Respuesta :
Answer:
ωf = 0.82 rad/s
Explanation:
By conservation of the angular momentum:
[tex]I_c * \omega_o = (I_c+I_m)*\omega_f[/tex]
Where [tex]I_c[/tex] is the inertia of the child, [tex]I_m[/tex] is the inertia of the merry-go-round.
Solving for ωf:
[tex]\omega_f = (m_c*R^2)*(V_o/R)/(m_c*R^2+M_m*R^2/2)[/tex]
[tex]\omega_f=0.82rad/s[/tex]
The angular speed of the many-go- round after the child has jumped on and is standing at its outer rim is 0.82 rad/s.
What is angular speed of a body?
The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,
[tex]\omega= \dfrac{\Delta \theta}{\Delta t}[/tex]
In a playground there is a small merry-go-round of radius 1.25 m and mass 175 kg. A child of mass 45 kg runs at a speed of 3.0 m/s tangent to the rim of the merry-go-round (initially at rest) and jumps on.
The initial angular momentum is,
[tex]L_i=m_cvR[/tex]
The final angular momentum is,
[tex]L_f=R^2(m_c+\dfrac{m_m^2}{2})\omega[/tex]
By the conservation of momentum,
[tex]L_i=L_f\\m_cvR=R^2(m_c+\dfrac{m_m^2}{2})\omega\\\omega=\dfrac{m_cv}{(m_c+\dfrac{m_m^2}{2})R}[/tex]
Put the values of mass velocity and radius as,
[tex]\omega=\dfrac{(45)(3)}{((45)+\dfrac{(175)^2}{2})(1.25)}\\\omega=0.82\rm \;rad/s[/tex]
Thus, the angular speed of the many-go- round after the child has jumped on and is standing at its outer rim is 0.82 rad/s.
Learn more about the angular speed here;
https://brainly.com/question/540174
