n=3390 rev/min=3390/60 rev/s
D= 0.52 m R= 0.26 m.
v=ω•R =2•π•n•R= 2• π•339•0.26/6 =92.3 m/s,
ω (fin)= ω - ε•t.
ω (fin)=0,
ε = ω /t = 339/6•3 =18.8 rad/s²
Consider a 52-cm-long lawn mower blade rotating about its center at 3390 rpm.
(a) Calculate the linear speed of the tip of the blade.
(b) If safety regulations require that the blade be stoppable within 3.0 s, what minimum angular acceleration will accomplish this? Assume that the angular acceleration is constant.
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