A solid cone is placed on a horizontal surfac | vedclass

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

Question

(D) The gravitational potential energy of a body is given by $U = mgh_{cm}$,where $h_{cm}$ is the height of the center of mass from the surface.In fig

Vedclass · Accepted Answer

$\frac{h}{R} = 2\sqrt 2$

Vedclass · Answer

(D) The gravitational potential energy of a body is given by $U = mgh_{cm}$,where $h_{cm}$ is the height of the center of mass from the surface.In figure $(A)$,the cone is standing on its base. The center of mass of a solid cone is at a height of $\frac{h}{4}$ from the base. Thus,$U_A = mg \left( \frac{h}{4} ight)$.In figure $(B)$,the cone is lying on its slant side. The center of mass is at a distance of $\frac{3h}{4}$ from the apex. Let $C$ be the center of mass. In the right-angled triangle formed by the apex $O$,the center of mass $C$,and the projection $M$ on the base,the height of the center of mass from the surface is $h_{cm} = \left( \frac{3h}{4} ight) \sin 	heta$.Since the potential energy does not change,$U_A = U_B$,so $\frac{h}{4} = \left( \frac{3h}{4} ight) \sin 	heta$.This gives $\sin 	heta = \frac{1}{3}$.We know that $	an 	heta = \frac{R}{h}$. Since $\sin 	heta = \frac{1}{3}$,the opposite side is $1$ and the hypotenuse is $3$. The adjacent side is $\sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$.Therefore,$	an 	heta = \frac{1}{2\sqrt{2}}$.Since $	an 	heta = \frac{R}{h}$,we have $\frac{R}{h} = \frac{1}{2\sqrt{2}}$,which implies $\frac{h}{R} = 2\sqrt{2}$.

$A$ solid cone is placed on a horizontal surface has height $h$,radius $R$ and apex angle $\theta$ as shown. If the gravitational potential energy of the cone does not change as the position of the cone is changed from figure $(A)$ to figure $(B)$,then,

Similar Questions

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

From a circular disc of radius $R$,a square is cut out with a radius as its diagonal. The center of mass of the remaining part is at a distance (from the centre) of:

The centre of mass of a solid hemisphere of radius $8 \, cm$ is $X \, cm$ from the centre of the flat surface. Then the value of $X$ is $......$

As shown in the figure,when a spherical cavity (centred at $O$) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$),the centre of mass of the remaining (shaded) part of the sphere is at $G$,i.e.,on the surface of the cavity. $R$ can be determined by the equation:

$A$ sphere of diameter $r$ is cut from a sphere of radius $r$ such that the centre of mass of the remaining mass is at the maximum distance from the original centre. What is this distance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module