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

• Liokh
In summary, the acceleration of a yo-yo if these are my datas:m= 10.456gr(small) = 0.7mmR= 1.9mmwould be 0.006m/s² to 930m/s².
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.

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

Liokh said:
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

Liokh said:
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.

~~

Last edited:
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.

Liokh said:
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

## 1. What is torque and how is it measured?

Torque is a measure of the twisting force on an object. It is measured in units of force multiplied by distance, such as newton-meters or foot-pounds. Torque is calculated by multiplying the magnitude of the force by the perpendicular distance from the axis of rotation to the point where the force is applied.

## 2. How is moment of inertia defined?

Moment of inertia is a measure of an object's resistance to rotational motion. It is calculated by taking the sum of the mass of each particle in the object multiplied by the square of its distance from the axis of rotation. This value is dependent on the object's shape and mass distribution.

## 3. How are torque and moment of inertia related?

Torque and moment of inertia are related by the equation τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration of the object. This equation shows that the amount of torque required to produce a certain angular acceleration is directly proportional to the object's moment of inertia.

## 4. How do you calculate the moment of inertia for a simple shape?

The moment of inertia for a simple shape, such as a rod or a sphere, can be calculated using specific formulas that take into account the shape and dimensions of the object. These formulas can be found in many physics textbooks or online resources. Alternatively, the moment of inertia can also be calculated by dividing the object into smaller, simpler shapes and then using the parallel axis theorem to add up their individual moments of inertia.

## 5. How does moment of inertia affect an object's rotational motion?

The moment of inertia of an object affects its rotational motion by determining how easily it can be rotated. Objects with higher moment of inertia require more torque to rotate, while objects with lower moment of inertia rotate more easily. This is because a higher moment of inertia means there is more mass distributed farther away from the axis of rotation, resulting in a larger resistance to rotational motion.

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