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

$I = \frac{mr^2}{12}$

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

$L_i = L_f$

$I_i \omega_1 = I_f \omega_2$

$(\frac{mr_{initial}^2}{12})(\omega_1)=(\frac{mr_{final}^2}{12})(\omega_2)$

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

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