The moment of inertia for point particles is given by . Rewriting this for L yields .
(a) The mass has three components, M = Mlid + Mshell + Mbottom = 50 g, where the lid and the bottom are identical. Mass is proportional to the surface area for uniform objects. The area of the lid and bottom is that of a circle, Acircle = πr2 . The surface area of a cylindrical shell is Ashell = 2πrL. So the total area of can is Atotal = 2Acircle + Ashell = 2πrL + 2πr2 = 2πr(r + L). So the mass of the lid or bottom is given by . Thus Mlid = 5.392 g.
(b) Similarly, the mass of the shell is given by . Thus Mshell = 39.216 g.
(c) The total moment of inertia of the beer can is given by the sum of the individual pieces, Itotal = Ilid + Ishell + Ibottom. Since each piece is revolving about its centre of mass, we do not need the parallel axis theorem. Looking up the moments of inertia of a flat solid disk and a thin cylindrical shell, we have Itotal = 2×½Mlidr2 + Mshellr2 = 4.86 × 10-5 kg-m2.
(a) The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, .For axis A, the rod is rotating about its centre of mass. Each sphere is a distance R+L/2 from the axis of rotation, so we must use the parallel axis theorem. Recall that the moment of inertia of a rod about its centre is and that the moment of inertia of a solid sphere about its centre is . Thus we have
(b) For axis B, the rod's centre is R+L/2 away from the axis of rotation. One sphere's centre is L+2R from the axis of rotation. The last sphere is rotating about axis B. Thus r
(c) The bar and both spheres are rotating about their own centres when rotating about axis c. However, note that the bar is a cylinder or radius r in this configuration.
We must treat the hole as an object of negative mass. The inertia of the object is then just. The plate and the hole are just disks and the inertia of a disk is . The hole is not rotating about its own centre of mass, so we must use the parallel axis theorem,
We must treat the hole at the centre as a sphere of negative mass. Since the moment of inertia of a sphere about its centre is , the moment of inertia of this object is
The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, . The rods are not rotating about their centre of mass, so we must use the parallel axis theorem. The centre of each rod is Rcyl+L/2 from the axis of rotation at the centre of the object. The moments of inertia for a cylindrical shell, a disk, and a rod are MR2, , and respectively. The moment of inertia of a point mass is . Thus the total moment of inertia is:
The moment of inertia of a composite body is equal to the sum of the moments of its individual pieces, . The cylinders are not rotating about their centre of mass, so we must use the parallel axis theorem. The centre of each rod is Rdisk- Rcyl from the axis of rotation at the centre of the object. The moments of inertia for a a disk or a cylindrical rod are is . The moment of inertia of a point mass is . Thus the total moment of inertia is:
Questions? mike.coombes@kpu.ca