Question

A uniform rod with a mass of 100 g and a length of 50.0 cm rotates in a horizontal plane about a fixed vertical, frictionless pin passing through the center of the rod. Two small beads, each having a mass of 30.0 g, are mounted on the rod so that they are able to slide without friction along its length. Initially, the beads are held by catches at positions 10.0 cm on each side of the rod's center, at this time, the system rotates at an angular speed of 20.0 rad/s. Suddenly, the catches are released, and the small beads slide outward along the rod. Find a) the angular speed of the system at eh instant the beads reach the end of the rod and b) the angular speed of the rod after the after the beads fly off the rod's ends.

Answer 1

(a) The angular speed of the system at the instant the beads reach the end of the rod is 9.26 rad/s.

(b) The angular speed of the rod after the after the beads fly off the rod's ends is 25.71 rad/s.

Moment of inertia through the center of the rod

I = ¹/₁₂ML²

I =  ¹/₁₂ (0.1)(0.5)²

I = 0.0021 kgm²

For the beads, I = 2Mr² = 2(0.03 x 0.1²) = 0.0006 kgm²

Total initial moment of inertia, Ii = 0.0021 kgm² + 0.0006 kgm²

I(i)= 0.0027 kgm²

When the beads reach the end, I = 2Mr² = 2(0.03)(0.25)² = 0.00373 kgm²

Total final moment of inertia, I(f) = 0.0021 kgm²  +  0.00373 kgm²

I(f) = 0.00583 kgm²

Speed of the system

The speed of the system at the moment the beads reach the end of the rod is calculated as follows;

$I_i \omega_i = I_f\omega _f\\\\\omega _f = \frac{I_i \omega_i}{I_f}$

$\omega_f = \frac{0.0027 \times 20}{0.00583} \\\\\omega _f = 9.26 \ rad/s$

Speed of the rod when the beads fly off

$\omega_f = \frac{0.0027 \times 20}{0.0021} \\\\\omega _f = 25.71 \ rad/s$

Learn more about moment of inertia of rods here: brainly.com/question/3406242