A) [tex]1789.4 kg m^2[/tex]
B) 1.76 rad/s
C) 1.26 rad/s
Explanation:
A)
The moment of inertia of a uniform circular disc rotating about its axis is given by
[tex]I=\frac{1}{2}MR^2[/tex]
where
M is the mass of the disc
R is the radius of the disc
For the merry-go-round in this problem, we have:
M = 317 kg is the mass of the disc
R = 3.36 m is its radius
Substituting, we find the moment of inertia:
[tex]I=\frac{1}{2}(317)(3.36)^2=1789.4 kg m^2[/tex]
B)
The angular velocity of the merry-go-round is the rate of change of its angular displacement, and it is given by:
[tex]\omega=\frac{2\pi}{T}[/tex]
where
[tex]\omega[/tex] is the angular velocity
T is the period of revolution (the time taken for completing 1 revolution)
For the merry-go-round in this problem, it takes 3.58 s to complete one revolution so:
[tex]T=3.58 s[/tex] is the period of revolution
Therefore, the angular velocity is:
[tex]\omega=\frac{2\pi}{3.58}=1.76 rad/s[/tex]
C)
When the man walks towards the edge, the angular momentum of the system, must be conserved, since there are no external torques acting on it.
The angurlar momentum is given by:
[tex]L=I\omega[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular velocity
Since the angular momentum must remain constant, we can write:
[tex]L_1 =L_2\\I_1\omega_1 = I_2 \omega_2[/tex]
where:
[tex]I_1=1789.4 kg m^2[/tex] is the initial moment of inertia of the system
[tex]\omega_1=1.76 rad/s[/tex] is the initial angular velocity
[tex]I_2=I_1+mR^2[/tex] is the final moment of inertia, where:
m = 63.6 kg is the mass of the man
R = 3.36 m is the final distance of the man from the axis
Substituting,
[tex]I_2=1789.4+(63.6)(3.36)^2=2507.4 kg m^2[/tex]
And [tex]\omega_2[/tex] is the final angular velocity. Substituting,
[tex]\omega_2=\frac{I_1 \omega_1}{I_2}=\frac{(1789.4)(1.76)}{2507.4}=1.26 rad/s[/tex]
