Yet another bizarre baton is created by taking four identical balls, each with m

Question

Yet another bizarre baton is created by taking four identical balls, each with mass 0.301 kg, and fixing them as before except that one of the rods has a length of 1.05 m and the other has a length of 1.64 m.

(a) Calculate the moment of inertia of this baton when oriented as shown in the figure.

b)Calculate the moment of inertia of this baton when oriented as shown in the figure, with the shorter rod vertical.

c)Calculate the moment of inertia of this baton when oriented as shown in the figure, but with longer rod vertical.

Yet another bizarre baton is created by taking four identical balls, each with mass 0.301 kg, and fixing them as before except that one of the rods has a length of 1.05 m and the other has a length of 1.64 m. Calculate the moment of inertia of this baton when oriented as shown in the figure. Calculate the moment of inertia of this baton when oriented as shown in the figure, with the shorter rod vertical. Calculate the moment of inertia of this baton when oriented as shown in the figure, but with longer rod vertical.

Explanation / Answer

A.) I = mr^2 = m1r^2 + m2r^2 + m3r^2 + m4r^2

(.301 kg)(.525 m)^2 + (.301 kg)(.82 m)^2 + (.301 kg)(.525 m)^2 + (.301 kg)(.82 m)^2=.5707 kg-m^2

B.) I = mr^2 = m1r^2 + m2r^2 + m3r^2 + m4r^2

(.301 kg)(0) + (.301 kg)(.82 m)^2 + (.301 kg)(0) + (.301 kg)(.82 m)^2 = .49367 kg m^2

C.) I = mr^2 = m1r^2 + m2r^2 + m3r^2 + m4r^2

(.301 kg)(.525 m)^2 + (.301 kg)(0) + (.301 kg)(.525 m)^2 + (.301 kg)(0) = .16592kg m^2

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