The string of a pendulum of length l is displ | vedclass

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

Question

(B) Let the mass of the bob be $m$ and the length of the string be $l$. When the pendulum is displaced by $90^o$ and released,the potential energy at

Vedclass · Accepted Answer

$3\,mg$

Vedclass · Answer

(B) Let the mass of the bob be $m$ and the length of the string be $l$. When the pendulum is displaced by $90^o$ and released,the potential energy at the initial position is converted into kinetic energy at the mean position. By the law of conservation of energy: $mgl = \frac{1}{2}mv^2$,which gives $v^2 = 2gl$. At the mean position,the forces acting on the bob are the tension $T$ (upwards) and weight $mg$ (downwards). The net centripetal force is $T - mg = \frac{mv^2}{l}$. Substituting $v^2 = 2gl$,we get $T - mg = \frac{m(2gl)}{l} = 2mg$. Therefore,$T = 2mg + mg = 3mg$. Thus,the minimum strength of the string must be $3mg$.

The string of a pendulum of length $l$ is displaced through $90^o$ from the vertical and released. Then,the minimum strength of the string in order to withstand the tension as the pendulum passes through the mean position is

Similar Questions

Who is known as the father of Indian Bryology?

There are $3$ sections in a question paper and each section contains $5$ questions. $A$ candidate has to answer a total of $5$ questions,choosing at least one question from each section. Then the number of ways,in which the candidate can choose the questions,is

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^\circ)$.

For the combination of logic gates shown in the figure,the equivalent logic gate is

If the lines $4x + 3y - 1 = 0$,$x - y + 5 = 0$ and $kx + 5y - 3 = 0$ are concurrent,then $k$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module