A carpenter has constructed a toy as shown in | vedclass

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

Question

(C) Let the density of the cone be $\rho$. Then its mass $m_1$ is given by:$m_1 = \frac{1}{3} \pi (2R)^2 (4R) \rho = \frac{16}{3} \pi R^3 \rho$The cen

Vedclass · Accepted Answer

at a distance of $4R$ from $O$

Vedclass · Answer

(C) Let the density of the cone be $\rho$. Then its mass $m_1$ is given by:$m_1 = \frac{1}{3} \pi (2R)^2 (4R) \rho = \frac{16}{3} \pi R^3 \rho$The centre of mass of the cone is at a height $h/4$ from its base,where $h = 4R$. Thus,$y_1 = \frac{4R}{4} = R$ from $O$.Similarly,the mass of the sphere $m_2$ with radius $R$ and density $12\rho$ is:$m_2 = \frac{4}{3} \pi R^3 (12\rho) = 16 \pi R^3 \rho = 3 m_1$The centre of mass of the sphere is at its centre $O_2$. The distance of $O_2$ from $O$ is $y_2 = 4R + R = 5R$.Taking the line of symmetry as the $y$-axis with the origin at $O$,the centre of mass $Y_{CM}$ of the toy is:$Y_{CM} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{m_1(R) + (3m_1)(5R)}{m_1 + 3m_1} = \frac{16 m_1 R}{4 m_1} = 4R$Thus,the centre of mass of the toy is at a distance of $4R$ from $O$.

$A$ carpenter has constructed a toy as shown in the adjoining figure. If the density of the material of the sphere is $12$ times that of the cone,the position of the centre of mass of the toy is given by:

Similar Questions

The variation of density of a solid cylindrical rod of cross-sectional area $\alpha$ and length $L$ is given by $\rho = \rho_0 \frac{x^2}{L^2}$, where $x$ is the distance from one end of the rod. The position of its centre of mass from that end $(x=0)$ is:

Two point masses of $1.5 \ g$ and $2.5 \ g$ are placed at a distance of $16 \ cm$ from each other. If the center of mass is at a distance $x$ from the $1.5 \ g$ mass,then $x = \dots \ cm$.

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

$A$ straight rod of length $L$ has one of its ends at the origin and the other at $x = L$. If the mass per unit length of the rod is given by $\lambda = Ax$ (where $A$ is a constant),then where is its center of mass from the origin?

$(n - 1)$ equal point masses each of mass $m$ are placed at the vertices of a regular $n$-polygon. The vacant vertex has a position vector $a$ with respect to the centre of the polygon. Find the position vector of the centre of mass.

Vedclass Test Series

Exam Paper Generator

Online Exam Module