Answer:
a) I = 0.42358 Kg*m^2
b) K = 1203.99 J
Explanation:
a) The moment of inertia is:
I = [tex]\frac{1}{2}M_dR_d^2+\frac{1}{2}M_a(R_e^2+R_i^2)[/tex]
where [tex]M_d[/tex] is the mass of the disk, [tex]R_d[/tex] is the radius of the disk, [tex]M_a[/tex] the mass of the annular cylinder, [tex]R_e[/tex] is the outer radius of the cylinder and [tex]R_i[/tex] the inner radius of the cylinder.
Replacing values, we get:
I = [tex]\frac{1}{2}(2.5kg)(0.52m)^2+\frac{1}{2}(1.1kg)(0.34m^2+0.2m^2)[/tex]
I = 0.42358 Kg*m^2
b) First, we will change the angular velocity from rev/s to rad/s as:
W = 12*2[tex]\pi[/tex]rad/s
W = 75.398 rad/s
Also the kinetic energy K is:
K = [tex]\frac{1}{2}IW^2[/tex]
Where I is the moment of inertia and W is the angular velocity in rad/s.
so, replacing values, we get:
K = [tex]\frac{1}{2}(0.42358)(75.398)^2[/tex]
K = 1203.99 J
(a) The moment of inertial of the given system is 0.424 kgm².
(b) The rotational kinetic energy of the system is 1,205.5 J.
The given parameters;
The moment of inertial of the given system is calculated as follows;
[tex]I = \frac{1}{2} m_d R_d^2 \ \ + \ \frac{1}{2}m_a (r_e^2 + r_i^2)\\\\I = \frac{1}{2} (2.5)(0.52)^2 \ + \ \frac{1}{2} (1.1)(0.34^2 + 0.2^2)\\\\I = 0.424 \ kgm^2[/tex]
The rotational kinetic energy of the system is calculated as follows;
[tex]K.E = \frac{1}{2} I \omega ^2\\\\K.E = \frac{1}{2} (0.424) (12 \ \frac{rev}{s} \times \frac{2 \pi \ rad}{1 \ rev} )^2\\\\K.E = 1,205.5 \ J[/tex]
"Your question is not complete, it seems to be missing the following information;"
(a) What is the moment of inertia of the system (in kg · m2)?
(b) What is its rotational kinetic energy (in J)?
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