A majorette takes two batons and fastens them together in the middle at right angles to make an "x" shape. Each baton was 0.90 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?Select one:a.0.075 kg·m2b.0.096 kg·m2c.0.064 kg·m2d.0.16 kg·m2
Question
A majorette takes two batons and fastens them together in the middle at right angles to make an "x" shape. Each baton was 0.90 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?Select one:a.0.075 kg·m2b.0.096 kg·m2c.0.064 kg·m2d.0.16 kg·m2
Solution
The moment of inertia (I) for a point mass is given by the equation I = m*r^2, where m is the mass and r is the distance from the axis of rotation.
In this case, there are four balls (two on each baton), each 0.45 m (half the length of a baton) from the axis of rotation.
So, the moment of inertia for each ball is I = mr^2 = 0.20 kg * (0.45 m)^2 = 0.0405 kgm^2.
Since there are four balls, the total moment of inertia is 4 * I = 4 * 0.0405 kgm^2 = 0.162 kgm^2.
So, the closest answer is d. 0.16 kg·m^2.
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