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296 Dynamics of Mechanical Systems
where n is a unit vector parallel to the support as in Figure 9.8.1. Let the velocity of B
before and after the application of F be V n and Vn, respectively. Then, from the linear
O
impulse–momentum principle, we have:
ˆ
I = ∆ L or F max ( t 2 n ) = mV n − mV O n (9.8.2)
or
V = V + F t 2 m (9.8.3)
ˆ
O max
Example 9.8.2: A Braked Flywheel
As a second example, consider a flywheel W rotating about its axis with an angular speed
as in Figure 9.8.3. Let W be supported by a shaft S, and let S be subjected to a brake that
exerts a moment M about the axis of S and W and whose magnitude M is depicted in
Figure 9.8.4. Specifically, let M(t) be sinusoidal such that:
(
M [ 1 − cos 2π t T)] 0 ≤≤
tT
Mt () = max (9.8.4)
≥
0 tT
Then, from Eq. (9.2.2), the angular impulse about O, the center of W, is:
T T
()
J = ∫ M t dt k = ∫ M max [ 1 − cos (2π t T)] k
O
0 0 (9.8.5)
= M [ t −( T 2π ) sin (2π t T)] T | k = M T k
max
0 max
where k is a unit vector parallel to the shaft as in Figure 9.8.3.
From Eq. (9.4.13) the angular momentum of W about O before and after braking is:
A 0 () = I ω k and A t () = I ω k (9.8.6)
O O O O O
FIGURE 9.8.3 FIGURE 9.8.4
A spinning flywheel. Profile of braking moment magnitude.