Answer:
[tex]\mathbf{\alpha = 16.06 \ rad/s^2}[/tex]
Explanation:[tex]A \ frictionless \ pulley, \ whic h\ can \ be \ modeled \ as \ a \ 0.83 k g \ solid \ cylinder \ with \ a \ 0.30 \ m \ radius,[/tex] [tex]\ has \ a \ rope \ going \ over \ it[/tex] [tex]The \ tensions \ in \ the \ rope \ are \[/tex] [tex]12 n \ and \ 10 n . \ What \ is \ the \ angular \ acceleration \ of \ the \ pulley?[/tex]
[tex]From \ the \ given \ information:[/tex]
[tex]The \ moment \ of \ inerti a \ of \ a \ solid \ cylinder[/tex] [tex]I = \dfrac{1}{2}mr^2[/tex]
[tex]I = \dfrac{1}{2}(0.83)(0.3)^2[/tex]
[tex]I = 0.03735 \ kg \ m^2[/tex]
[tex]The \ net \ torque = F_1(r) - F_2(r) \\ \\ = 12(0.3) - 10(0.3) \\ \\ = 0.6 \ N.m[/tex]
[tex]Torque (T) = moment \ of \inertia (I) \times angular \ acceleration ( \alpha)[/tex]
[tex]\alpha = \dfrac{T}{I}[/tex]
[tex]\alpha = \dfrac{0.6 \ Nm}{0.03735 \ kg \ m^2}[/tex]
[tex]\mathbf{\alpha = 16.06 \ rad/s^2}[/tex]
