The distance between the vertex and the centr | vedclass

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

Question

(a)

Refer to figure below,

Consider a differentiable are of angle $d 	heta$.

Area of $	riangle O P Q=\frac{1}{2} 	imes P Q 	imes O P$

Vedclass · Accepted Answer

$\frac{4}{3} R \frac{\sin (	heta / 2)}{	heta}$

Vedclass · Answer

(a)

Refer to figure below,

Consider a differentiable are of angle $d 	heta$.

Area of $	riangle O P Q=\frac{1}{2} 	imes P Q 	imes O P$

$=\frac{1}{2} R d 	heta 	imes R=\frac{1}{2} R^2 d 	heta$

Centre of mass of this $	riangle O P Q$ is at a distance of $\frac{2}{3} R \cos 	heta$ from $O$.

So, position of centre of mass of complete segment is

$\bar{y}=\int y d m / \int d m$

$\bar{y}=\frac{\int \limits_0^{	heta / 2} \frac{2 R}{3} \cos 	heta \cdot ho \frac{r^2}{2} d 	heta}{\int \limits_0^{	heta / 2} ho \frac{r^2}{2} d 	heta}$

where, $ho=$ mass density.

$	herefore \quad \bar{y} =\frac{\frac{2 r}{3} \int \limits_0^{	heta / 2} \cos 	heta d 	heta}{\int \limits_0^{	heta / 2} d 	heta}=2 \frac{R}{3} \cdot \frac{\sin 	heta / 2}{	heta / 2}$

$=\frac{4}{3} R \frac{\sin (	heta / 2)}{	heta}$

The distance between the vertex and the centre of mass of a uniform solid planar circular segment of angular size $\theta$ and radius $R$ is given by

Similar Questions

Three point particles of masses $1.0\; \mathrm{kg} .1 .5 \;\mathrm{kg}$ and $2.5\; kg$ are placed at three comers of a right angle triangle of sides $4.0\; \mathrm{cm}, 3.0 \;\mathrm{cm}$ and $5.0\; \mathrm{cm}$ as shown in the figure. The center of mass of the system is at a point

A uniform rectangular thin sheet $ABCD$ of mass $M$ has length $a$ and breadth $b$, as shown in the figure. If the shaded portion $HBGO$ is cut off, the coordinates of the centre of mass of the remaining portion will be

A uniform thin rod $AB$ of length $L$ has linear mass density $\mu \left( x \right) = a + \frac{{bx}}{L}$ , where $x$ is measured from $A$. If the $CM$ of the rod lies at a distance of $\left( {\frac{7}{12}} \right)L$ from $A$, then $a$ and $b$ are related as

Distance of the centre of mass of a solid uniform cone from its vertex is $z_0$ . If the radius of its base is $R$ and its height is $h$ then $z_0$ is equal to

Obtain an expression for the position vector of centre of mass of a system of $n$ particles in one dimension.