Respuesta :
Okay here initially child walks on the outer edge of the disc and then started to move inside
So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside
As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.
But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system
Here as we know that
[tex]\tau = \frac{dL}{dt}[/tex]
where L = angular momentum
since here torque is ZERO
[tex]0 = \frac{dL}{dt}[/tex]
L = constant
so here we can write initial angular momentum of the system as
[tex]L = (I_1 + I_2)*\omega[/tex]
here we know that
[tex]I_1 [/tex] = moment of inertia of merry go round
[tex]I_2 [/tex] = moment of inertia of child
so here we can say
[tex](I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2[/tex]
so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as
[tex]I_2 = mr^2[/tex]
here m = mass of child
r = distance of child from center
Since child is moving inside so his distance from center is decreasing
so here moment of inertia of child is decreasing as he starts moving inside
so final angular speed of merry go round will increase as child go inside
[tex]\omega_2 = \frac{(I_1 + I_2)*\omega}{(I_1 + I_2')}[/tex]
so here as
[tex]I_2' < I_2[/tex]
final angular speed will be more than initial speed as child moves inside
