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.

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

To solve this problem, we'll need to use some formulas and conversions. We'll start with part (a) and then move on to part (b).

(a) To calculate the linear speed of the tip of the blade, we can use the formula for linear speed (v) given the radius (r) and angular velocity (ω):
v = r * ω

Given:
- Radius (r) of the blade = 52 cm = 0.52 m
- Angular velocity (ω) = 3390 revolutions per minute (rpm)

First, we need to convert the angular velocity from rpm to radians per second (rad/s) because the formula requires angular velocity in radians per second.
To convert rpm to rad/s, we'll use the conversion factor: 1 revolution = 2π radians

ω = 3390 rpm * (2π rad/1 revolution) * (1 min/60 s)
= 356.7π rad/s (approximately)

Now, we can substitute the values into the formula for linear speed:
v = (0.52 m) * (356.7π rad/s)
≈ 588.9 m/s

Therefore, the linear speed of the tip of the blade is approximately 588.9 m/s.

(b) To find the minimum angular acceleration (α) that will stop the blade within 3.0 seconds, we can use the formula:
ω = ω₀ + αt

Given:
- Initial angular velocity (ω₀) = 356.7π rad/s
- Time (t) = 3.0 s

We're trying to find α, so we rearrange the formula:
α = (ω - ω₀) / t

Substituting the given values:
α = (0 rad/s - 356.7π rad/s) / 3.0 s
= -356.7π rad/s² / 3.0 s
= -118.9π rad/s²

Therefore, the minimum angular acceleration that will stop the blade within 3.0 seconds is approximately -118.9π rad/s². The negative sign indicates that the blade is decelerating.