Chapter 8, Problem 41P

To determine

To find: The acceleration of the masses.

Answer to Problem 41P

  $a=\frac{\left(m_2-m_1\right)g}{\left(m_1+m_2\right)}$

Explanation of Solution

Given information:

An Atwood’s machine consists of two masses, $m_1$ and $m_2$ , which are connected by a masslessinelastic cord that passes over a pulley.

Thepulley has radius R and moment ofinertia I about its axle.

Formula used:

Tangential acceleration:

  $a=a_{\tan }=\alpha R_0$

Calculation:

On assuming ${\text{m}}_{\text{2}}>{\text{ m}}_{\text{l}}$

The positive direction is taken as the direction of acceleration of each object.As the linear acceleration ofthe masses is the tangential acceleration ofthe rim ofthe pulley,

  $a=a_{\tan }=\alpha R_0$

Consider the block of mass m₂${\displaystyle \sum F_y}=m_{2}g-F_{T_2}=m_{2}a$

For block of mass m₁${\displaystyle \sum F_y}=F_{T_1}-m_{1}g=m_{1}a$

Write ${\displaystyle \sum \tau }=I\alpha$ for the pulley about its axle

  $\begin{array}{l}{F}_{T_2}{R}_0-F_{T_1}{R}_0=I\alpha \\ Or{F}_{T_2}-F_{T_1}=Ia/R_0^2\mathrm{............}\left(1\right)\end{array}$

Adding the two force equations,

  $F_{T_1}-F_{T_2}=\left(m_1+m_2\right)a+\left(m_1-m_2\right)g\mathrm{...........}\left(2\right)$

Comparing equation (1) and (2)

  $\begin{array}{l}\left(m_1+m_2\right)a+\left(m_1-m_2\right)g=Ia/R_0^2\\ Ora=\frac{\left(m_2-m_1\right)g}{\left(m_1+m_2+I/R_0^2\right)}\end{array}$

If the moment of inertia of the pulley is ignored, from the torque equation, the two tensions are equal. Setting I = 0

  $a=\frac{\left(m_2-m_1\right)g}{\left(m_1+m_2\right)}$

Thus, the value is $a_0>a$

Conclusion:

The value is $a=\frac{\left(m_2-m_1\right)g}{\left(m_1+m_2\right)}$ .