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kkurt [141]
3 years ago
5

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

7 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?
Physics
1 answer:
Vikki [24]3 years ago
6 0
The formula for the rotational kinetic energy is

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

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.

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

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

Then,

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

KE_{rot} =18,671.31 \ J

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Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

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From this expression and using the given values we obtain the following:

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3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

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Explanation:

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