A thin smooth rod of length L and mass M is rotating freely with angular speed ω0 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 :

Question

A thin smooth rod of length L and mass M is rotating freely with angular speed ω0 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 : (A) (Mω0/M +3 m)  (B) (Mω0/M + m)  (C) (Mω0/M + 2 m)  (D) (Mω0/M + 6 m)

Tardigrade Admin · Accepted Answer

Applying angular momentum conservation, about axis of rotation Li = Lf (ML2/12) ω0 = ( (ML2/12) + m ( (L/2) )2 × 2 ) ω ⇒ ω = (M ω0/M + 6m)

$\frac{M ω _{0}}{M + 3 m}$

$\frac{M ω _{0}}{M + m}$

$\frac{M ω _{0}}{M + 2 m}$

$\frac{M ω _{0}}{M + 6 m}$

Applying angular momentum conservation, about axis of rotation
$L_{i} = L_{f}$
$\frac{M L ^{2}}{12} ω_{0} = (\frac{M L ^{2}}{12} + m (\frac{L}{2})^{2} \times 2) ω$
$\Rightarrow ω = \frac{M ω _{0}}{M + 6 m}$

M+3mMω0​​

M+mMω0​​

M+2mMω0​​

M+6mMω0​​

Applying angular momentum conservation, about axis of rotation Li​=Lf​ 12ML2​ω0​=(12ML2​+m(2L​)2×2)ω ⇒ω=M+6mMω0​​

$\frac{M ω _{0}}{M + 3 m}$

$\frac{M ω _{0}}{M + m}$

$\frac{M ω _{0}}{M + 2 m}$

$\frac{M ω _{0}}{M + 6 m}$

Applying angular momentum conservation, about axis of rotation
$L_{i} = L_{f}$
$\frac{M L ^{2}}{12} ω_{0} = (\frac{M L ^{2}}{12} + m (\frac{L}{2})^{2} \times 2) ω$
$\Rightarrow ω = \frac{M ω _{0}}{M + 6 m}$