Question

A system consists of a disk of mass 2.5 kg and radius 52 cm upon which is mounted an annular cylinder of mass 1.1 kg with inner radius 20 cm and outer radius 34 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 12 rev/s.

Answer 1

a) I = 0.42358 Kg*m^2

b) K = 1203.99 J

Step-by-step explanation:

a) The moment of inertia is:

I =
`(1)/(2)M_dR_d^2+(1)/(2)M_a(R_e^2+R_i^2)`

where
`M_d` is the mass of the disk,
`R_d` is the radius of the disk,
`M_a` the mass of the annular cylinder,
`R_e` is the outer radius of the cylinder and
`R_i` the inner radius of the cylinder.

Replacing values, we get:

I =
`(1)/(2)(2.5kg)(0.52m)^2+(1)/(2)(1.1kg)(0.34m^2+0.2m^2)`

I = 0.42358 Kg*m^2

b) First, we will change the angular velocity from rev/s to rad/s as:

W = 12*2
`\pi`rad/s

W = 75.398 rad/s

Also the kinetic energy K is:

K =
`(1)/(2)IW^2`

Where I is the moment of inertia and W is the angular velocity in rad/s.

so, replacing values, we get:

K =
`(1)/(2)(0.42358)(75.398)^2`

K = 1203.99 J

Answer 2

(a) The moment of inertial of the given system is 0.424 kgm².

(b) The rotational kinetic energy of the system is 1,205.5 J.

The given parameters;

• mass of the disk,
  `m_d` = 2.5 kg
• radius of the disk,
  `R_d` = 52 cm
• outer annular radius,
  `r_e` = 34 cm
• inner annular radius,
  `r_i` = 20 cm

The moment of inertial of the given system is calculated as follows;

`I = (1)/(2) m_d R_d^2 \ \ + \ (1)/(2)m_a (r_e^2 + r_i^2)\\\\I = (1)/(2) (2.5)(0.52)^2 \ + \ (1)/(2) (1.1)(0.34^2 + 0.2^2)\\\\I = 0.424 \ kgm^2`

The rotational kinetic energy of the system is calculated as follows;

`K.E = (1)/(2) I \omega ^2\\\\K.E = (1)/(2) (0.424) (12 \ (rev)/(s) * (2 \pi \ rad)/(1 \ rev) )^2\\\\K.E = 1,205.5 \ J`

"Your question is not complete, it seems to be missing the following information;"

(a) What is the moment of inertia of the system (in kg · m2)?

(b) What is its rotational kinetic energy (in J)?

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