High School

Two uniform solid cylinders, each rotating about its central (longitudinal) axis at 235 rad/s, have the same mass of 1.25 kg but differ in radius. What is the rotational kinetic energy of each cylinder?

Answer :

The rotational kinetic of the two rotating solid cylinders is 44.2 (r₁²+r₂²). The result is obtained by inserting total inertia value in rotational kinetic equation.

How to get moment of inertia and rotational kinetic value?

The moment of inertia of a solid cylinder can be expressed as

[tex]I = \frac{1}{2} mr^{2}[/tex]

Where

  • I = moment of inertia (kg.m²)
  • m = mass (kg)
  • r = radius (m)

If there are two of more cylinders, the total moment of inertia would be

[tex]I = I_{1} + I_{2} + ...[/tex]

The kinetic energy of a rotating body can be expressed as

[tex]E_{rot} = \frac{1}{2} I \omega^{2}[/tex]
Where

  • Erot = rotational kinetic energy (Joule)
  • I = moment of inertia (kg.m²)
  • ω = angular velocity (rad/s)

If given

  • ω = 235 rad/s
  • m₁ = m₂ = m = 1.25 kg

What is the rotational kinetic energy?

We first calculate the moment of inertia of the two cylinders.

[tex]I = \frac{1}{2} m_{1} r_{1}^{2} + \frac{1}{2} m_{2}r_{2}^{2}[/tex]

[tex]I = \frac{1}{2} m r_{1}^{2} + \frac{1}{2} mr_{2}^{2}[/tex]

[tex]I = \frac{1}{2} m (r_{1}^{2} + r_{2}^{2})[/tex]

[tex]I = \frac{1}{2} \times 1,25 (r_{1}^{2} + r_{2}^{2})[/tex]

[tex]I = 0,625 (r_{1}^{2} + r_{2}^{2})[/tex]

Then, the rotational kinetic energy is

[tex]E_{rot} = \frac{1}{2} I \omega^{2}[/tex]

[tex]E_{rot} = \frac{1}{2} \times 0,625 (r_{1}^{2} + r_{2}^{2}) \times \omega^{2}[/tex]

[tex]E_{rot} = \frac{1}{2} \times 0,625 (r_{1}^{2} + r_{2}^{2}) \times 235^{2}[/tex]

[tex]E_{rot} = 117.5 \times 0.625 (r_{1}^{2} + r_{2}^{2}) \times 235[/tex]

[tex]E_{rot} = 44.2 (r_{1}^{2} + r_{2}^{2}) Joule[/tex]

Hence, the rotational kinetic energy of the two rotating solid cylinders is 44.2 (r₁²+r₂²).

Learn more about rotational kinetic energy here:

https://brainly.com/question/20261989

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