91Ó°ÊÓ

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:

$$\begin{gathered} \text{F}=\text{ma} \\ \mathrm{Ï„}=\text{I}\mathrm{Î\pm }=\text{F}×\text{r} \end{gathered}$$

The magnitude of the acceleration of the blocks:

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

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

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

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

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

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

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

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

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

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

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