Question

Consider a 62.7-cm long lawn mower blade rotating about its center at 4070 rpm.

a) Calculate the linear speed of the tip of the blade.

b) If safety regulations require that the blade be stoppable within 3.33 s, what minimum angular acceleration will accomplish this? Assume that the angular acceleration is constant.

Answer 1

Given that :

angular speed of blade, = 4070 rpm

rpm convert into rad/s

= [(4070) x (2 / 60)] rad/s 425.9 rad/s

length of the lawn mower blade, L = 62.7 cm

radius of the lawn mower blade, r = (62.7 cm) / 2 0.3135 m

(a) The linear speed of the tip of the blade which will be given as :

using an equation,   v = r                                                    { eq.1 }

inserting the values in above eq.

v = ( 425.9 rad/s) ( 0.3135 m)

v = 133.5 m/s

(b) If safety regulations require that the blade be stoppable within 3.33 s, then the minimum angular acceleration will be given as :

using equation of rotational motion,

_f = ₀ + t                                                              { eq.2 }

where, _f = final angular speed = 0 rad/s

inserting the values in eq.2,

(0 rad/s) = (425.9 rad/s) + (3.33 s)

= - (425.9 rad/s) / (3.33 s)

= -127.9 rad/s²