Determine the moment of inertia of a 19-kg door that is 2.5 m high and 1.3 m wide and is hinged along long side. ignore the thickness of the door.

I = 11 kg*m^2 The moment of inertia is the sum of the mass of every particle multiplied by the square of the radius. This can be done using the calculus process of integration. Thankfully, these have been a lot of physicists who've already done this for a lot of common shapes and the moment of inertia for a rectangle hinged on one side is: I = mr^2/3 where I = moment of inertia m = mass r = radius which is the width of the rectangle extending from the axis of rotation. For this problem, we can ignore the long side since all we need is the total mass of the door and the radius which is 1.3 m. So substitute the known values into the formula and calculate the moment of inertia. I = mr^2 / 3 I = 19 kg*(1.3 m)^2 / 3 I = 19 kg*1.69 m^2 / 3 I = 32.11 kg*m^2 / 3 I = 10.70333333 kg*m^2 Rounding to 2 significant figures gives I = 11 kg*m^2

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