In Fig.$\mathbf{10}\mathbf{-}\mathbf{41}$, two blocks, of mass ${\mathbf{\text{m}}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{400}\mathit{g}$and${\mathbf{\text{m}}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{600}\mathit{g}$, are connected by a massless cord that is wrapped around a uniform disk of mass$\text{M}=500\text{g}$and radius $\mathbf{\text{R}}\mathbf{=}\mathbf{}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{cm}}$. The disk can rotate without friction about a fixed horizontal axis through its centre; the cord cannot slip on the disk. The system is released from rest. Find (a) the magnitude of the acceleration of the blocks, (b) the tension in the cord at the left, and (c) the tension${\text{T}}_{\text{2}}$in the cord at the right.

1. Mass of block 1 is${\text{m}}_1=400\text{g}$
2. Mass of block 2 is${\text{m}}_2=600\text{g}$
3. Mass of the disk is$\text{M}=500\text{g}$
4. Radius of disk is$\text{R}=12.0\text{cm}$

Use the concept of Newton’s second law. Find acceleration of the system by applying Newton’s second law for both masses and a pulley. Write three equations because we have three unknowns. Solving these equations simultaneously, we get acceleration. Once acceleration is found, tensions can be found.

Formulae:

$$\begin{gathered} \text{F}=\text{ma} \\ \tau =\text{I}\alpha =\text{F}\times \text{r} \end{gathered}$$

The magnitude of the acceleration of the blocks:

First we write the equation for net force on each block and pulley:

$$\begin{gathered} {\text{m}}_1\text{a}={\text{T}}_1-{\text{m}}_1\text{g}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right) \\ {\text{m}}_2\text{a}={\text{m}}_2\text{g}-{\text{T}}_2\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ...\left(2\right) \\ {\text{T}}_2\text{R}-{\text{T}}_1\text{R}=\text{I}\alpha \end{gathered}$$

We know moment of inertia of disk is
$\text{I}=\frac{1}{2}{\text{MR}}^2$

also
$$\begin{gathered} \alpha =\frac{\text{a}}{\text{R}} \\ \Rightarrow {\text{T}}_2\text{R}-{\text{T}}_1\text{R}=\frac{1}{2}{\text{MR}}^2\left(\frac{\text{a}}{\text{R}}\right) \end{gathered}$$

R will get cancelled out from both sides, so we get

${\text{T}}_2-{\text{T}}_1=\frac{1}{2}\text{Ma}$

We can write this equation as

${\text{T}}_2=\frac{1}{2}\text{Ma}+{\text{T}}_1$

Using this value in equation (2), we get

${\text{m}}_2\text{a}={\text{m}}_2\text{g}-\frac{1}{2}\text{Ma}-{\text{T}}_1$

$({\text{m}}_2+\frac{1}{2}\text{M})\text{a}={\text{m}}_2\text{g}-\Rightarrow {\text{T}}_1\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(3\right)$

Adding equation (1) and (3), we get

$$\begin{gathered} \left({\text{m}}_1+{\text{m}}_2+\frac{1}{2}\text{M}\right)\text{a}=\left({\text{m}}_2-{\text{m}}_1\right)\text{g} \\ \Rightarrow \text{a}=\frac{\left({\text{m}}_2-{\text{m}}_1\right)}{\left({\text{m}}_1+{\text{m}}_2+\frac{1}{2}\text{M}\right)}\text{g} \\ \Rightarrow \text{a}=\frac{\left(600-400\right)}{\left(400+600+\frac{1}{2}500\right)}\left(9.8\right) \\ \Rightarrow \text{a}=1.57\text{m}/{\text{s}}^2 \end{gathered}$$

The tension in the cord at left:

Plugging the value of ‘a’ in equation (1), we get

$$\begin{gathered} 0.400\left(1.57\right)={\text{T}}_1-0.400\left(9.8\right) \\ {\text{T}}_1=0.628+3.92 \\ {\text{T}}_1=4.55\text{N} \end{gathered}$$

The tension in the cord at right:

Using value of acceleration in equation (2), we get

$\begin{array}{rcl}0.600\left(1.57\right)& =& 0.600\left(9.8\right)-{\text{T}}_2\\ & \Rightarrow & {\text{T}}_2=5.88-0.942\\ & \Rightarrow & {\text{T}}_2=4.94\text{N}\end{array}$