Question

Suppose that the rod is held at rest horizontally and then released. (Throughout the remainder of this problem, your answer may include the symbol , the moment of inertia of the assembly, whether or not you have answered the first part correctly.) What is the angular acceleration of the rod immediately after it is released? Take the counterclockwise direction to be positive. Express in terms of some or all of the variables , , , , , and .

Answer 1

Answer:

$\alpha=\frac { gx(m_1-m_2)}{(\frac {m_r x^{2}}{3} + m_1 x^{2}+ m_2 x^{2})}$

Explanation:

Mass of rod is $m_r$ and its moment of inertia $I_r$ is given by

$I_r=\frac {(2x)^{2}m_r}{12}=\frac {m_r x^{2}}{3}$

Moment of inertia of particle of mass $m_1$ is given by

$I_1=m_1 x^{2}$

Moment of inertia of particle of mass $m_2$ is given by

$I_2=m_2 x^{2}$

Total moment of inertia for the system is given by

$I=\frac {m_r x^{2}}{3} + m_1 x^{2}+ m_2 x^{2}$

Torque on the pivot is given by

$\tau=gxm_1-gxm_2=gx(m_1-m_2)$

Net torque is given by

$\tau=I\alpha$

$gx(m_1-m_2)=I\alpha$

$gx(m_1-m_2) =(\frac {m_r x^{2}}{3} + m_1 x^{2}+ m_2 x^{2})\alpha$

Making $\alpha the subject then [tex]\alpha=\frac { gx(m_1-m_2)}{(\frac {m_r x^{2}}{3} + m_1 x^{2}+ m_2 x^{2})}$ or simply  

$\alpha=\frac { gx(m_1-m_2)}{I}$