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    Rotational inertia and angular velocity of a mud ball mass that hits a solid door

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    A solid door of mass M = 20kg and width l= 1m is hit at a right angle by a mud ball of mass m = .5kg, which hits the door at the edge with speed v= 10m/s and sticks.

    (a) What is the rotational inertia of the door about the hinges?
    (b) What is the angular velocity of the door after having been struck?
    (c) What fraction of the initial energy does the moving door-mud ball sytem retain?

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    Solution Preview

    Let's start with (a)

    To calculate the rotational inertia, I, of an object, you have to integrate (r^2)*dm, where r is the distance from the pivot point to the mass element dm. In this case, we have a fairly simple integral to solve: the hard part is usually working out WHAT to integrate. This involves looking at the geometry of the object and dividing up its mass in such a way that we can write "dm" as a function of r.

    In this case, we have a door of fixed height h and fixed thickness t. The volume of the door can then be written as V = r*t*h where r is the length of the door. The best way to divide up the door is into strips of height h, thickness t, and width dr, where dr is the width of each little ...

    Solution Summary

    With complete calculations and explanations, the problem is solved.


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