Moment Of Inertia Of A Hinged Door

A uniform thin solid door has height 2 2 m width 87m and mass 23kg.

Moment of inertia of a hinged door.

Find the location of its center of mass x cm if the end masses m 1 and m 2 are 5 0 kg and 10 0 kg respectively. Find the moment of inertia for rotation on its hinges. Homework equations i x 2dm the attempt at a solution b the way i set it up was that i made into strips of length h height and width dr. Moment of inertia denoted by i measures the extent to which an object resists rotational acceleration about a particular axis and is the rotational analogue to mass mass moments of inertia have units of dimension ml 2 mass length 2 it should not be confused with the second moment of area which is used in beam calculations the mass moment of inertia is often also known as the.

The moment of inertia of an object depends on where the axis of rotation is. The width of the door is unnecessary. Homework equations i 1 12 m a 2 b 2 mr 2 parallel axis theorem the attempt at a solution i initially used the door height and width as a and b i m sure this is right and used the door thickness as the distance between axis. A uniform thin solid door has a height of 2 2 m a width of 0 87 m and a mass of 23 kg.

A dumbbell has a connecting bar of negligible mass. The moment of inertia can be found by breaking up the object into little pieces multiplying the mass of each little piece by the square of the distance it is from the axis of rotation and adding all these products up. Dm λhdr where λ is mass over the length h. Find its moment of inertia for rotation on its hinges.

Are any of the data unnecessary. R 0 95 2 16. The moment of inertia of an infinitesimal mass element d m is d m r 2 in which r represents the distance from the mass element to the axis about which we are considering the moment of inertia. What is the moment of inertia of the door about the axis through its hinges.