Two disks with radii $r_{1}$ = 0.3 m and $r_{2}$ = 0.2 m are fixed together so that their centres are above each other. The rotational inertia of the disk-system is 0.25 kg- $m^{2}$ . The larger disk stands on a frictionless table. Massless chords that are wrapped around the larger and smaller disks pass over pulleys and are attached to small objects of masses $m_{1}$ = 5 kg and $m_{2}$ respectively as shown. The value of $m_{2}$ at which the axis of symmetry of the disk-system remains stationary is $α$ kg. Find the value of $α$ .

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Solution

Given: Two disks with radii $r_{1} = 0.3 m$ and $r_{2} = 0.2 m$ are fixed together so that their centres are above each other. The rotational inertia of the disk-system is $0.25 k g - m^{2}$ . The larger disk stands on a frictionless table. Massless chords that are wrapped around the larger and smaller disks pass over pulleys and are attached to small objects of masses $m_{1} = 5 k g$ and $m_{2}$ respectively as shown. The value of $m_{2}$ at which the axis of symmetry of the disk-system remains stationary is $α k g$ .
To find the value of $α$
Solution:
As per given criteria,
$r_{1} = 0.3 m$
$r_{2} = 0.2 m$
$m_{1} = 5 k g$
Consider FBD of the disc:
For axis of symmetry remains stationary, $T_{1} = T_{2} . . . . . . . (i)$
Let $T_{1} = T_{2} = T$
If axis of symmetry is at rest then there will be no rotation and translational possible for $T_{1} = T_{2} = T$ as $r_{1} \neq r_{2}$
So here it is asked to consider axis of rotation at rest
Now,
$T (r_{1} - r_{2}) = I α ⟹ T (0.3 - 0.2) = I α ⟹ 0.1 T = I α . . . . . . . . . . (i i)$
For $m_{1} \to T - m_{1} g = - m_{1} a_{1} ⟹ T - 5 g = - 5 a_{1} . . . . . . . . . (i i i)$
Similarly for $m_{2} \to T - α g = α a_{2} . . . . . . . . . . . . (i v)$
$\frac{a_{1}}{r_{1}} = \frac{a_{2}}{r_{2}} = α$
$\frac{a_{1}}{a_{2}} = \frac{r_{1}}{r_{2}} ⟹ \frac{a_{1}}{a_{2}} = \frac{0.3}{0.2} = \frac{3}{2} . . . . . . . . (v)$
From eqn(ii) and (iii) with $a_{1} = r_{1} α$
$a_{1} = \frac{3 g}{θ}$
Puting this values in eqn(v), we get,
$⟹ a_{2} = \frac{2}{3} a_{1} ⟹ a_{2} = \frac{g}{4}$
$T = \frac{25 g}{θ}$
Put all the values in eqn(iv) , we get
$T - α g = α a_{2} ⟹ \frac{25 g}{θ} - α g = α (\frac{g}{4}) ⟹ α = 2.5 k g$

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