91Ó°ÊÓ

Two blocks of masses m₁and m₂ are attached with a mass less string passing over a pulley:
(a) The pulley is mass less
(b) The pulley has mass m_p

First draw the diagram to understand and solve the problem .

Now equate forces in both the blocks, we get

T = m₁a .........................................................(1)

m₂g - T = m₂a .............................................(2)

Sow substitute this value either in (1) or in (2) , and solve for T we get

$\mathrm{T}=\frac{m_1{m}_2}{m_1+m_2}g$

Use this in equation (1) we get

$$\begin{gathered} \frac{m_1{m}_2}{m_1+m_2}g=m_{1}a \\ so, \\ a=\frac{m_2}{m_1+m_2}g \end{gathered}$$

Two blocks of masses m₁and m₂ are attached with a mass less string passing over a pulley:
(a) The pulley is mass less
(b) The pulley has mass m_p

Draw the diagram to solve the problem

Equate the force , we get

T₁ = m₁a ..................................................(4)

m₂g - t₂ = m₂a ......................................(5)

As pulley is in disc shape so its moment of inertia is $I=\frac{m_p{R}^2}{2}$

Moment of inertia is the difference of torque multiplied by angular acceleration

We can find the angular acceleration by

$\mathrm{Î\pm }=\left(\frac{a}{R}\right)m_{p}g$

Now find the difference of torque

${\mathrm{Ï„}}_{net}=T_{2}R-T_{1}R=I\mathrm{Î\pm }$

Substitute the value of moment of inertia and angular acceleration, we get

$$\begin{gathered} T_2-T_1=\left(\frac{m_p{R}^{2}a}{2R^2}\right) \\ simplify, \\ {T}_2-T_1=\frac{m_p}{2}a................................\left(6\right) \end{gathered}$$

From equation 4, 5 and 6, we get

$a=\frac{2m_2}{2\left(m_1+m_2\right)+m_p}$

Substitute the value of a in equation (2) and (3) we get

$$\begin{gathered} T_1=\frac{2m_1{m}_2}{2\left(m_1+m_2\right)+m_p}g \\ {T}_2=\frac{\left(2m_1+m_2\right)}{2\left(m_1+m_2\right)+m_p}g \end{gathered}$$