Use the rotational energy equation, E = 0.5Iω², where I is the moment of inertia for a long thin rod with an axis through it's midpoint (I = (1/12)mL²).
E = 0.5Iω²
E = 0.5 [1/12)mL²] ω²
E = (1/24) mL² ω²
Find ω by multiplying the rpm by 2π and then dividing by 60 to put units into rad/sec (or multiply ω by π/30)
3450 * 2π = 21 676.9... rad/min
(21 676.9... rad/min)/60 = 361.28... rad/sec
Insert into energy eqn and solve
E = (1/24) mL² (361.28...rad/s)²
E = (1/24)(0.65kg)(0.55m²)(361.28...rad/s)²
E = ~1069 J
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65kg and its length is 0.55m. What is the rotational energy of the blade at its operating angular speed of 3450 rpm? If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
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