A baton twirler in a marching band complains that her baton is defective. The ma

Question

A baton twirler in a marching band complains that her baton is defective. The manufacturer specifies that the baton should have an overall length of L= 60.0 cm and a total mass between 940.0 and 950.0 g (there is one 350-g objects on each end). Also according to the manufacturer, the moment of inertia about the central axis passing through the baton should fall between 0.0750 and 0.0800 kg/m2. The twirler (who has completed a class in physics) claims this is impossible.

What is the minimum moment of inertia?

What is the maximum moment of inertia?

Explanation / Answer

We know that the moment of inertia of the rod is given by:

I = 1/12 M L^2

tat of point masses is:

I = m r^2

So the total moment of inertia of the given system will be:

I = 1/12 M L^2 + 2 m r^2

m(min) = 940 g = 0.940 kg ; L = 60 cm = 0.6 m ; m = 350 g = 0.35 kg ; r = L/2 = 30 cm = 0.3 m

I(min) = 1/12 x 0.940 x 0.6^2 + 2 x 0.35 x 0.3^2 = 0.0912 kg-m^2

Hence, I(min) = 0.0912 kg-m^2

For I(max) ; m(max) = 950 g = 0.95 kg

I(max) = 1/12 x 0.950 x 0.6^2 + 2 x 0.35 x 0.3^2 = 0.0915 kg-m^2

Hence, I(max) = 0.0915 kg-m^2

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