Answer :
Answer:
[tex]KE=194.9750\ J[/tex]
Explanation:
Given:
mass of disk, [tex]m=8\ kg[/tex]
radius of the disk, [tex]r=0.55\ m[/tex]
time taken by the disk to complete on rotation, [tex]T=0.35\ s[/tex]
We know that rotational kinetic energy is given as:
[tex]KE=\frac{1}{2}\times I.\omega^2[/tex] ............................(1)
Now the angular speed of the disk:
[tex]\omega=\frac{2\pi}{T}[/tex]
[tex]\omega=\frac{2\pi}{0.35}[/tex]
[tex]\omega=17.952\ rad.s^{-1}[/tex]
The moment of inertia of a disk is given as:
[tex]I=\frac{1}{2} m.r^2[/tex]
[tex]I=0.5\times 8\times 0.55^2[/tex]
[tex]I=1.21\ kg.m^2[/tex]
Now using eq. (1):
[tex]KE=\frac{1}{2} \times 1.21\times 17.952^2[/tex]
[tex]KE=194.9750\ J[/tex]
Answer:
Explanation:
mass of disc, m = 8 kg
radius of disc, r = 0.55 m
Time period, t = 0.35 s
Angular velocity, ω = 2π/T
ω = 2 x 3.14 / 0.35 = 17.94 rad/s
moment of inertia of the disc, I = 0.5 mr²
I = 0.5 x 8 x 0.55 x 0.55 = 1.21 kgm²
The rotational kinetic energy is given by
K = 0.5 x I x ω²
K = 0.5 x 1.21 x 17.94 x 17.94 = 194.72 J