Question

In preparation for building a space station, an astronaut removes a self-telescoping uniform rod from the cargo bay and releases it, not noticing that he gave it an angular speed of 0.0500 radians per second. The pole is 3.00 meters long. A catch slips, and the pole spring retracts it into a shorter 1.50 meter long uniform pole about a minute after the astronaut releases it. Find the angular speed after the catch slips.

Answer 1

To solve this problem it is necessary to apply the concepts related to the conservation of angular momentum. This can be expressed mathematically as a function of inertia and angular velocity, that is:

`L = I\omega`

Where,

I = Moment of Inertia

`\omega`= Angular Velocity

For the given object the moment of inertia is equivalent to

`I = (mr^2)/(12)`

Considering that the moment of inertia varies according to distance, and that there are two of these without altering the mass we will finally have to

`L_i = L_f`

`I_i \omega_1 = I_f \omega_2`

`((mr_(initial)^2)/(12))(\omega_1)=((mr_(final)^2)/(12))(\omega_2)`

`(r_(initial)^2})(\omega_1)=(r_(final)^2)(\omega_2)`

Our values are given as,

`r_(initial) = 3m\\\omega_1 = 0.05rad/s \\r_(final)=1.5m`

Replacing we have,

`(3^2})(0.05)=(1.5^2)(\omega_2)`

`\omega_2 = 0.2rad/s`

Therefore the angular speed after the catch slips is 0.2rad/s