a) linear speed v = w*r (where w is the angular speed in radians/sec and r is the radial distance from the center)
Here, w = 2990 rpm = 49.83 rev.per sec.
= 49.83*2*pi rad/sec
=313.11 rad/sec
& r = 0.59/2 = 0.295 m
So, v = 313.11*0.295 = 92.4m/s
b) w(final) = w(initial) + alpha*t
0 = 313.11 + alpha*3
alpha = -313.11/3 = -104rad/s^2
Consider a 59-cm-long lawn mower blade rotating about its center at 2990 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.
The answers are a) 92.4 m/s and b) -104 rad/s^2. How do I get these answers? Thank you in advance.
2 answers
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
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
1 answer