Question

Consider a uniform rod of length L, bent into an uniform are making an angle 0 to the origin. Find the distance of the center of mass from the origin

Answer 1

\frac{2L \sin(\theta/2)}{\theta^2}

Answer 2

Let the uniform rod of length L be bent into a uniform circular arc of radius R. The arc subtends an angle $\theta$ at the origin, which is the center of the circle. The length of the arc is related to the radius and the angle by $L = R\theta$. Thus, the radius of the arc is $R = L/\theta$.

To find the center of mass of the arc, we can place the arc in a coordinate system. Let the origin be at the center of the circle. Due to the symmetry of the uniform arc, the center of mass must lie on the angle bisector of the arc. Let's align the angle bisector with the x-axis. The arc then extends from angle $-\theta/2$ to $+\theta/2$ in polar coordinates.

Consider a small element of the arc of length $dl$ at an angle $\phi$ from the x-axis. The mass of this element is $dm = \lambda dl$, where $\lambda$ is the linear mass density of the rod. Since the rod is uniform, $\lambda = M/L$, where M is the total mass. In polar coordinates, $dl = R d\phi$. So $dm = \lambda R d\phi$. The coordinates of this element are $x = R \cos\phi$ and $y = R \sin\phi$.

The coordinates of the center of mass $(x_{CM}, y_{CM})$ are given by: $x_{CM} = \frac{\int x dm}{\int dm}$ and $y_{CM} = \frac{\int y dm}{\int dm}$ The total mass is $M = \int dm = \int_{-\theta/2}^{\theta/2} \lambda R d\phi = \lambda R [\phi]_{-\theta/2}^{\theta/2} = \lambda R \theta$. Since $L = R\theta$, $M = \lambda L$, which is consistent.

Now, let's calculate the numerator for $x_{CM}$: $\int x dm = \int_{-\theta/2}^{\theta/2} (R \cos\phi) (\lambda R d\phi) = \lambda R^2 \int_{-\theta/2}^{\theta/2} \cos\phi d\phi$ $\int_{-\theta/2}^{\theta/2} \cos\phi d\phi = [\sin\phi]_{-\theta/2}^{\theta/2} = \sin(\theta/2) - \sin(-\theta/2) = \sin(\theta/2) - (-\sin(\theta/2)) = 2\sin(\theta/2)$. So, $\int x dm = \lambda R^2 (2\sin(\theta/2))$.

Now, calculate $x_{CM}$: $x_{CM} = \frac{\int x dm}{M} = \frac{\lambda R^2 (2\sin(\theta/2))}{\lambda R\theta} = \frac{R (2\sin(\theta/2))}{\theta} = \frac{R \sin(\theta/2)}{\theta/2}$.

Next, let's calculate the numerator for $y_{CM}$: $\int y dm = \int_{-\theta/2}^{\theta/2} (R \sin\phi) (\lambda R d\phi) = \lambda R^2 \int_{-\theta/2}^{\theta/2} \sin\phi d\phi$ $\int_{-\theta/2}^{\theta/2} \sin\phi d\phi = [-\cos\phi]_{-\theta/2}^{\theta/2} = -\cos(\theta/2) - (-\cos(-\theta/2)) = -\cos(\theta/2) + \cos(\theta/2) = 0$. So, $\int y dm = 0$.

Now, calculate $y_{CM}$: $y_{CM} = \frac{\int y dm}{M} = \frac{0}{M} = 0$.

The center of mass is located at $(x_{CM}, y_{CM}) = \left(\frac{R \sin(\theta/2)}{\theta/2}, 0\right)$. The origin is at $(0,0)$. The distance of the center of mass from the origin is $\sqrt{x_{CM}^2 + y_{CM}^2} = \sqrt{\left(\frac{R \sin(\theta/2)}{\theta/2}\right)^2 + 0^2} = \left|\frac{R \sin(\theta/2)}{\theta/2}\right|$. Since $\theta$ is the angle subtended by the arc, $0 < \theta \le 2\pi$. For this range, $\theta/2 \in (0, \pi]$. $\sin(\theta/2) \ge 0$ for $\theta/2 \in (0, \pi]$, and $\theta/2 > 0$. Also $R>0$. So $\frac{R \sin(\theta/2)}{\theta/2} \ge 0$. The distance is $d = \frac{R \sin(\theta/2)}{\theta/2}$.

The question provides the length L and the angle $\theta$. We need to express the distance in terms of L and $\theta$. We use the relation $R = L/\theta$. Substitute R into the distance formula: $d = \frac{(L/\theta) \sin(\theta/2)}{\theta/2} = \frac{L \sin(\theta/2)}{\theta (\theta/2)} = \frac{L \sin(\theta/2)}{\theta^2/2} = \frac{2L \sin(\theta/2)}{\theta^2}$.

The distance of the center of mass from the origin is $\frac{2L \sin(\theta/2)}{\theta^2}$.