(a) [tex]0.2888 kg m^2[/tex]
The moment of inertia of a uniform-density disk is given by
[tex]I=\frac{1}{2}MR^2[/tex]
where
M is the mass of the disk
R is its radius
In this problem,
M = 16 kg is the mass of the disk
R = 0.19 m is the radius
Substituting into the equation, we find
[tex]I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2[/tex]
(b) 142.5 J
The rotational kinetic energy of the disk is given by
[tex]K=\frac{1}{2}I\omega^2[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular velocity
We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s[/tex]
And so, the rotational kinetic energy is
[tex]K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J[/tex]
(c) [tex]9.07 kg m^2 /s[/tex]
The rotational angular momentum of the disk is given by
[tex]L=I\omega[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular velocity
Substituting the values found in the previous parts of the problem, we find
[tex]L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s[/tex]
The moment of inertia of the disk is equal to [tex]0.2888kgm^2[/tex]. The rotational kinetic energy is [tex]142.5 J[/tex] and the rotational angular momentum is [tex]9.07kgm^2/s[/tex].
[tex]I=\frac{1}{2} MR^2[/tex]
In this equation, the "M" will be the mass of the disk, while the "R" will be the radius. Therefore, we can calculate the equation as follows:
[tex]I=\frac{1}{2} 16*(0.19)^2 = 0.2888kgm^2[/tex]
[tex]K=\frac{1}{2} IW^2[/tex]
In this equation, the "I" is the moment of inertia, while the "W" is the angular velocity.
To calculate the angular velocity, we will use the equation:
[tex]W= \frac{2\pi }{t}\\W=\frac{2\pi }{0.2} = 31.4rad/s[/tex]
Now, we can calculate rotational kinetic energy as follows:
[tex]K=\frac{1}{2} 0.2888*(31.4)^2=142.5J[/tex]
[tex]L=IW\\L=0.2888*31.4=9.07kgm^2/s[/tex]
More information about inertia is in the link:
https://brainly.com/question/3268780
