Question

A 1.00 kg solid, uniform disk rolls without slipping across a level surface, translating at 3.00 m/s. If the disk's radius is 0.390 m, find the following. (a) the disk's translational kinetic energy (in J) (b) the disk's rotational kinetic energy (in J)

Answer 1

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 :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 1\times 3^2$

K = 4.5 J

(b) Let K' is the rotational kinetic energy of the disk. Its formula is given by :

$K'=\dfrac{1}{2}I\omega^2$

$K'=\dfrac{1}{2}\dfrac{mr^2}{2}(\dfrac{v}{r})^2$

$K'=\dfrac{1}{4}mv^2$

$K'=\dfrac{1}{4}\times 1\times 3^2$

K' = 2.25 J

Hence, this is the required solution.