Answer to Question #138923 in Classical Mechanics for dein

Question

A baton consists of a nearly massless rod of length L and two spheres attached to the ends of the rod, whose masses are each M and whose centers are separated by L. The spheres have radius R. If the sphere radius R is not negligible compared to the length L, what is the moment of inertia of the baton about an axis perpendicular to its length and passing through its center?


A. 2MR2


B. 2ML2


C. 2ML2 + 2MR2


D. (1/2) ML2


E. (1/2) ML2 + 2MR2

Answer 1

According to the conditions of the problem, the distance between the centers of gravity of the spheres is L.The length of the rod connecting the sphere is also L.Consequently, the value of R tends to zero and the spheres can be taken as material points.It is necessary to calculate the moment of inertia of two bodies located at a distance $\frac{L}{2}$ from the axis of rotation.

$I= M*(\frac{L}{2})^2+M*(\frac{L}{2})^2=\frac{ML^2}{2}$

Answer:$D.\ \frac{1}{2}ML^2$