Question 8.12: THE BATON TWIRLER GOAL Calculate a moment of inertia. PROBLE...
THE BATON TWIRLER
GOAL Calculate a moment of inertia.
PROBLEM In an effort to be the star of the halftime show, a majorette twirls an unusual baton made up of four balls fastened to the ends of very light rods (Fig. 8.24). Each rod is 1.0 \mathrm{~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 O O^{\prime}, as shown in Figure 8.25 on page 242 . Calculate the moment of inertia of the baton about this axis.
STRATEGY In Figure 8.24, all four balls contribute to the moment of inertia, whereas in Figure 8.25, with the new axis, only the two balls on the left and right contribute. Technically, the balls on the top and bottom in Figure 8.25 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.
(a) Calculate the moment of inertia of the baton when oriented as in Figure 8.24.
Apply Equation 8.12,
I ≡ \sum mr^2 [8.12]
neglecting the mass of the connecting rods:
\begin{aligned}I=& \sum m r^{2}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}+m_{4} r_{4}^{2} \\=&(0.20 \mathrm{~kg})(0.50 \mathrm{~m})^{2}+(0.30 \mathrm{~kg})(0.50 \mathrm{~m})^{2} \\&+(0.20 \mathrm{~kg})(0.50 \mathrm{~m})^{2}+(0.30 \mathrm{~kg})(0.50 \mathrm{~m})^{2} \\I=& 0.25 \mathrm{~kg} \cdot \mathrm{m}^{2}\end{aligned}
(b) Calculate the moment of inertia of the baton when oriented as in Figure 8.25.
Apply Equation 8.12 again, neglecting the radii of the 0.20-\mathrm{kg} balls.
\begin{aligned}I=& \sum m r^{2}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}+m_{4} r_{4}^{2} \\=&(0.20 \mathrm{~kg})(0)^{2}+(0.30 \mathrm{~kg})(0.50 \mathrm{~m})^{2}+(0.20 \mathrm{~kg})(0)^{2} \\&+(0.30 \mathrm{~kg})(0.50 \mathrm{~m})^{2} \\I=& 0.15 \mathrm{~kg} \cdot \mathrm{m}^{2}\end{aligned}
REMARKS The moment of inertia is smaller in part (b) because in this configuration the 0.20-kg balls are essentially located on the axis of rotation.