Question

A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. T?
A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s2. What is the moment of inertia of the door about the hinges?

Answer 1

-- The torque he applies to the door is

                         (force) x (distance from the hinge)

                     =    (5 N)  x  (0.8 m)  =    4 N-m  .

-- Torque  = (moment of inertia) x (angular acceleration)

     4 N-m  =  (moment of inertia)  x  (2 /s²)

     Moment of inertia  =  (4 N-m) / (2 /s²)

                                   = (4 kg-m²/s²) /  (2 /s²)

                                   =       2 kg-m²

Note that the pusher's mass makes no difference, only the force exerted
perpendicular to the door.  It could be an elephant pushing lightly, or even
a mosquito if he could get up the required 5 N (about 18 ounces of force).

Answer 2

The moment of inertia of the door about the hinges is $\boxed{2\,{\text{kg - }}{{\text{m}}^{\text{2}}}}$ .

Further Explanation:

Since the force is applied perpendicular to the door and the door is hinged at the other end, the force will produce a torque in the door.

The torque produced in the door due to the applied force is given by:

$\tau  = F\times d$

Here, $F$  is the force applied and $d$  is the distance of application of force from the axis of rotation.

Substitute $5\,{\text{N}}$  for $F$  and $0.8\,{\text{m}}$  for $d$   in above expression.

$\begin{aligned}\tau=5\times0.8\\=4\,{\text{N}}\cdot{\text{m}}\\\end{aligned}$

The torque acting on the door produces an angular acceleration in the door. The torque is represented in terms of angular acceleration of the door as:

$\tau=I\times\alpha$  

Here, $I$  is the moment of inertia of the door and $\alpha$  is the angular acceleration of the door.

Substitute the values of $\tau$  and $\alpha$  in above expression.

$\begin{aligned}4\,{\text{N}}\cdot{\text{m}}&=I\times 2\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\I&=\frac{4}{2}\,\\&=2\,{\text{kg-}}{{\text{m}}^{\text{2}}}\\\end{aligned}$

Thus, the moment of inertia of the door about the hinge is $\boxed{2\,{\text{kg - }}{{\text{m}}^{\text{2}}}}$ .

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Moment of Inertia

Keywords:

Moment of inertia, door, hinges, 72 kg person, small doorknob, 5.00 N, frictionless hinges, torque, angular acceleration, 2.0 rad/s^2.