Two semicircular rings of linear mass densiti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the mass of the semicircular ring with density $\lambda$ be $M_1 = \lambda (\pi R) = M$. Then,the mass of the semicircular ring with density $

Vedclass · Accepted Answer

$\frac{R}{\pi}$

Vedclass · Answer

(B) Let the mass of the semicircular ring with density $\lambda$ be $M_1 = \lambda (\pi R) = M$. Then,the mass of the semicircular ring with density $3\lambda$ is $M_2 = 3\lambda (\pi R) = 3M$. The center of mass of a semicircular ring of radius $R$ is at a distance $\frac{2R}{\pi}$ from its center along the axis of symmetry. Let the geometrical center be at the origin $(0,0)$. Place the first ring (mass $M$) in the left half,so its center of mass is at $x_1 = -\frac{2R}{\pi}$. Place the second ring (mass $3M$) in the right half,so its center of mass is at $x_2 = \frac{2R}{\pi}$. The center of mass of the combined system is $X_{cm} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}$. $X_{cm} = \frac{M(-\frac{2R}{\pi}) + 3M(\frac{2R}{\pi})}{M + 3M} = \frac{\frac{4MR}{\pi}}{4M} = \frac{R}{\pi}$. Thus,the distance from the geometrical center is $\frac{R}{\pi}$.

Two semicircular rings of linear mass densities $\lambda$ and $3\lambda$ and of radius $R$ each are joined to form a complete ring. The distance of the center of mass of the complete ring from its geometrical center is

Similar Questions

Find the centre of mass of a triangular lamina.

Two particles of masses $m_1$ and $m_2$ are separated by a distance $d$. The shift in the centre of mass when the two particles are interchanged is:

If the linear density of a rod of length $3 \ m$ varies as $\lambda = 2 + x$,then the position of the centre of mass of the rod is:

$A$ rod of length $3 \; m$ has a linear mass density that is directly proportional to the distance $x$ from one of its ends. Find the position of its centre of gravity from that end. (in $; m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module