p11_071.pdf

71. The Hint given in the problem would make the computation in part (a) very straightforward (without

doing the integration as we show here), but we present this further level of detail in case that hint is not

obvious or – simply – in case one wishes to see how the calculus supports our intuition.

(a) The (centripetal) force exerted on an inﬁnitesimal portion of the blade with mass dm located a

distance r from the rotational axis is (Newton’s second law) dF =( dm ) ω 2 r ,where dm can be

written as ( M/L ) dr and the angular speed is ω = (320)(2 π/ 60) = 33 . 5 rad/s. Thus for the entire

blade of mass M and length L the total force is given by

F =

dF =

ω 2 rdm

ω 2 rdr

Mω 2 r 2

2 L

= Mω 2 L

(110kg)(33 . 5rad / s) 2 (7 . 80m)

=4 . 8

10 5 N .

(b) About its center of mass, the blade has I = ML 2 / 12 according to Table 11-2(e), and using the

parallel-axis theorem to “move” the axis of rotation to its end-point, we ﬁnd the rotational inertia

becomes I = ML 2 / 3. Using Eq. 11-37, the torque (assumed constant) is

τ = Iα

3 ML 2 ∆ ω

∆ t

3 (110kg)(7 . 8m) 2 33 . 5rad / s

6 . 7s

=1 . 1

10 4 N

m .

W =∆ K = 1

2 Iω 2

−

3 ML 2 ω 2

6 (110kg)(7 . 80m) 2 (33 . 5rad / s) 2

=1 . 3

10 6 J .

= 1