### Chapter Nine#### Problem 9.11For the sliding rod in Problem 9.10, find the angle \( \theta \) at which the rod loses contact with the wall, assuming the rod starts sliding with \( \theta(0) = 0 \) and \( \dot{\theta}(0) \) small.#### Explanation of DiagramThe diagram shows a thin rigid cylindrical rod leaned against a vertical wall. The rod is inclined at an angle \( \theta \) with the wall. The length of the rod is labeled as \( l \). The rod is depicted as resting on a horizontal floor and leaning against a vertical wall, forming a right-angled triangle with the floor and wall. The problem involves analyzing the motion of the rod as it slides down without friction under the influence of gravity.

Given:From figure x=l2cosθx˙=dxdt=-l2θ˙sinθx¨=-l2θ¨sinθ-l2θ˙2cosθ When the rod losses the contact with the wall there is no sliding motion along x-direction and therefore, x¨=0-l2θ¨sinθ-l2θ˙2cosθ=0θ¨sinθ+θ˙2cosθ=0 ..... 1 From the law of conservation of energy. Change in potential energy is equal to kinetic energy. PE=KEwl2cosθ0-wl2cosθ=12MIθ˙2Here, MI is moment of inertia of rod, and θ0is the initial angle.wl2cosθ0-wl2cosθ=1wl212g+wl22θ˙2wl2cosθ0-wl2cosθ=1wl212g+3wl212gθ˙2wl2cosθ0-wl2cosθ=13θ˙2θ˙=3glcosθ0-cosθ from initial condition θ0=0θ˙=3gl1-cosθ ......(2). Now, when the rod leaves the horizontal velocity is zero and the change in it with respect to angle is zero. x˙=-l2θ˙cosθ=-l23gl1-cosθ cosθdx˙dθ=0ddθl23gl1-cosθ cosθ=0ddθ1-cosθ cosθ=0sinθ21-cosθ-1-cosθsinθ=0cosθ=23θ=48.19o

Answered: Figure 9.30 Problem 9.10. of gravity.…