Answer:
The answer is "[tex]- 0.5747\ \frac{rad}{s^2}[/tex]"
Explanation:
Let
[tex]M_L = 12\ kg\\\\M_R = 5.8\ kg[/tex]
While it is in balance, its net force mostly on a machine is zero where the board will rotate an upward torques were given by:
[tex]\to M_L \times g \times R_L - M_R \times g \times R_R = \tau[/tex]
[tex]\to 12 \times 9.8 \times 0.6 - 5.8 \times 9.8 \times 1.4 = \tau[/tex]
[tex]\to \tau= 70.56-79.576\\\to \tau = -9.016[/tex]
let,
[tex]\tau = I \alpha[/tex]
where
[tex]\alpha =[/tex]angular acceleration
I = moment of inertia of the system
[tex]\to I = M_L \times r \times L_2 + M_R \times r \times R_2\\\\\to I = 12 \times 0.6 \times 0.6 + 5.8 \times 1.4 \times 1.4\\\\\to I= 4.32+11.368\\\\\to I = 15.688\ kg\ m2\\\\[/tex]
Calculating the angular acceleration:
[tex]\alpha = \frac{\tau}{I}\\\\[/tex]
[tex]= \frac{-9.016}{15.688}\\\\=- 0.5747\ \frac{rad}{s^2}\\\\[/tex]
