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

Step-by-step 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=(1)/(2)mv^2`

`K=(1)/(2)* 1* 3^2`

K = 4.5 J

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

`K'=(1)/(2)I\omega^2`

`K'=(1)/(2)(mr^2)/(2)((v)/(r))^2`

`K'=(1)/(4)mv^2`

`K'=(1)/(4)* 1* 3^2`

K' = 2.25 J

Hence, this is the required solution.