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A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.715 m0.715 m and 2.51 kg2.51 kg , respectively. When the propellor rotates at 527 rpm527 rpm (revolutions per minute), what is its rotational kinetic energy?

Respuesta :

Answer:

3260.33 J

Explanation:

[tex]n [/tex]  = number of rods = 5

[tex]L [/tex]  = length of each rod = 0.715 m

[tex]m [/tex]  = mass of rod = 2.51 kg

[tex]I [/tex]  = total moment of inertia

Total moment of inertia is given as

[tex]I = \frac{nmL^{2}}{3}[/tex]

[tex]I = \frac{(5)(2.51)(0.715)^{2}}{3}[/tex]

[tex]I [/tex] = 2.14 kgm²

[tex]w [/tex]  = angular speed = 527 rpm = 55.2 rad/s

Rotational kinetic energy is given as

E = (0.5) [tex]I [/tex]  ([tex]w [/tex] )²

E = (0.5) (2.14) (55.2)²

E = 3260.33 J

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