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3 years ago

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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 inne

r 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

1 answer:

Gekata [30.6K]3 years ago

5 0

Answer:

Kinetic energy of the system = 2547.41 Joules.

Explanation:

Given:

Disk:

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

Radius of the disk (r) = $50$ cm = $\frac{50}{100} =0.5$ 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 = $15.1\times 2\pi =94.876$ rad/sec

Formula to be used:

Rotational Kinetic energy, $(KE)_r$ = $\frac{I\times \omega^2}{2}$

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) = $\frac{mr^2}{2} +\frac{M(R_i+R_o)^2}{2}$

⇒ MOI (system) = $\frac{2.33\times (0.5)^2}{2} + \frac{2.20\times (0.2+0.3)^2}{2}$

⇒ MOI (system) = $0.566$ kg.m^2

Now

The rotational Kinetic energy.

⇒ $(KE)_r =\frac{I\omega^2}{2}$

Plugging the values.

⇒ $(KE)_r=\frac{0.566\times (94.876)^2}{2}$

⇒ $(KE)_r=2547.41$ Joules

Then

The kinetic energy of the rotational system is 2547.41 J.

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