Question

You turned the knob of a 196.3 cm by 109.1 cm door (with a mass of 9.88 kg) and pushed it with a perpendicular force of 9.46N to open it. What is the angular acceleration of the door (in rad/s²)?

Answer 1

$\tau=I\epsilon \to F\cdot d=I\epsilon \to \epsilon=F\cdot d/I=$

$=F\cdot d/(ma^2/3)=3F\cdot d/(ma^2)=$

$=3\cdot9.46\cdot1.091/(9.88\cdot 1.091^2)=2.63\ (rad/s^2)$ . Answer