Question 266: Centre of Mass and Rotational Motion

Question

A thin smooth rod of length L and mass M is
rotating freely with angular speed $$\omega $$0 about an
axis perpendicular to the rod and passing
through its center. Two beads of mass m and
negligible size are at the center of the rod
initially. The beads are free to slide along the
rod. The angular speed of the system , when
the beads reach the opposite ends of the rod,
will be :-

Accepted Answer

Initial angular momentum = Final Angular
Momentum
$${{M{L^2}} \over {12}}{\omega _0} = \left( {{{M{L^2}} \over {12}} + 2{{m{L^2}} \over 4}} \right)\omega $$
$$ \Rightarrow \omega  = {{M{\omega _0}} \over {M + 6m}}$$

Answer

$${{M{\omega _0}} \over {M + 3m}}$$

Answer

$${{M{\omega _0}} \over {M + m}}$$

Answer

$${{M{\omega _0}} \over {M + 6m}}$$

Answer

$${{M{\omega _0}} \over {M + 2m}}$$

Centre of Mass and Rotational Motion

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