Question 9.7: Stability of a plank on a log A uniform thin rigid plank is ...

Suppose that the plank is disturbed from its equilibrium position and is tilted by an angle θ as shown in Figure 9.6. The plank is known to roll on the log, which means that the distance GC from the centre G of the plank to the contact point C must always be equal to the arc length of the log that has been traversed. If the radius of the log is a, then this arc length is aθ.

We are not yet able to calculate the kinetic energy of the plank in terms of the coordinate θ. This is done in the next section. However, we do not need it to investigate stability.

The only contribution to the potential energy of the plank comes from uniform gravity. This is given by V = MgZ, where Z is the vertical displacement of the centre of mass G of the plank. Elementary trigonometry (see Figure 9.6) shows that Z = a cos θ + aθ sin θ − a, so that

V=M g a(\cos \theta+\theta \sin \theta-1)

We must now show that the constraint forces do no work. The rate of working of the constraint force R that the log exerts on the plank is R \cdot v ^{C}, \text { where } v ^{C} is the velocity of the particle C of the plank that is instantaneously in contact with the log.
But, since the plank rolls on the log, v ^{C}= 0 so that the rate of working of R is zero.
Also, the internal constraint forces that enforce the rigidity of the plank do no work in total. Hence, the constraint forces do no work in total.
Energy conservation therefore applies in the form

T+M g a(\cos \theta+\theta \sin \theta-1)=E.

It follows that the equilibrium position (with the plank on top of the log) will be stable if V has a minimum at θ = 0. Now V′ = Mgaθ cos θ and V^{\prime \prime} = Mga(cos θ −θ sin θ ) so that, when θ = 0, V′ = 0 and V^{\prime \prime}=1 .. Hence V has a minimum at θ = 0 and so the equilibrium position is stable.