# Torque and Moment of Inertia

1. Apr 27, 2005

### 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.

2. Apr 27, 2005

### Hippo

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

3. Apr 28, 2005

### Andrew Mason

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

4. Apr 28, 2005

### xanthym

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.

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Last edited: Apr 28, 2005
5. Apr 28, 2005

### 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.

6. Apr 28, 2005

### Andrew Mason

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