Question

The tensions in the rope are 12 n and 10 n . What is the angular acceleration of the pulley?

Answer 1

Answer:

$\mathbf{\alpha = 16.06 \ rad/s^2}$

Explanation:$A \ frictionless \ pulley, \ whic h\ can \ be \ modeled \ as \ a \ 0.83 k g \ solid \ cylinder \ with \ a \ 0.30 \ m \ radius,$  $\ has \ a \ rope \ going \ over \ it$ $The \ tensions \ in \ the \ rope \ are$ $12 n \ and \ 10 n . \ What \ is \ the \ angular \ acceleration \ of \ the \ pulley?$

$From \ the \ given \ information:$

$The \ moment \ of \ inerti a \ of \ a \ solid \ cylinder$ $I = \dfrac{1}{2}mr^2$

$I = \dfrac{1}{2}(0.83)(0.3)^2$

$I = 0.03735 \ kg \ m^2$

$The \ net \ torque = F_1(r) - F_2(r) \\ \\ = 12(0.3) - 10(0.3) \\ \\ = 0.6 \ N.m$

$Torque (T) = moment \ of \inertia (I) \times angular \ acceleration ( \alpha)$

$\alpha = \dfrac{T}{I}$

$\alpha = \dfrac{0.6 \ Nm}{0.03735 \ kg \ m^2}$

$\mathbf{\alpha = 16.06 \ rad/s^2}$