# A yo-yo is made of 3 disks (same material) and the inner one is 3times smaller than the two outer ones (R/r=3). A string (small thickness) is wrapped around the center disk. What will its acceleration be equal to?

Aug 4, 2018

$a = \frac{38}{201} \textcolor{w h i t e}{l} g$

#### Explanation:

Let the mass of the center disk be $m$; each of the two outer disks has a lateral area of ${3}^{2} = 9$ times that of the center disk and would thus be $9 \textcolor{w h i t e}{l} m$ in mass.

The yo-yo would experience a downward gravitational pull of magnitude $G = 19 \textcolor{w h i t e}{l} m \cdot g$ where $g$ the gravitational acceleration.

Let $T$ resembles the magnitude of the tension force the string applies on the yo-yo. Apply Newton's Second Law of Motion:

$G - T = \Sigma F = 19 \textcolor{w h i t e}{l} m \cdot a$

The three disk revolves around the center of the yo-yo. The equation

$I = \frac{1}{2} \cdot m \cdot {r}^{2}$

gives the moment of inertia for each of the disk. The yo-yo would, therefore, have a moment of inertia of

$\Sigma I = 2 \times \frac{1}{2} \cdot 9 \textcolor{w h i t e}{l} m \cdot {\left(3 \textcolor{w h i t e}{l} r\right)}^{2} + \frac{1}{2} \cdot m \cdot {r}^{2} = \frac{163}{2} \textcolor{w h i t e}{l} m \cdot {r}^{2}$

The tension force the string exerts on the yo-yo is applied at a distance $r$ away from the center of rotation. Thus the torque:

$\tau = T \cdot r$

The string would drive the yo-yo to spin at an angular acceleration $\alpha$ of

$\alpha = \frac{\tau}{\Sigma I} = \frac{T \cdot r}{163 / 2 \cdot m \cdot {r}^{2}}$

The equation $a = \alpha \cdot r$ relates the angular acceleration $\alpha$ to its linear counterpart $a$:

$a = \alpha \cdot r$
$\textcolor{w h i t e}{a} = r \cdot \frac{\tau}{\Sigma I}$
$\textcolor{w h i t e}{a} = r \cdot \frac{T \cdot r}{163 / 2 \cdot m \cdot {r}^{2}}$
$\textcolor{w h i t e}{a} = \frac{T}{163 / 2 \cdot m}$

Therefore

$T = \frac{163}{2} \cdot m \cdot a$

Substituting $T$ back to the $\Sigma F = m \cdot a$ expression and solve for $a$:

$19 \textcolor{w h i t e}{l} m \cdot g - \frac{163}{2} \cdot m \cdot a = 19 \textcolor{w h i t e}{l} m \cdot a$
$a = \frac{38}{201} \textcolor{w h i t e}{l} g$

Does this result make sense? For reference, a yo-yo consisting of a single disk would experience a liner acceleration equal to $2 / 3$ that of the gravitational acceleration under such configurations. [1] The addition of the two disks- despite adding to the weigh of the yo-yo- increase its moment of inertia making it harder to spin.

Reference
[1] "Physics of a Yo-Yo", CK-12 Foundation, https://www.ck12.org/physics/physics-of-a-yo-yo/lesson/Yo-Yo-Type-Problems-PPC/

Aug 5, 2018

Based on earlier answer by @jacob-t-3

#### Explanation:

Let the mass of the central disk be $= m$.
Given is radius of outer disks $= \frac{R}{r} = 3$.
The disks are made of same material. It is assumed that thickness of all three is same. Consequently, mass of each outer disk is proportional to its ${\text{radius}}^{2}$.

$\implies$ Mass of each outer disk $= 9 \setminus m$
Total weight of yo-yo $M = \left(2 \times 9 + 1\right) \setminus m = 19 \setminus m$

Total weight of yo-yo acts downwards $= M g$

$\therefore$ Downwards force acting on yo-yo$= 19 m g$ .......(1)
where $g$ is acceleration due to gravity.

Let $T$ be the magnitude of the tension force in the string due to the yo-yo. Net downwards force

${F}_{\text{net}} = M g - T$

If $a$ is acceleration produced in the yo-yo, from Newtons second Law of motion

$M g - T = M a$ ......(2)

The three disks revolve around the center of the yo-yo. The moment of inertia of a disk rotating about its center of mass is given as

$I = \frac{1}{2} {\text{mass"* "radius}}^{2}$

Total moment of inertia of yo-yo is sum of moments inertia of three disks.

$\Sigma I = 2 \left(\frac{1}{2} \cdot 9 m \cdot {\left(3 r\right)}^{2}\right) + \frac{1}{2} m \cdot {r}^{2}$
$\implies \Sigma I = \frac{162}{2} \setminus m \cdot {r}^{2} + \frac{1}{2} \cdot m \cdot {r}^{2}$
$\implies \Sigma I = \frac{163}{2} \setminus m \cdot {r}^{2}$

Weight, which is acting at the center of mass, donot produce any torque. Therefore, total torque $\tau$ produced is by tension $T$ only.

$\tau = T \cdot r$

Let this torque produce an angular acceleration $\alpha$ in the yo-yo

$\therefore \alpha = \frac{\tau}{\Sigma I} = \frac{T \cdot r}{\frac{163}{2} \setminus m \cdot {r}^{2}}$ ....(3)

Expression relating the angular acceleration $\alpha$ to its acceleration $a$ is

$a = r \cdot \alpha$

Using (3) we get

$a = r \cdot \frac{T \cdot r}{\frac{163}{2} \cdot m \cdot {r}^{2}}$
$\implies T = \frac{163}{2} \cdot m \cdot a$

Substituting $T$ in (2), using (1) and solving for $a$ we get

$19 \setminus m \cdot g - \frac{163}{2} \setminus m \cdot a = 19 \setminus m \cdot a$
$\implies 19 \setminus m \cdot g = \frac{163}{2} \setminus m \cdot a + 19 \setminus m \cdot a$
$\implies 19 \setminus m \cdot g = \frac{201}{2} \setminus m \cdot a$
$\implies a = \frac{38}{201} \setminus g$