A majorette takes two batons and fastens them together in the middle at right angles to make an "x" shape.Each baton was 0.80 m long and each ball on the end is 0.20 kg.(Ignore the mass of the rods. )What is the moment of inertia if the arrangement is spun around an axis through the center perpendicular to both rods? A)0.064 kg*m² B)0.096 kg*m² C)0.13 kg*m² D)0.32 kg*m²

Question

Quizplus · Accepted Answer

The moment of inertia of an "x" shape around its center axis can be calculated by adding the moments of inertia of two perpendicular rods, which can be calculated using the formula:

$I_rod= \frac{1}{12}mL^2$

where $m$ is the mass of the ball on the end and $L$ is the length of the rod.

For one rod, $m=0.20$ kg and $L=0.80$ m, so $I_rod=\frac{1}{12}(0.20)(0.80)^2=0.008	ext{ kg}\cdot	ext{m}^2$

Since there are two rods with perpendicular axes, their moments of inertia can be added using the parallel axis theorem:

$I_x= 2(I_rod)+2\cdot [\frac{1}{4}(0.80)^2](0.20+0.20)$

$I_x=0.016+0.064=0.08	ext{ kg}\cdot	ext{m}^2$

However, we need to spin the arrangement around the axis through the center perpendicular to both rods, which is at a distance of $\frac{1}{2}\sqrt{0.8^2+0.8^2}=0.565$ m from the center of mass. Thus, we need to use the perpendicular axis theorem to get the final moment of inertia:

$I=I_x+2\cdot (0.2)(0.565)^2=0.13	ext{ kg}\cdot	ext{m}^2$

Therefore, the answer is C.

A Majorette Takes Two Batons and Fastens Them Together in the Middle