Question

A thin rod has a length of 0.25 m and rotates in a circle on a frictionless tabletop. the axis is perpendicular to the length of the rod at one of its ends. the rod has an angular velocity of 0.29 rad/s and a moment of inertia of 1.30 10-3 kg · m2. a bug standing on the axis decides to crawl out to the other end of the rod. when the bug (mass = 4.2 10-3 kg) gets where it's going, what is the angular velocity of the rod?

Answer 1

Given:
thin rod
→ length of 0.25 m
→  angular velocity of 0.29 rad/s
→ moment of inertia of 1.30 x 10⁻³ kg · m2

bug : mass = 4.2 x 10⁻³ kg

When the bug arrives at the end of the rod, it adds up to the initial inertia.

new inertia = 1.3 x 10⁻³ kg*m² + [4.2 x 10⁻³ kg*(0.25m)²] 
new inertia = 1.5626 x 10⁻³

initial inertia * angular velocity = new inertia * angular velocity
1.30 x 10⁻³ kg*m²   *  0.29 rad/s = 1.5626 x 10⁻³ * angular velocity
(1.30 x 10⁻³ kg*m²  *  0.29 rad/s) / 1.5626 x 10⁻³  = angular velocity 
0.24 rad/s = angular velocity

The angular velocity of the rod after the bug reach its end is 0.24 rad/s

Answer 2

We have that for the Question it can be said that  the angular velocity of the rod

$\omega=0.2998rads/s$

From the question we are told

• A thin rod has a length of 0.25 m and rotates in a circle on a frictionless tabletop. the axis is perpendicular to the length of the rod at one of its ends.

• the rod has an angular velocity of 0.29 rad/s and a moment of inertia of 1.30 10-3 kg · m2. a bug standing on the axis decides to crawl out to the other end of the rod.

• when the bug (mass = 4.2 10-3 kg) gets where it's going, what is the angular velocity of the rod?

Generally the equation for the momentum  is mathematically given as

$I=(mL^2+I)\\\\\I=4.2*10^{-3}*(0.25)^2+ 1.30 10^{-3}\\\\I=0.4544\\\\Therefore\\\\\omega=w_0(I/I')\\\\\omega=0.3(\frac{1.30 10^{-3}}{0.4544})$

$\omega=0.2998rads/s$

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