Page 353 - Dynamics of Mechanical Systems

0593_C10_fm Page 334 Monday, May 6, 2002 2:57 PM

334 Dynamics of Mechanical Systems

O

T
θ

P

P(m) mg

FIGURE 10.8.1 FIGURE 10.8.2
A simple pendulum. Free body diagram of the pendulum bob.
Because the pendulum is released from rest its initial kinetic energy is zero. Its kinetic
energy K as it passes through the equilibrium position may be expressed as:

˙
K = 1 mv = 1 m( ) 2 (10.8.2)
lθ
2
2 2
From the work–energy principle of Eq. (10.6.5), we then have:

W = ∆ K or mg = 1 mv = 1 mlθ 2 ˙ (10.8.3)
2
l
2 2
or

v = mgl (10.8.4)

Observe that this speed is the same as that of an object freely falling through a distance
(see Eq. (10.7.4)), even though the direction of the velocity is different.
As a generalization of this example, consider a pendulum released from rest at an angle
θ with the objective of determining the speed of P when it falls to an angle θ , as in Figure
i
f
10.8.3. The work W done by gravity as the pendulum falls from θ to θ is:
f
i
i
f
(
∆
W = mg h = mgl cosθ −cosθ i) (10.8.5)
i f f
where ∆h is the change in elevation of P as the pendulum falls (see Eq. (10.2.20)).
The kinetic energies K and K of P when θ is θ and θ are:
i
f
f
i
2
K = 12 m ( lθ ˙ i) = 0 and K = 1 m lθ ˙ f ( ) 2 (10.8.6)
i f
2
˙
θ
˙
θ
θ
˙
where and are the values of when θ is θ and θ , respectively.
i
f
f
i
The work–energy principle then leads to:
W = ∆ K = K − K (10.8.7)
i f f i