Explanation:
Given that,
Mass of the disk, m = 1 kg
Translational speed of the disk, v = 3 m/s
Radius of the disk, r = 0.39 m
(a) Let K is the tranlational kinetic energy of the disk. Its formula is given by :
[tex]K=\dfrac{1}{2}mv^2[/tex]
[tex]K=\dfrac{1}{2}\times 1\times 3^2[/tex]
K = 4.5 J
(b) Let K' is the rotational kinetic energy of the disk. Its formula is given by :
[tex]K'=\dfrac{1}{2}I\omega^2[/tex]
[tex]K'=\dfrac{1}{2}\dfrac{mr^2}{2}(\dfrac{v}{r})^2[/tex]
[tex]K'=\dfrac{1}{4}mv^2[/tex]
[tex]K'=\dfrac{1}{4}\times 1\times 3^2[/tex]
K' = 2.25 J
Hence, this is the required solution.
