Calculating Acceleration of a Yo-Yo Using Torque and Moment of Inertia

[Liokh]

What would be the acceleration of a yo-yo if these are my datas:

m= 10.456g
r(small) = 0.7mm
R= 1.9mm

Now on I've had accelerations varying from 0.006m/s² to 930m/s² -yeah right...

Worked on that prrety all of the time today and all I got left is a few so that I'm really in a hurry right now.

 

[Hippo]

You haven't enough data, as far as I can tell.

 

[Andrew Mason]

What would be the acceleration of a yo-yo if these are my datas:

m= 10.456g
r(small) = 0.7mm
R= 1.9mm

Now on I've had accelerations varying from 0.006m/s² to 930m/s² -yeah right...

Worked on that prrety all of the time today and all I got left is a few so that I'm really in a hurry right now.

Think of the torque of the centre of mass of the yo-yo about the point of tangential contact with the string:

\tau = mgr_{small} = I\alpha

where I is the moment of inertia of the yo-yo about that small radius. Use the parallel axis theorem to work that out.

Use the relationship between \alpha and acceleration to find the acceleration of the centre of mass.

Are you sure these dimensions are mm and not cm? That is one awfully small yo-yo!

AM

 

[xanthym]

What would be the acceleration of a yo-yo if these are my datas:

m= 10.456g
r(small) = 0.7mm
R= 1.9mm

Now on I've had accelerations varying from 0.006m/s² to 930m/s² -yeah right...

Worked on that prrety all of the time today and all I got left is a few so that I'm really in a hurry right now.

SOLUTION HINTS:
{Yoyo Small Radius} = r = (0.7 mm) = (7e(-4) meters)
{Yoyo Large Radius} = R = (1.9 mm) = (1.9e(-3) meters)
{Yoyo Mass} = (10.456 g) = (1.0456e(-2) kg)
{Yoyo Moment of Inertia about CM Axis} = I = (1/2)*m*(R^2)
{Force Acting Thru CM} = F = m*a
{Gravitational Force on Yoyo} = m*g
{String Tension on Yoyo} = T
{Torque from String Tension about CM Axis} = τ = r*T

For Linear Motion of CM:
F = m*a =
= T - mg
::: ⇒ a = (T/m) - g ::: <---- Eq #1

For Rotational Motion about CM axis:
τ = I*α = (1/2)*m*(R^2)*{-a/r} =
= r*T ::: <---- Torque from tension "T" acting at radius "r"
::: ⇒ T = -(1/2)*m*a*(R/r)^2 ::: <---- Eq #2

Placing Eq #2 into Eq #1:
a = -(1/2)*a*(R/r)^2 - g
::: ⇒ a*{1 + (1/2)*(R/r)^2} = -g

Solve for "a" in terms of "R" and "r" (and "g"), values for which are given in the problem statement.


~~

 

[Liokh]

something is wrong with your calculation
I think T is confused with τ and R^2*-a/r is derived into a*(R/r)^2
otherwise the answer makes sense either way (r² or r)
lastly, you're corect sir Mason, I did translate my cm in mm AND add the mm sign therefore doing the process twice.

 

[Andrew Mason]

lastly, you're corect sir Mason, I did translate my cm in mm AND add the mm sign therefore doing the process twice.

Ok. So here is how I see it (which is equivalent to xanthym's approach):

\tau = mgr = I\alpha = I(a/r) = (\frac{1}{2}mR^2 + mr^2)\frac{a}{r}

a = \frac{gr^2}{(\frac{1}{2}R^2 + r^2)} = \frac{2g}{(\frac{R^2}{r^2} + 2)}

I get a = 2.1 m/sec^2

AM