Physics- rotating rod?

Question

A uniform rod of mass m_1 and length L rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass m_2, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance r on each side from the center of the rod, and the system is rotating at an angular velocity w (omega). Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

a) What is the angular speed of the system at the instant when the rings reach the ends of the rod?
b) What is the angular speed of the rod after the rings leave it?

Answer 1

This is solved by conservation of angular momentum, which is I*w, where I=moment of inertia, w = angular velocity. Calculate I1, with the rings in the first postion (that is the sum of the I of the rod Ir, plus the I from the rings, which for each ring is m_2*r^2; so the initial momentum is

(Ir + 2*m_2*r^2)*w1.

At the time the rings are about to leave the rod, they are at L/2 from the center of rotation, so the angular momentum then is

[Ir + 2*m_2*(L/2)^2]*w2

Equate these and solve for w2.

When the rings leave the rod, the only moment is that of the rod, and its angular momentum is

Ir*w3,

which must equal the initial momentum (Ir + 2m_2*r^2)*w1

Set these equal and solve for w3.

Look up the angular momentum Ir of a rod of length L rotating about its centerpoint (I think it's m_1 * (L^2)/12, but you should check it.)

Answer 2

The answer for part b is the same as it is for part a. This is because the rings leave the system. Think of it this way: if you were skateboarding behind a car which propelled you, if you were to let go, the speed of your skateboard would remain the same.

Hope that helps!

Answer 3

use conservation of angular momentum
( I+2*m(2)*r^2 ) w(1) = ( I+2*m(2)*L^2 ) *w(2)

where I=(1/12) * m(1) * L^2

for second part i think you just remove the m(2) part from right hand side of above expression

Answer 4

Rotating Rod? I like the sound of that.....=)




Sorry, I couldn't resist.