Question

Yet another bizarre baton is created by taking four identical balls, each with mass 0.261 kg, and fixing them as before except that one of the rods has a length of 1.18 m and the other has a length of 1.58 m.

(a) Calculate the moment of inertia of this baton when oriented as shown in the figure.
I =  kg · m²

(b) Calculate the moment of inertia of this baton when oriented as shown in the figure, with the shorter rod vertical.
I =  kg · m²

(c) Calculate the moment of inertia of this baton when oriented as shown in the figure, but with longer rod vertical.
I =  kg · m²

Answer 1

the moment of inertia of this baton when oriented is

I = m1 r1^2 + m2 r2^2 + m3 r3^2 + m4 r4^2

= m ( 2 r1^2 + 2 r2^2)

=2m ( r1^2 + r2^2)

= 2(0.261) ((1.18/2)^2 + (1.58/2)^2)

=0.5074 kg m^2

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

I = 2 m ( r2^2)

= 2 (0.261) ( 1.58/2)^2

=0.3257802 kg m^2

(c)

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

I = 2 m ( r2^2)

= 2 (0.261) ( 1.18/2)^2

=0.181 kg m^2