Answer:
The final angular speed of the carousel after the person climbs aboard is 1.70 rad/s
Explanation:
Conservation of angular moment states that if there are not external torques on a system its angular moment will be conserved i.e. final angular momentum is equal to initial angular momentum, mathematically:
[tex] \overrightarrow{L_{i}}=\overrightarrow{L_{f}} [/tex] (1)
With Li the initial angular momentum and Lf the final angular momentum, we are interested only on the magnitude of the angular momenta. Initially, the total angular momentum of the system carousel-person is only the angular momentum of the playground carousel because the person isn’t moving, so it is the product of its moment of inertia (I) and angular velocity ( [tex] \omega_{i}) [/tex]:
[tex] L_{i}=I\omega_{i}=124*3.48=431.52\,\frac{kg\,m^{2}}{s} [/tex] (2)
But the final angular momentum of the system is the angular momentum of the carousel plus the angular momentum of the person moving on a circle, again the angular momentum of the carousel is the product of its moment of inertia and angular velocity, but the angular momentum of the person is the product of the radius (R) of the carousel, the person’s mass (m) and its tangential velocity [tex] v_{tan} [/tex]:
[tex] L_{person}=Rmv_{tan} [/tex] (3)
The angular velocity of the person is the same of the carousel because the person is on the carousel, and its related with the tangential velocity by:
[tex] v_{tan}=\omega_{f}R [/tex] (4)
Replacing (4) on (3) and adding this to the angular moment of the carousel we obtain the final angular momentum of the system:
[tex] L_{f}=I\omega_{f}+mR^{2}\omega_{f}=\omega_{f}(I+mR^{2}) [/tex] (5)
Equating this with the initial angular momentum an solving for [tex] \omega_{f} [/tex] we obtain:
[tex] \omega_{f}=\frac{431.52}{(I+mR^{2})}=\frac{431.52}{(124+46.8*1.66^{2})}\simeq1.70\,\frac{rad}{s} [/tex]
