Chapter 8, Problem 49P

To determine

To Find: The speed of $m_2$ .

Answer to Problem 49P

The speed of ${\text{m}}_{\text{2}}$is $3.22\text{m/s}$ .

Explanation of Solution

Given Information:

First mass, ${\text{m}}_{\text{1}}=18\text{kg}$

Second mass ${\text{m}}_{\text{2}}=26.5\text{kg}$

Height, $\text{h}=3\text{m}$

Radius of pulley, $\text{r}=0.260\text{m}$

Mass of pulley, $\text{M}=7.50\text{kg}$

Concept used:

By law of conservation of energy

  $\begin{array}{l}{E}_I=E_F\\ E_I=m_{2}gh\\ E_F=m_{1}gh+\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+\frac{1}{2}I{\omega }^2\\ m_{2}gh=m_{1}gh+\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+\frac{1}{2}I{\omega }^2\\ \text{and}I=\frac{M{r}^2}{2},\omega =\frac{v}{r}\\ m_{2}gh-m_{1}gh=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+\frac{1}{4}M{v}^2\\ v=\sqrt{\frac{2\left(m_2-m_1\right)gh}{m_1+m_2+\frac{1}{2}M}}\end{array}$

Here,

  $v$ is the velocity,

  $M$ is the mass of pully,

  $r$ is the radius of pully,

  $m_1,m_2$ are the masses of object,

  $I$ is the moment of inertia of pulley,

  $\omega$ is the angular velocity of the pulley.

Calculation:

  $v=\sqrt{\frac{2(m_2-m_1)gh}{m_1+m_2+\frac{1}{2}M}}$

  $\begin{array}{l}v=\sqrt{\frac{2(26.5-18)(9.8)(3)}{18+26.5+\frac{1}{2}(7.50)}}\\ =3.22\text{m/s}\end{array}$

  $v=3.22\text{m/s}$

Conclusion:

The speed of

  $m_2$ is $v=3.22\text{m/s}$ .