Question

A bob of mass m is suspended by a light string of length l. It is imparted a horizontal velocity $v _ { 0 }$ at the lowest point. A such that it completes a semicircular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point C.

Match the Column I with Column II:

Column I		Column II
(A)	Velocity $v _ { 0 }$ is	(p)	3
(B)	Velocity at point B is	(q)	$\sqrt { g L }$
(C)	Velocity at point C is	(r)	$\sqrt { 5 g L }$
(D)	Ratio of kinetic energy at B and C is	(s)	$\sqrt { 3 g L }$

• A. (1)(p), (2)(q), (3)(s), (4)(r)
• B. (1)(q), (2)(r), (3)(p), (4)(s)
• C. (1)(r), (2)(s), (3)(q), (4)(p)
• D. (1)(s), (2)(p), (3)(r), (4)(q)

Answer 1

Answer: (C)

(1)(r), (2)(s), (3)(q), (4)(p)

Answer 2

There are two external forces on the bob : gravity (mg) and tension (T) in the string. The tension does no work as displacement is always perpendicular to the string. Total mechanical energy (5) of the system is conserved.

If we take potential energy of the system to be zero at the lowest point A. then

….(i)

Form Newton’s second law

…(ii)

Where is the tension in the string at A.

At the highest point C, the string slackens as the tension in the string $\left( T _ { C } \right)$ becomes zero.

At C, …(iii)

From Newton’s second law

….(iv)

Where is the velocity at C.

From (iv),

C-q

From (iii), $\mathrm { E } = \frac { 1 } { 2 } \mathrm {~m} ( \mathrm { gL } ) + 2 \mathrm { mgL } = \frac { 5 } { 2 } \mathrm { mgL }$…(v)

Using (i),

….(vi)

A – r

At B,

Where is the velocity at B.

(Using (v))

B – S

The ratio of kinetic energies at B and C is

$\therefore \frac { \mathrm { K } _ { \mathrm { B } } } { \mathrm { K } _ { \mathrm { C } } } = \frac { \frac { 1 } { 2 } \mathrm { mv } _ { \mathrm { B } } ^ { 2 } } { \frac { 1 } { 2 } \mathrm { mv } _ { \mathrm { C } } ^ { 2 } } = \frac { 3 \mathrm { gL } } { \mathrm { gL } } = \frac { 3 } { 1 }$

D – P