Calculate a moment of inertia. 0.20 kg 0.20 kg 0.30 kg 0.30 kg 0.30 kg 0.20...
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Calculate a moment of inertia. 0.20 kg 0.20 kg 0.30 kg 0.30 kg 0.30 kg 0.20 kg 0.50 m 0.30 kg 0.20 kg Figure (a): Four objects connected to light rods rotating in the plane of the page. Figure (b): A double baton rotating about the axis 00'. PROBLEM In an effort to be the star of the half-time show, a majorette twirls an unusual baton made up of four spheres fastened to the ends of very light rods (Figure (a)). Each rod is 1.0 m long. (a) Find the moment of inertia of the baton about an axis perpendicular to the page and passing through the point where the rods cross. (b) The majorette tries spinning her strange baton about the axis 00', as shown in Figure (b). Calculate the moment of inertia of the baton about this axis. STRATEGY In Figure (a), all four balls contribute to the moment of inertia, whereas in Figure (b), with the new axis, only the two balls on the left and right contribute. Technically, the balls on the top and bottom in Figure (b) still make a small contribution because they're not really point particles. However, their contributions can be neglected because the distance from the axis of rotation of the balls on the horizontal rod is much greater than the radii of the balls on the vertical rod. SOLUTION (A) Calculate the moment of inertia of the baton when oriented as in Figure (a). Apply the equation to the right, neglecting the mass of the connecting rods. I = mr2 = m11 2 + 2 m2r2m3r3 2 2 +m4r 4 = (0.20 kg)(0.50 m) + (0.30 kg)(0.50 m) + (0.20 kg)(0.50 m) + (0.30 kg)(0.50 m) I = 0.25 kg. m (B) Calculate the moment of inertia of the baton when oriented as in Figure (b). Apply the same equation as in part (A), neglecting the radii of the 0.20-kg spheres. I = mr = m 2 2 2 = mr (0.20 kg)(0 m)2 + (0.30 kg)(0.50 m) + (0.20 kg)(0 m) + (0.30 kg)(0.50 m) +m2r2 +m3r3 + m4r4 2 2 I = 0.15 kg m m LEARN MORE REMARKS The moment of inertia is smaller in part (B) because in this configuration the 0.20-kg spheres are essentially located on the axis of rotation. QUESTION If one of the rods is lengthened, which one would cause the larger change in the moment of inertia? the rod connecting balls two and four the rod connecting the bars one and three PRACTICE IT Use the worked example above to help you solve this problem. In an effort to be the star of the half-time show, a majorette twirls an unusual baton made up of four spheres fastened to the end of very light rods (see figure). Each rod is 1.20 m long. (a) Find the moment of inertia of the baton about an axis perpendicular to the page and passing through the point where the rods cross. kgm (b) The majorette tries spinning her strange baton about the axis 00', as shown in the figure. Calculate the moment of inertia of the baton about this axis. EXERCISE kgm HINTS: GETTING STARTED | I'M STUCK! Yet another bizarre baton is created by taking four identical balls, each with mass 0.279 kg, and fixing them as before except that one of the rods has a length of 1.06 m and the other has a length of 1.64 m. (a) Calculate the moment of inertia of this baton when oriented as shown in the figure. kgm I = (b) Calculate the moment of inertia of this baton when oriented as shown in the figure, with the shorter rod vertical. I = . kg m m (c) Calculate the moment of inertia of this baton when oriented as shown in the figure, but with longer rod vertical. I = kg m m Calculate a moment of inertia. 0.20 kg 0.20 kg 0.30 kg 0.30 kg 0.30 kg 0.20 kg 0.50 m 0.30 kg 0.20 kg Figure (a): Four objects connected to light rods rotating in the plane of the page. Figure (b): A double baton rotating about the axis 00'. PROBLEM In an effort to be the star of the half-time show, a majorette twirls an unusual baton made up of four spheres fastened to the ends of very light rods (Figure (a)). Each rod is 1.0 m long. (a) Find the moment of inertia of the baton about an axis perpendicular to the page and passing through the point where the rods cross. (b) The majorette tries spinning her strange baton about the axis 00', as shown in Figure (b). Calculate the moment of inertia of the baton about this axis. STRATEGY In Figure (a), all four balls contribute to the moment of inertia, whereas in Figure (b), with the new axis, only the two balls on the left and right contribute. Technically, the balls on the top and bottom in Figure (b) still make a small contribution because they're not really point particles. However, their contributions can be neglected because the distance from the axis of rotation of the balls on the horizontal rod is much greater than the radii of the balls on the vertical rod. SOLUTION (A) Calculate the moment of inertia of the baton when oriented as in Figure (a). Apply the equation to the right, neglecting the mass of the connecting rods. I = mr2 = m11 2 + 2 m2r2m3r3 2 2 +m4r 4 = (0.20 kg)(0.50 m) + (0.30 kg)(0.50 m) + (0.20 kg)(0.50 m) + (0.30 kg)(0.50 m) I = 0.25 kg. m (B) Calculate the moment of inertia of the baton when oriented as in Figure (b). Apply the same equation as in part (A), neglecting the radii of the 0.20-kg spheres. I = mr = m 2 2 2 = mr (0.20 kg)(0 m)2 + (0.30 kg)(0.50 m) + (0.20 kg)(0 m) + (0.30 kg)(0.50 m) +m2r2 +m3r3 + m4r4 2 2 I = 0.15 kg m m LEARN MORE REMARKS The moment of inertia is smaller in part (B) because in this configuration the 0.20-kg spheres are essentially located on the axis of rotation. QUESTION If one of the rods is lengthened, which one would cause the larger change in the moment of inertia? the rod connecting balls two and four the rod connecting the bars one and three PRACTICE IT Use the worked example above to help you solve this problem. In an effort to be the star of the half-time show, a majorette twirls an unusual baton made up of four spheres fastened to the end of very light rods (see figure). Each rod is 1.20 m long. (a) Find the moment of inertia of the baton about an axis perpendicular to the page and passing through the point where the rods cross. kgm (b) The majorette tries spinning her strange baton about the axis 00', as shown in the figure. Calculate the moment of inertia of the baton about this axis. EXERCISE kgm HINTS: GETTING STARTED | I'M STUCK! Yet another bizarre baton is created by taking four identical balls, each with mass 0.279 kg, and fixing them as before except that one of the rods has a length of 1.06 m and the other has a length of 1.64 m. (a) Calculate the moment of inertia of this baton when oriented as shown in the figure. kgm I = (b) Calculate the moment of inertia of this baton when oriented as shown in the figure, with the shorter rod vertical. I = . kg m m (c) Calculate the moment of inertia of this baton when oriented as shown in the figure, but with longer rod vertical. I = kg m m
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