To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.
By definition Newton's second law is described as
F= ma
Where,
m= mass
a = Acceleration
Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,
[tex]\sum F = m_b a[/tex]
[tex]m_Bg-T_B = m_Ba[/tex]
[tex]T_B = m_Bg-m_Ba[/tex]
In the case of mass A,
[tex]\sum F = m_A a[/tex]
[tex]T_A = m_Ag-m_Aa[/tex]
Making summation of Torques in the Pulley we have to
[tex]\sum\tau = I\alpha[/tex]
[tex]T_BR_0-T_AR_0=I\alpha[/tex]
[tex]T_B-T_A=I\frac{a}{R^2_0}[/tex]
Replacing the values previously found,
[tex](m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}[/tex]
[tex](m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}[/tex]
[tex]a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}[/tex]
[tex]a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}[/tex]
[tex]a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}[/tex]
Replacing with our values
[tex]a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}[/tex]
[tex]a=0.7736m/s^2[/tex]
PART B) Ignoring the moment of inertia the acceleration would be given by
[tex]a' =\frac{(m_B-m_A)g}{(m_B+m_A)}[/tex]
[tex]a' =\frac{(8-6.8)(9.8)}{(8+6.8)}[/tex]
[tex]a' = 0.7945[/tex]
Therefore the error would be,
[tex]\%error = \frac{a'-a}{a}*100[/tex]
[tex]\%error = \frac{0.7945-0.7736}{0.7736}*100[/tex]
[tex]\%error = 2.7%[/tex]
