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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 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Respuesta :

MissPhiladelphia
The formula for the rotational kinetic energy is

[tex]KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2} [/tex]

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

[tex]I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2} [/tex]

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

[tex]KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}[/tex]

[tex]KE_{rot} =18,671.31 \ J[/tex]

crab0001

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> [tex]573*2\pi *\frac{1}{60} =60[/tex]rad/s

I=[tex]\frac{1}{3} mL^2=\frac{1}{3} (2.67kg)(0.777m)^2=0.537[/tex]kgm^2

K=1/2(number of rods)(I)(ω)=[tex]\frac{1}{2} *(5)(0.537)(60)^2=4833[/tex]J

I know it's very late, but hope this helps anyone else trying to find the answer.

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