Respuesta :
Answer:
[tex]W_s = 19.05 rpm[/tex]
Explanation:
For answer this we will use the law of the conservation of the angular momentum L where:
[tex]L_i = L_f[/tex]
First, the important data is:
[tex]M_t = 300 kg[/tex] (mass of the merry-go-round)
[tex]R_ t = 1.5 m[/tex] (radius of the merry-go-round)
[tex]W_t = 16 rpm = 1.675 rad/s[/tex] (angular velocity of the merry-go-round)
[tex]V_j =5.4m/s [/tex] (velocity od jhon)
[tex]M_j = 30 kg[/tex] (mass of jhon)
Then,
[tex]L_i = L_f[/tex]
[tex]I_tW_t + M_jV_jR_t = I_sW_s[/tex]
where [tex]I_s[/tex] is the moment of inerta after jhon jumps on, [tex]I_t[/tex] is the moment of inertia of the merry-go-round and [tex]W_s[/tex] is the angular velocity of the merry-go-round after jhon jumps on.
So, we have to find the moment of inertia of the merry-go-round as:
[tex]I_t = \frac{1}{2}M_tR_t^2[/tex]
[tex]I_t = \frac{1}{2}(300)(1.5)^2[/tex]
[tex]I_t = 337.5 kg*m^2[/tex]
Now we have to find[tex]I_s[/tex] as:
[tex]I_s = I_t + M_jR_t^2[/tex]
[tex]I_s = 337.5 + (30)(1.5)^2[/tex]
[tex]I_s = 405 kg*m^2[/tex]
Then, we replace the data on the initial equation and we get:
[tex](337.5)(1.675) + (30)(5.4)(1.5) = (405)W_s[/tex]
Solving for [tex]W_s[/tex]:
[tex]W_s = 1.995 rad/s[/tex]
Finally, we change it to rpm as:
[tex]W_s = 19.05 rpm[/tex]
