A uniform chain of length l and mass m lies o | vedclass

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

Question

(B) Let the angle subtended by the chain at the center of the hemisphere be $\alpha = l/R$. Consider an element of the chain of length $dl$ at an angl

Vedclass · Accepted Answer

$\frac{m g R^2}{l} \sin \left(\frac{l}{R}\right)$

Vedclass · Answer

(B) Let the angle subtended by the chain at the center of the hemisphere be $\alpha = l/R$. Consider an element of the chain of length $dl$ at an angle $	heta$ from the base. The height of this element from the base is $h = R \sin 	heta$. The length of the element is $dl = R d	heta$. The mass of this element is $dm = \frac{m}{l} dl = \frac{m}{l} R d	heta$. The gravitational potential energy of this element is $dU = (dm)gh = \left(\frac{m}{l} R d	hetaight) g (R \sin 	heta) = \frac{mgR^2}{l} \sin 	heta d	heta$. The chain extends from the top $(	heta = \pi/2)$ to an angle $	heta = \pi/2 - \alpha = \pi/2 - l/R$. Integrating $dU$ from $\pi/2 - l/R$ to $\pi/2$: $U = \int_{\pi/2 - l/R}^{\pi/2} \frac{mgR^2}{l} \sin 	heta d	heta = \frac{mgR^2}{l} [-\cos 	heta]_{\pi/2 - l/R}^{\pi/2}$ $U = \frac{mgR^2}{l} [-\cos(\pi/2) - (-\cos(\pi/2 - l/R))]$ $U = \frac{mgR^2}{l} [0 + \sin(l/R)] = \frac{mgR^2}{l} \sin \left(\frac{l}{R}ight)$.

$A$ uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R$ $(R > l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is

Similar Questions

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity,the work required to pull the hanging part onto the table is:

The potential energy of a body of mass $m$ is given by $U = ax + by$,where $x$ and $y$ are the position coordinates of the particle. The acceleration of the particle is:

The potential energy of a particle is given by $U = 8x^2 - 4x + 400 \, J$. Find the position $x$ where the force acting on the particle is zero.

If a body has a potential energy of $(4x^2 + 2x) \,J$ at a position $x = 2 \,m$, then the force acting on the body is: (in $\,N$)

$A$ stick of mass $m$ and length $l$ is pivoted at one end and is displaced through an angle $\theta$. The increase in potential energy is ............

Vedclass Test Series

Exam Paper Generator

Online Exam Module