Question

show answer No Attempt 50% Part (b) Now consider a disk of mass 8 kg with radius 0.55 m. Disk rotates around itself once every 0.35 sec. What is the rotational energy of this disk?

Answer 1

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

Answer 2

Answer:

$KE=194.9750\ J$

Explanation:

Given:

mass of disk, $m=8\ kg$

radius of the disk, $r=0.55\ m$

time taken by the disk to complete on rotation, $T=0.35\ s$

We know that rotational kinetic energy is given as:

$KE=\frac{1}{2}\times I.\omega^2$ ............................(1)

Now the angular speed of the disk:

$\omega=\frac{2\pi}{T}$

$\omega=\frac{2\pi}{0.35}$

$\omega=17.952\ rad.s^{-1}$

The moment of inertia of a disk is given as:

$I=\frac{1}{2} m.r^2$

$I=0.5\times 8\times 0.55^2$

$I=1.21\ kg.m^2$

Now using eq. (1):

$KE=\frac{1}{2} \times 1.21\times 17.952^2$

$KE=194.9750\ J$