Question

A professor's office door is 0.99 m wide, 2.2 m high, 4.2 cm thick; has a mass of 27 kg, and pivots on frictionless hinges. A "door closer" is attached to door and the top of the door frame. When the door is open and at rest, the door closer exerts a torque of 5.6 N*m. What is the moment of inertia of the door? If you let go of the open door, what is its angular acceleration immediately afterward?

Answer 1

Answer:

$I=8.8209\ kg.m^2$

$\alpha=0.6348\ rad.s^{-2}$

Explanation:

Given:

• width of door, $w=0.99\ m$

• height of the door, $h=2.2\ m$

• thickness of the door, $t=4.2\ cm$

• mass of the door, $m=27\ kg$

• torque on the door, $\tau=5.6\ N.m$

∵Since the thickness of the door is very less as compared to its other dimensions, therefore we treat it as a rectangular sheet.

• For a rectangular sheet we have the mass moment of inertia inertia as:

$I=\frac{1}{3} m.w^2$

$I=\frac{1}{3}\times 27\times 0.99^2$

$I=8.8209\ kg.m^2$

We have a relation between mass moment of inertia, torque and angular acceleration as:

$\alpha=\frac{\tau}{I}$

$\alpha=\frac{5.6}{8.8209}$

$\alpha=0.6348\ rad.s^{-2}$