Answer to Question #81312 in Mechanics | Relativity for Sarada prasan mandal

Question

A child of mass 50kg is standing on the edge of a merry go round of mass 250 kg and radius 3.0m which is rotating with an angular velocity of 3.0 rad s^-1 . The child then starts walking towards the centre of the merry ho round. What will be the final angular velocity of the merry go round when the child reaches the centre ?

Answer 1

We can find the final angular velocity of the merry-go-round from the law of conservation of angular momentum:
L_i=L_f,
I_i ω_i=I_f ω_f,
here, I_i is the initial rotational inertia of the system, I_f is the final rotational inertia of the system, ω_i is the initial angular velocity of the merry-go-round, ω_fis the final angular velocity of the merry-go-round.
We can find the initial rotational inertia of the system as follows:
I_i=(I_(disk,i)+I_(child,i) )=(1/2 m_disk r_(disk,i)^2+m_child r_(child,i)^2 ),
here, I_(disk,i)=1/2 m_disk r_(disk,i)^2 is the initial rotational inertia of the merry-go-round, I_(child,i)=m_child r_(child,i)^2 is the initial rotational inertia of the child, m_disk is the mass of the marry-go-round, m_child is the mass of the child, r_disk is the radius of the merry-go-round, r_child is the distance from the centre of the merry-go-round to the child.
Then, we can calculate I_i:
I_i=(1/2 m_disk r_(disk,i)^2+m_child r_(child,i)^2 )==(1/2∙250 kg∙(3.0 m)^2+50 kg∙(3.0 m)^2 )=1575 kg∙m^2.
Similarly, we can find the final rotational inertia of the system:
I_f=(I_(disk,f)+I_(child,f) )=(1/2 m_disk r_(disk,f)^2+m_child r_(child,f)^2 ),
here, I_(disk,f)=1/2 m_disk r_(disk,f)^2 is the final rotational inertia of the merry-go-round, I_(child,f)=m_child r_(child,f)^2 is the final rotational inertia of the child, m_disk is the mass of the marry-go-round, m_child is the mass of the child, r_disk is the radius of the merry-go-round, r_child is the distance from the centre of the merry-go-round to the child.
Then, we can calculate I_f:
I_f=(1/2 m_disk r_(disk,f)^2+m_child r_(child,f)^2 )==(1/2∙250 kg∙(3.0 m)^2+50 kg∙(0.0 m)^2 )=1125 kg∙m^2.
Finally, we can calculate the final angular velocity of the merry-go-round from the law of conservation of angular momentum:
I_i ω_i=I_f ω_f,
ω_f=ω_i I_i/I_f =3.0 rad/s∙(1575 kg∙m^2)/(1125 kg∙m^2 )=4.2 rad/s.