Respuesta :
Answer:
(a) 20,154.1 J
(b) 95,223.5 J
Explanation:
The expression for the moment of inertia for the uniform solid cylinder is as follows;
[tex]I= \frac{1}{2}mr^{2}[/tex]
Here, I is the moment of inertia, r is the radius and m is the mass of the object.
The expression for the rotational kinetic energy is as follows;
[tex]K= \frac{1}{2}I\omega ^{2}[/tex]
Here, K is the rotational kinetic energy and [tex]\omega[/tex] is the angular velocity.
(a)
Calculate the moment of inertia of the smaller solid cylinder.
[tex]I= \frac{1}{2}mr^{2}[/tex]
Put m= 3.44 kg and r= 0.356 m.
[tex]I= \frac{1}{2}(3.44)(0.356)^{2}[/tex]
[tex]I= 0.218 kg m^{2}[/tex]
Calculate the rotational kinetic energy of the smaller cylinder.
[tex]K= \frac{1}{2}I\omega ^{2}[/tex]
Put [tex]I= 0.218 kg m^{2}[/tex] and [tex]\omega = 430 rads^{-1}[/tex].
[tex]K= \frac{1}{2}(0.218)(430) ^{2}[/tex]
K= 20,154.1 J
Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.
(b)
Calculate the moment of inertia of the larger solid cylinder.
[tex]I= \frac{1}{2}mr^{2}[/tex]
Put m= 3.44 kg and r= 0.775 m.
[tex]I= \frac{1}{2}(3.44)(0.775)^{2}[/tex]
[tex]I= 1.03 kg m^{2}[/tex]
Calculate the rotational kinetic energy of the smaller cylinder.
[tex]K= \frac{1}{2}I\omega ^{2}[/tex]
Put [tex]I= 1.03 kg m^{2}[/tex] and [tex]\omega = 430 rads^{-1}[/tex].
[tex]K= \frac{1}{2}(1.03)(430) ^{2}[/tex]
K= 95,223.5 J
Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.
