1. **Problem Statement:** We want to sketch the graph of the time period $T$ of a uniform rod pendulum as a function of the distance $x$ from the rod's center of mass to the pivot point, considering both cases where the pivot is above or below the center of mass. 2. **Formula for Time Period:** The time period for small oscillations of the rod about the pivot is given by $$T = 2\pi \sqrt{\frac{I_x}{Mg d}}$$ where: - $I_x$ is the moment of inertia of the rod about the pivot point, - $M$ is the mass of the rod, - $g$ is the acceleration due to gravity, - $d$ is the distance from the pivot to the center of mass, which equals $|x|$. 3. **Moment of Inertia $I_x$:** Using the parallel axis theorem, $$I_x = I_{cm} + M x^2 = \frac{1}{12} M L^2 + M x^2$$ where $I_{cm} = \frac{1}{12} M L^2$ is the moment of inertia about the center of mass. 4. **Substitute $I_x$ and $d$ into $T$:** $$T = 2\pi \sqrt{\frac{\frac{1}{12} M L^2 + M x^2}{M g |x|}} = 2\pi \sqrt{\frac{\frac{1}{12} L^2 + x^2}{g |x|}}$$ 5. **Behavior of $T$ as a function of $x$:** - For $x > 0$ (pivot below center of mass), $T$ is defined and positive. - For $x < 0$ (pivot above center of mass), $T$ is also defined because of the absolute value in the denominator. - As $x \to 0$, $T \to \infty$ because the denominator approaches zero. - For large $|x|$, $T \approx 2\pi \sqrt{\frac{x^2}{g |x|}} = 2\pi \sqrt{\frac{|x|}{g}}$, so $T$ grows roughly like $\sqrt{|x|}$. 6. **Sketching the Graph:** - Draw the $x$-axis representing the distance $x$ from the center of mass (negative to the left, positive to the right). - Draw the $T$-axis vertically. - At $x=0$, draw a vertical asymptote since $T$ tends to infinity. - For $x>0$, plot $T$ starting from very large near zero, decreasing to a minimum, then increasing slowly. - For $x<0$, do the same symmetric behavior because of the absolute value. - The graph is symmetric about $x=0$ but with a vertical asymptote there. This qualitative sketch helps visualize how the time period changes with pivot position. **Final note:** The graph is continuous for all $x \neq 0$ and diverges at $x=0$.