Respuesta :
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia, [tex]I_1=900\ kg.m^2[/tex]
Initial angular velocity of the platform, [tex]\omega=0.95\ rad/s[/tex]
Part A,
Let [tex]\omega_2[/tex] is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :
[tex]I_1\omega_1=I_2\omega_2[/tex]
Here, [tex]I_2=I_1+mr^2[/tex]
[tex]I_1\omega_1=(I_1+mr^2)\omega_2[/tex]
[tex]900\times 0.95=(900+70\times (2.9)^2)\omega_2[/tex]
Solving the above equation, we get the value as :
[tex]\omega_2=0.574\ rad/s[/tex]
Part B,
The initial rotational kinetic energy is given by :
[tex]k_i=\dfrac{1}{2}I_1\omega_1^2[/tex]
[tex]k_i=\dfrac{1}{2}\times 900\times (0.95)^2[/tex]
[tex]k_i=406.12\ rad/s[/tex]
The final rotational kinetic energy is given by :
[tex]k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2[/tex]
[tex]k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2[/tex]
[tex]k_f=245.24\ rad/s[/tex]
Hence, this is the required solution.
