Question

A system consists of a disk of mass 2.33 kg and radius 50 cm upon which is mounted an annular cylinder of mass 2.20 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 15.1 rev/s. What is the systems rotational kinetic energy

Answer 1

Kinetic energy of the system = 2547.41 Joules.

Step-by-step explanation:

Given:

Disk:

Mass of the disk (m) =
`2.33` kg

Radius of the disk (r) =
`50` cm =
`(50)/(100) =0.5` m

Cylinder:

Mass of the annular cylinder (M) =
`2.20` kg

Inner radius of the cylinder
`(R_i)` =
`0.2` m

Outer radius of the cylinder
`(R_o)` =
`0.3` m

The angular speed of the system
`(\omega)` =
`15.1` rev/s

Angular speed in in terms of Rad/sec =
`15.1* 2\pi =94.876` rad/sec

Formula to be used:

Rotational Kinetic energy,
`(KE)_r` =
`(I* \omega^2)/(2)`

So, before that we have to work with the moment of inertia (MOI) of the system.

⇒ MOI of the system = MOI of the disk + MOI of the cylinder

⇒ MOI (system) =
`(mr^2)/(2) +(M(R_i+R_o)^2)/(2)`

⇒ MOI (system) =
`(2.33* (0.5)^2)/(2) + (2.20* (0.2+0.3)^2)/(2)`

⇒ MOI (system) =
`0.566` kg.m^2

Now

The rotational Kinetic energy.

⇒
`(KE)_r =(I\omega^2)/(2)`

Plugging the values.

⇒
`(KE)_r=(0.566* (94.876)^2)/(2)`

⇒
`(KE)_r=2547.41` Joules

Then

The kinetic energy of the rotational system is 2547.41 J.