Question 9.13: The sliding ladder A uniform ladder of length 2a is supporte...
The sliding ladder
A uniform ladder of length 2a is supported by a smooth horizontal floor and leans against a smooth vertical wall.* The ladder is released from rest in a position making an angle of 60° with the downward vertical. Find the energy conservation equation for the ladder.
Let θ be the angle that the ladder makes with the downward vertical after time t. The (x,z)-coordinates of the centre of mass G are then given by
X=a \sin \theta, \quad Z=a \cos \theta,
and the corresponding velocity components by
\dot{X}=(a \cos \theta) \dot{\theta}, \quad \dot{Z}=-(a \sin \theta) \dot{\theta}.
The angular velocity of the ladder at time t is simply \dot{\theta} (see Figure 9.11). The kinetic energy of the ladder is therefore given by
T=\frac{1}{2} M\left(\dot{X}^{2}+\dot{Y}^{2}\right)+\frac{1}{2} I \dot{\theta}^{2}=\frac{1}{2} M a^{2} \dot{\theta}^{2}+\frac{1}{2} I \dot{\theta}^{2},
where M is the mass of the ladder and I is its moment of inertia about the horizontal axis through G. From the Appendix, we find that I = Ma²/3 so that the kinetic energy of the ladder is given by T=\left(2 M a^{2} / 3\right) \dot{\theta}^{2} .
The gravitational potential energy of the ladder is given by V = MgZ = Mga cos θ.
We must now dispose of the constraint forces. The reaction forces that the smooth floor and wall exert on the ladder are both perpendicular to the particles of the ladder on which they act. These reaction forces therefore do no work. Also, the internal forces that keep the ladder rigid do no work in total. Hence the constraint forces do no work in total.
Conservation of energy therefore applies in the form
\frac{2}{3} M a^{2} \dot{\theta}^{2}+M g a \cos \theta=E,
where E is the total energy. From the initial conditions \dot{\theta}=0 and θ = π/3 when t = 0, it follows that E=\frac{1}{2} M g a . The energy conservation equation for the ladder is therefore
\dot{\theta}^{2}=\frac{3 g}{4 a}(1-2 \cos \theta).
Since the system has only one degree of freedom, this equation is sufficient to determine the motion.
A curious feature of this problem (not proved here) is that the ladder does not maintain contact with the wall all the way down, but leaves the wall when θ becomes equal to \cos ^{-1}(1 / 3) \approx 71^{\circ} .
