Question

A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.1 kg · m2 . A student tosses a 1.5-kg mass with a speed of 2.7 m/s to the professor, who catches it at a distance of 0.40 m from the axis of rotation. What is the resulting angular speed of the professor and the stool? (Assume that when the professor catches the mass, their arm is extended along a line radially outward from the axis of rotation, and the velocity of the mass is perpendicular to that line.)

Answer 1

`\omega=0.37 [rad/s]`

Step-by-step explanation:

We can use the conservation of the angular momentum.

`L=mvR`

`I\omega=mvR`

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

`(I_(proffesor - stool)+mR^(2))\omega=mvR`

Now, we just need to solve it for ω.

`\omega=(mvR)/(I_(proffesor-stool)+mR^(2))`

`\omega=(1.5*2.7*0.4)/(4.1+1.5*0.4^(2))`

`\omega=0.37 [rad/s]`

I hope it helps you!