*b
can i write the formula like this?
2mg-T=2ma---------first formula
Tb=m*(R^2)*(a/b)--------second formula
a-acceleration T-tension
then solve those equations,
is that right?

 

[Galileo]

Yes, that looks correct.

 

[adrian116]

but i can't get the answer
my director have the answer like that:
a=(b^2/b^2+R^2)g α=a/b T=(2MR^2/b^2+R^2)g
the answers have confused me so much...

 

[Galileo]

Yeah, there's a factor of two missing somewhere...
If you take your second equation:
Tb=m(R^2)(a/b)
and replace it with
Tb=2m(R^2)(a/b)
you get the same answer. But I think your equation is correct. The total moment of inertia is 2(1/2mR^2)=mR^2 not 2mR^2. I don't see the error.

 

[adrian116]

exactly, i think he give me the wrong answer
thank you so much~