A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.74 k

Question

A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.74 kg and its length is 0.56 m.
(a) What is the rotational energy of the blade at its operating angular speed of 3500 rpm? kJ
(b) If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise? m

Answer 1

the moment of inertia of rod (m. L) about an axis through centre of mass = I = (1/12) m L^2

in 1 rotation angle turned = 2 pi
in 3500 rots = 2 pi*3500

this is in 60 secs
angular speed w= 2 pi 3500/60

Rot. KE = 0.5* I * w^2
= 0.5 [(1/12) 0.74*0.56*0.56] [2 pi 3500/60]^2

= 0.5 [0.0193] [366.33]^2
=1295.03 joule

b) if rotational KE >>>> PE gravitational

if this rotating blade is raised to height (h) against gravity then it will do work at the cost of its rot-KE so its speed will diminsh.

the height (h) at which KE is converted fully in PE

m g h = 1295.03

h = 1295.03 / 0.74 * 9.8 = 178.57 meters above its rotating position (origin)