A bob of mass m is suspended at a point O by a light strig of length left to perform vertical motion (circular) as shown in figure. Initially, by aplying horizontal velocity at the point 'A'. The string becomes slack when, the bob reaches at the point '

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A bob of mass m is suspended at a point O by a light string of length $l$ and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity $v_{0}$ at the point 'A'. The string becomes slack when, the bob reaches at the point 'D'. The ratio of the kinetic energy of the bob at the points B and C is [2025]

1 2

2 1

3 4

4 3

Ans.
(1)

Applying conservation of mechanical energy,

$\frac{1}{2} m v_{A}^{2} = \frac{1}{2} m v_{B}^{2} + m g h$

$\Rightarrow \frac{1}{2} m (5 g l) = \frac{1}{2} m v_{B}^{2} + m g \frac{l}{2}$

$\Rightarrow \frac{5 m g l}{2} - \frac{m g l}{2} = {KE}_{B}$

$\Rightarrow {KE}_{B} = 2 m g l$

$\frac{1}{2} m v_{C}^{2} = \frac{1}{2} m v_{D}^{2} + m g \frac{l}{2}$

$\Rightarrow \frac{1}{2} m g l + m g \frac{l}{2} = m g l$

${KE}_{C} = m g l$

$\Rightarrow \frac{{KE}_{B}}{KEc} = 2$

Want Us to Email you About Special Offers & Updates?