Question

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?

Answer 1

`a = 38/201 color(white)(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 color(white)(l) m` in mass.

The yo-yo would experience a downward gravitational pull of magnitude `G = 19 color(white)(l) m * 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 color(white)(l) m * a`

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

`I = 1/2 * m* 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 xx 1/2 * 9 color(white)(l) m * (3 color(white)(l) r)^2 + 1/2 * m * r^2 = 163/2 color(white)(l) m * 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 * r`

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

`alpha = tau / (Sigma I) = (T * r)/(163//2 * m * r^2)`

The equation `a = alpha * r` relates the angular acceleration `alpha` to its linear counterpart `a`:

`a = alpha * r`
`color(white)(a) = r * tau / (Sigma I)`
`color(white)(a) = r * (T * r)/(163//2 * m * r^2)`
`color(white)(a) = (T)/(163//2 * m )`

Therefore

`T = 163/2 * m * a`

Substituting `T` back to the `Sigma F = m * a` expression and solve for `a`:

`19 color(white)(l) m * g - 163/2 * m * a = 19 color(white)(l) m * a`
`a = 38/201 color(white)(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/

Answer 2

Based on earlier answer by @jacob-t-3

Explanation:

Let the mass of the central disk be `=m`.
Given is radius of outer disks `=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 `"radius"^2`.

`=>` Mass of each outer disk `=9\ m`
Total weight of yo-yo `M=(2xx9+1)\ m=19\ m`

Total weight of yo-yo acts downwards `=Mg`

`:.` Downwards force acting on yo-yo`=19mg` .......(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_"net"=Mg-T`

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

`Mg-T=Ma` ......(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 = 1/2 "mass"* "radius"^2`

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

`Sigma I = 2 (1/2 * 9 m * (3 r)^2) + 1/2 m * r^2`
`=>Sigma I = 162/2\ mcdot r^2 + 1/2 * m * r^2`
`=>Sigma I = 163/2 \ m * 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 * r`

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

`:.alpha = tau / (Sigma I) = (T * r)/(163/2\ m * r^2)` ....(3)

Expression relating the angular acceleration `alpha` to its acceleration `a` is

`a = r*alpha`

Using (3) we get

`a = r * (T * r)/(163/2 * m * r^2)`
`=>T = 163/2 * m * a`

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

`19\ m * g - 163/2 \ m * a = 19\ m * a`
`=>19\ m * g= 163/2 \ m * a + 19\ m * a`
`=>19\ m * g= 201/2 \ m * a`
`=>a = 38/201\ g`