A uniform rod of mass 2.55×10−2 kg and length 0.380 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.200 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.80×10−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 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. What is the angular speed of the system at the instant when the rings reach the ends of the rod?

What is the angular speed of the rod after the rings leave it?

[tex]L_{system}=I_{system}\omega \\\\I_{system}=I_{rod}+2\times m\times r^{2}\\\\I_{system}=\frac{1}{12}ml^{2}+2\times 0.2\times (4.8\times 10^{-2})^{2}\\\\I_{system}=1.22845\times 10^{-3}kgm^{2}\\\\L_{system}=1.22845\times 10^{-3}kgm^{2}\times 3.66rad/sec[/tex]

[tex]\L_{system}=4.496\times 10^{-3}kgm^{2}[/tex]

Now the final angular momentum of the system

[tex]L_{fianl}=I_{final}\omega \\\\I_{final}=I_{rod}+2\times m\times r_{f}^{2}\\\\I_{final}=\frac{1}{12}ml^{2}+2\times 0.2\times (0.19)^{2}\\\\I_{final}=0.01447kgm^{2}\\\\L_{final}=0.01447kgm^{2}\times \omega _{final}rad/sec[/tex]

Thus equating initial and final angular momentum we solve for final angular velocity of rod as

[tex]\omega _{f}=\frac{4.496\times 10^{-3}}{0.01447}\\\\\omega _{f}=0.3107rad/sec[/tex]

part b)

When the rings leave the system we again conserve the angular momentum just before the rings leave the system and the instant when they just leave

[tex]\therefore 0.01447=\frac{1}{12}ml^{2}w\\\\\therefore w=\frac{4.582\times 10^{-3}}{3.068\times10^{-4} }\\\\w_{final}=14.93rad/sec[/tex]

migueletorallam migueletorallam · Answer 2 · 2019-12-21T19:11:01+01:00

Answer:

A)[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].

B)[tex]\omega_{f}=140\frac{rev}{min}[/tex].

Explanation:

A)

For this problem, we will use the conservation of angular momentum.

[tex]L_{0}=L_{f}\\[/tex].

In the beginning, we have that

[tex]L_{0}=I_{0}\omega_{0}\\[/tex]

where [tex]I_{0}[/tex] is the inertia moment of all the system at the starting position, this is the inertia moment of the rod plus the inertia moment of each ring ([tex]mr^{2}[/tex], with [tex]r[/tex] the distance from the ring to the fixed axis and, [tex]m[/tex] its mass) at the starting position and, [tex]\omega_{0}[/tex] is the initial angular velocity. So

[tex]L_{0}=(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}[/tex].

When the rings are at the ends of the rod the angular momentum becomes

[tex]L_{f}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex],

where [tex]r_{f}[/tex] is the distance from the fixed axis to the end of the rod (the final position of the rings).

Using conservation of angular momentum we get

[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex].

thus

[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2}+2mr_{f}^{2})}[/tex]

computing this last expresion we get

[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2+2(0.200)(0.19)^{2})}[/tex]

[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].

B)

Again we use the conservation of angular momentum. The initial angular momentum if the same as before. The final angular momentum will be

[tex]L_{f}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex],

this time we will not take into account the inertia moment of the rings because they are no longer part of the system (they leave the rod).

[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex].

thus

[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2})}[/tex]

computing this last expresion we get

[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2})[/tex]

[tex]\omega_{f}=140\frac{rev}{min}[/tex].