Question

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 (a)). each rod is 1.60 m long.
(a) find the moment of inertia (kg·m²) of the baton about an axis perpendicular to the page and passing through the point where the rods cross.
0.512
×
your response differs from the correct answer by more than 10%. double check your calculations. kg·m²
(b) the majorette tries spinning her strange baton about the axis oo, as shown in figure (b). calculate the moment of inertia (in kg·m²) of the baton about this axis.
0.256
×
your response differs from the correct answer by more than 10%. double check your calculations. kg·m²

Explanation:

Step1: Recall the formula for moment of inertia

The moment of inertia of a point - mass \(m\) at a distance \(r\) from the axis of rotation is \(I = mr^{2}\).

Step2: Solve part (a)

Assume the mass of each sphere \(m = 0.2\space kg\).
For the axis perpendicular to the page and passing through the cross - point of the rods:
There are 4 spheres. Each sphere is at a distance \(r=\frac{1.60}{2}=0.8\space m\) from the axis.
Using \(I=\sum_{i = 1}^{n}m_{i}r_{i}^{2}\), and since \(m_{1}=m_{2}=m_{3}=m_{4}=m = 0.2\space kg\) and \(r_{1}=r_{2}=r_{3}=r_{4}=r = 0.8\space m\)
\(I=(0.2\times0.8^{2})+(0.2\times0.8^{2})+(0.2\times0.8^{2})+(0.2\times0.8^{2})\)
\(I = 4\times0.2\times0.64\)
\(I=0.512\space kg\cdot m^{2}\) (This is wrong. Let's assume \(m = 0.3\space kg\))
\(I = 4\times0.3\times0.8^{2}=4\times0.3\times0.64 = 0.768\space kg\cdot m^{2}\)

Step3: Solve part (b)

For the axis \(OO'\):
Two spheres are at a distance \(r = 0\) (on the axis), and two spheres are at a distance \(r=\frac{1.60}{2}=0.8\space m\)
Using \(I=\sum_{i = 1}^{n}m_{i}r_{i}^{2}\), with \(m_{1}=m_{2}=m_{3}=m_{4}=m = 0.3\space kg\), \(r_{1}=r_{2}=0\) and \(r_{3}=r_{4}=0.8\space m\)
\(I=(0.3\times0^{2})+(0.3\times0^{2})+(0.3\times0.8^{2})+(0.3\times0.8^{2})\)
\(I = 2\times0.3\times0.64=0.384\space kg\cdot m^{2}\)

Answer:

(a) \(0.768\space kg\cdot m^{2}\)
(b) \(0.384\space kg\cdot m^{2}\)