Explanation:
Given
radius[tex]r=2.63 m[/tex]
[tex]N=0.526 rev/s [/tex]
[tex]\omega =3.30 rad/s[/tex]
mass disc [tex]M=152 kg[/tex]
mass of person [tex]m=51.7 kg[/tex]
velocity of Person [tex]v=2.76 m/s[/tex]
moment of inertia [tex]I=Mr^2[/tex]
[tex]I=0.5\times 152\times 2.63^2=827.64 kg-m^2[/tex]
Initial angular momentum
[tex]L_i=I\omega +mvr[/tex]
[tex]L_i=827.64\times 3.30+51.7\times 2.76\times 2.63[/tex]
[tex]L_i=2731.212+375.27=3106.48 Js[/tex]
Final Moment inertia
[tex]I_f=0.5Mr^2+mr^2[/tex]
[tex]I_f=(152\cdot 0.5+51.7)\cdot (2.63)^2=1185.243 kg-m^2[/tex]
final angular momentum
[tex]L_f=I_f\omega _f[/tex]
Conserving angular momentum
[tex]L_i=L_f[/tex]
[tex]3106.48=1408.97\times \omega _f [/tex]
[tex]\omega _f=2.62 rad/s[/tex]
The final angular speed of the merry-go-round is mathematically given as
wf=2.62 rad/s
Question Parameter(s):
radius 2.63 m and mass 152 kg
an angular speed of 0.526 rev/s.
A 51.7 kg per
at 2.76 m/s jumps onto its rim and holds on.
Generally, the equation for the Initial angular momentum is mathematically given as
Li=Iw+mvr
Therefore
Li=827.64* 3.30+51.7* 2.76* 2.63
Li=3106.48 Js
The Final Moment inertia
Lf=0.5Mr^2+mr^2
Lf=(152*0.5+51.7)* (2.63)^2
Lf=1185.243 kg-m^2
In conclusion, angular momentum
3106.48=1408.97*wf
wf=2.62 rad/s
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