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.
\(\begin{aligned}4\,{\text{N}}\cdot{\text{m}}&=I\times 2\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-
ulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\I&=\frac{4}{2}\,\\&=2\,{\text{kg-}}{{\text{m}}^{\text{2}}}\\\end{aligned}\)
Thus, the moment of inertia of the door about the hinge is \(\boxed{2\,{\text{kg - }}{{\text{m}}^{\text{2}}}}\) .
Learn More:
1. A toy train rolls around a horizontal 1.0-m-diameter track https://brainly.com/question/9575487
2. On a highway a car is driven 80 https://brainly.com/question/504792
3. Assume that, at a certain angular speed ω2, the radius r becomes twice l. Find ω2 https://brainly.com/question/5813257
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.
