Respuesta :
Answer:
[tex]\alpha=\frac { gx(m_1-m_2)}{(\frac {m_r x^{2}}{3} + m_1 x^{2}+ m_2 x^{2})}[/tex]
Explanation:
Mass of rod is [tex]m_r[/tex] and its moment of inertia [tex]I_r[/tex] is given by
[tex]I_r=\frac {(2x)^{2}m_r}{12}=\frac {m_r x^{2}}{3}[/tex]
Moment of inertia of particle of mass [tex]m_1[/tex] is given by
[tex]I_1=m_1 x^{2}[/tex]
Moment of inertia of particle of mass [tex]m_2[/tex] is given by
[tex]I_2=m_2 x^{2}[/tex]
Total moment of inertia for the system is given by
[tex]I=\frac {m_r x^{2}}{3} + m_1 x^{2}+ m_2 x^{2}[/tex]
Torque on the pivot is given by
[tex]\tau=gxm_1-gxm_2=gx(m_1-m_2) [/tex]
Net torque is given by
[tex]\tau=I\alpha[/tex]
[tex]gx(m_1-m_2)=I\alpha[/tex]
[tex]gx(m_1-m_2) =(\frac {m_r x^{2}}{3} + m_1 x^{2}+ m_2 x^{2})\alpha[/tex]
Making [tex]\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})}[/tex] or simply
[tex]\alpha=\frac { gx(m_1-m_2)}{I}[/tex]
For better understanding refer the below solution.
Solution :
Let mass of rod is [tex]\rm m_r[/tex] and inrertia be [tex]\rm I_r[/tex] is given by
[tex]\rm I_r = \dfrac{(2x)^2 m_r}{12}=\dfrac{x^2m_r}{3}[/tex]
Let moment of inertia of particle whose mass [tex]\rm m_1[/tex] is [tex]\rm I_1[/tex] and is given by
[tex]\rm I_1 = m_1x^2[/tex]
Let moment of inertia of particle whose mass [tex]\rm m_2[/tex] is [tex]\rm I_2[/tex] and is given by
[tex]\rm I_2 = m_2x^2[/tex]
Now for the system toatal moment of inertia is
[tex]\rm I = I_r+I_1+I_2 = \dfrac{x^2m_r}{3}+m_1x^2+m_2x^2[/tex]
Torque on the pivot can be written as
[tex]\rm \tau = g xm_1-gxm_2=gx(m_1-m_2)[/tex]
Now net torque is
[tex]\rm \tau = I\alpha[/tex] --- (1)
Now put the value of [tex]\rm \tau \;and \; I[/tex] in equation (1)
[tex]\rm gx(m_1-m_2)=(\dfrac{m_rx^2}{3}+m_1x^2+m_2x^2)\alpha[/tex]
[tex]\alpha = \dfrac{gx(m_1-m_2)}{\dfrac{x^2m_r}{3}+m_1x^2+m_2x^2}[/tex]
For more information, refer the link given below
https://brainly.com/question/19247046?referrer=searchResults
