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# Chapter 10 Problems - page 5 / 19

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### Figure P10.24

25.A uniform thin solid door has height 2.20 m, width 0.870 m, and mass 23.0 kg.  Find its moment of inertia for rotation on its hinges.  Is any piece of data unnecessary?

26.Attention! About face!  Compute an order-of-magnitude estimate for the moment of inertia of your body as you stand tall and turn about a vertical axis through the top of your head and the point half way between your ankles.  In your solution state the quantities you measure or estimate and their values.

27.The density of the Earth, at any distance r from its center, is approximately

= [14.2 – 11.6 r/R] 103 kg/m3

where R is the radius of the Earth. Show that this density leads to a moment of inertia I = 0.330MR2 about an axis through the center, where M is the mass of the Earth.

28.Calculate the moment of inertia of a thin plate in the shape of a right triangle, about an axis that passes through one end

of the hypotenuse and is parallel to the opposite leg of the triangle, as in Figure P10.28a.  Let M represent the mass of the triangle and L the length of the base of the triangle perpendicular to the axis of rotation. Let h represent the height of the triangle and w the thickness of the plate, much smaller than L or h. Do the calculation in either or both of the following ways, as your instructor assigns:

(a) Use Equation 10.17.  Let an element of mass consist of a vertical ribbon within the triangle, of width dx, height y, and thickness w.  With x representing the location of the ribbon, show that y = hx/L.   Show that the density of the material is given by = 2M/Lwh.   Show that the mass of the ribbon is dm = yw dx = 2Mx dx/L2.  Proceed to use Equation 10.17 to calculate the moment of inertia.

(b) Let I represent the unknown moment of inertia about an axis through the corner of the triangle.  Note that Example 9.15 demonstrates that the center of mass of the triangle is two-thirds of the way along the length L, from the corner toward the side of height h.  Let ICM represent the moment of inertia of the triangle about an axis through the center of mass and parallel to side h.  Demonstrate that I = ICM + 4ML2/9.  Figure P10.28b shows the same object in a different orientation.  Demonstrate that the moment of inertia of the triangular plate, about the y axis is Ih = ICM + ML2/9.   Demonstrate that the sum of the moments of inertia of the triangles shown in parts (a) and (b) of the figure must be the moment of inertia of a rectangular sheet of mass 2M and length L, rotating like a door about an axis along its edge of height h.  Use information in Table 10.2 to write down the moment of inertia of the rectangle, and set it equal to the sum of the moments of inertia of the two triangles.  Solve the equation to find the moment of inertia of a triangle about an axis through its center of mass, in terms of

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