Question

A uniform rod of mass 3.00×10−2 kg and length 0.350 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.240 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.50×10−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 25.0 rev/min . Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

Answer 1

Answer:

The angular speed of the system at the instant when the rings reach the ends of the rod is 2.93 rev/min.

Explanation:

Given that,

Mass of rod $m= 3.00\times10^{-2}\ kg$

Length = 0.350 m

Mass of small ring = 0.240 kg

Distance $d= 5.50\times10^{-2}\ m$

Angular velocity = 25.0 rev/min

Suppose we calculate the angular speed of the system at the instant when the rings reach the ends of the rod

We need to calculate the moment of inertia of rod

Using formula of moment of inertia

$I_{rod}=\dfrac{1}{12}ML^2$

Put the value into the formula

$I_{rod}=\dfrac{1}{12}\times3.00\times10^{-2}\times0.350^2$

$I_{rod}=3.0625\times10^{-4}\ kg-m^2$

We need to calculate the moment of inertia of the rings when they are at a distance r₁

Using formula of moment of inertia

$I_{ring}=2mr_{1}^2$

Put the value into the formula

$I_{ring}=2\times0.240\times(5.50\times10^{-2})^2$

$I_{ring}=14.52\times10^{-4}\ kg-m^2$

We need to calculate the moment of inertia of the rings when they are at a distance r₂

Using formula of moment of inertia

$I_{ring}=2mr_{2}^2$

Put the value into the formula

$I_{ring}'=2\times0.240\times(\dfrac{0.350}{2})^2$

$I_{ring}'=0.0147\ kg-m^2$

We need to calculate the angular speed of the system at the instant when the rings reach the ends of the rod

Using conservation of angular momentum

$(I_{rod}+I_{ring})\omega_{0}=(I_{rod}+I_{ring}')\omega$

Put the value into the formula

$\omega=\dfrac{(I_{rod}+I_{ring})\omega_{0}}{(I_{rod}+I_{ring}')}$

$\omega=\dfrac{3.0625\times10^{-4}+14.52\times10^{-4}}{3.0625\times10^{-4}+0.0147}\times25.0$

$\omega=2.93\ rev/min$

Hence, The angular speed of the system at the instant when the rings reach the ends of the rod is 2.93 rev/min.