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).
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