Respuesta :
Answer:
Kinetic energy of the system = 2547.41 Joules.
Explanation:
Given:
Disk:
Mass of the disk (m) = [tex]2.33[/tex] kg
Radius of the disk (r) = [tex]50[/tex] cm = [tex]\frac{50}{100} =0.5[/tex] m
Cylinder:
Mass of the annular cylinder (M) = [tex]2.20[/tex] kg
Inner radius of the cylinder [tex](R_i)[/tex] = [tex]0.2[/tex] m
Outer radius of the cylinder [tex](R_o)[/tex] = [tex]0.3[/tex] m
The angular speed of the system [tex](\omega)[/tex] = [tex]15.1[/tex] rev/s
Angular speed in in terms of Rad/sec = [tex]15.1\times 2\pi =94.876[/tex] rad/sec
Formula to be used:
Rotational Kinetic energy, [tex](KE)_r[/tex] = [tex]\frac{I\times \omega^2}{2}[/tex]
So, before that we have to work with the moment of inertia (MOI) of the system.
⇒ MOI of the system = MOI of the disk + MOI of the cylinder
⇒ MOI (system) = [tex]\frac{mr^2}{2} +\frac{M(R_i+R_o)^2}{2}[/tex]
⇒ MOI (system) = [tex]\frac{2.33\times (0.5)^2}{2} + \frac{2.20\times (0.2+0.3)^2}{2}[/tex]
⇒ MOI (system) = [tex]0.566[/tex] kg.m^2
Now
The rotational Kinetic energy.
⇒ [tex](KE)_r =\frac{I\omega^2}{2}[/tex]
Plugging the values.
⇒ [tex](KE)_r=\frac{0.566\times (94.876)^2}{2}[/tex]
⇒ [tex](KE)_r=2547.41[/tex] Joules
Then
The kinetic energy of the rotational system is 2547.41 J.
