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 semi-circular trajectory in the vertical plane with the B string becoming slack only on reaching Q the topmost point, C.
Column-I Column-II
(i) Velocity $v_{0}$ is (p) 3
(ii) Velocity at point B is (q) $\sqrt{gL}$
(iii) Velocity at point C is r) $\sqrt{5 gL}$
(iv) Ratio of kinetic energy at B and C is (s) $\sqrt{3 gL}$
Now match the Column-I with Column-II and mark the correct choice from the codes given below.
Codes

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 semi-circular trajectory in the vertical plane with the B string becoming slack only on reaching Q the topmost point, C.

	Column-I		Column-II
(i)	Velocity $v_{0}$ is	(p)	3
(ii)	Velocity at point B is	(q)	$\sqrt{gL}$
(iii)	Velocity at point C is	r)	$\sqrt{5 gL}$
(iv)	Ratio of kinetic energy at B and C is	(s)	$\sqrt{3 gL}$

Now match the Column-I with Column-II and mark the correct choice from the codes given below.

Codes

Infinity Learn · Accepted Answer

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

If we take potential energy of the system to be zero at the lowest point A, then
At A, E = $\frac{1}{2} m v_{0}^{2}$ ---------(i)
From Newton's second law
$T_{A} - mg = \frac{{mv}_{0}^{2}}{L}$ -----------(ii)

Where $T_{A}$ is the tension in the string at A.

At the highest point C, the string slackens as the tension in the string ( $T_{C}$ ) becomes zero

$C, E = \frac{1}{2} {mv}_{C}^{2} + mg (2 L) - - - - - (iii)$

From Newton's second law

$mg = \frac{{mv}_{C}^{2}}{L} - - - - - - - (iv)$

Where $v_{C} is the velocity at C,$

From (iv), $v_{C} = \sqrt{gL}$

(iii)-q

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

(i)-r

At B, B , E = $\frac{1}{2} {mv}_{B}^{2} + mg (L)$

Where, $v_{B}$ is the velocity at B.

or $\frac{1}{2}, {v_{B}}^{2} = E - mg (L) = \frac{5}{2} mgL - mgL (Using (v))$

$\frac{1}{2} {mv}_{B}^{2} = \frac{3}{2} mgL$

$∴ v_{B} = \sqrt{3 gL}$

(ii)-s

The ratio of kinetic energies at B and C is

$∴ \frac{K_{B}}{K_{C}} = \frac{\frac{1}{2} {mv}_{B}^{2}}{\frac{1}{2} {mv}_{C}^{2}} = \frac{3 gL}{gL} = \frac{3}{1}$

(iv)-p

Vertical circular motion

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