## A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to a position ….

Q: A bob is attached to one end of a string other end of which is fixed at peg A. The bob is taken to a position where string makes an angle of 30° with the horizontal. On the circular path of the bob in vertical plane there is a peg B’ at a symmetrical position with respect to the position of release as shown in the figure. If Vc and Va be the minimum tangential velocity in clockwise and anticlockwise directions respectively, given to the bob in order to hit the peg B’ then ratio Vc : Va is equal (A) 1 : 1

(B) 1 : √2

(C) 1 : 2

(D) 1 : 4

Click to See Solution :
Ans: (C)
Sol: For anti-clockwise motion, speed at the highest
point should be $\sqrt{gR}$ . Conserving energy at (1) & (2) : $\displaystyle \frac{1}{2}m v_a^2 = m g\frac{R}{2} + \frac{1}{2}m g R$

$\displaystyle v_a^2 = g R + g R$

$\displaystyle v_a = \sqrt{2 g R}$

For clock-wise motion, the bob must have atleast that much speed initially, so that the string must not
become loose any where until it reaches the peg B. At the initial position :

$T + m g cos60^o = \frac{m v_c^2}{R}$ VC being the initial speed in clockwise direction. For VC min : Put T = 0 ;

$\displaystyle V_c = \sqrt{\frac{gR}{2}}$

$\displaystyle \frac{V_c}{V_a} \frac{\frac{gR}{2}}{\sqrt{2gR}} = \frac{1}{2}$

Vc : Va = 1 : 2

## A section of fixed smooth circular track of radius R in vertical plane is shown in the figure …..

Q: A section of fixed smooth circular track of radius R in vertical plane is shown in the figure . A block is released
from position A and leaves the track at B . The radius of curvature of its trajectory when it just leaves the track
at B is (A) R

(B) R/4

(C) R/2

(D) none of these

Click to See Solution :
Ans: (C)

Sol: By energy conservation between

$\displaystyle M g \frac{2R}{5} + 0 = M g \frac{R}{5} + \frac{1}{2}M v^2$

$\displaystyle v = \sqrt{\frac{2gR}{5}}$

$\displaystyle \frac{v_{perp}^2}{a_r} = \frac{2gR/5}{g cos37^o} = \frac{R}{2}$

## A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire…..

Q: A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire. The circle is rotated about vertical diameter AB as axis with a constant speed ω as shown in figure. The bead P is at rest w.r.t. the wire in the position shown. Then ω2 is equal to : (A) $\frac{2g}{a}$

(B) $\frac{2g}{a\sqrt{3}}$

(C) $\frac{g\sqrt{3}}{a}$

(D) $\frac{2 a}{g\sqrt{3}}$

Click to See Solution :
Ans: (B)
Sol: $cos\theta = \frac{a/2}{a} = \frac{1}{2}$ $\theta = 60^o$

$N sin60^o = m g$

$N cos60^o = m \frac{\omega^2 a}{2}$

$tan60^o = \frac{2 g}{\omega^2 a}$

$\omega^2 = \frac{2g}{a\sqrt{3}}$

## A particle is revolving in a circle increasing its speed uniformly. Which of the following is constant ?

Q: A particle is revolving in a circle increasing its speed uniformly. Which of the following is constant?

(A) centripetal acceleration

(B) tangential acceleration

(C) angular acceleration

(D) none

Click to See Solution :
Ans: (C)

Sol: Angular acceleration $\alpha = \frac{a_t}{r}$

Since $|\vec{a_t}| = |\frac{d\vec{v}}{dt}| = constant$

magnitude of α is constant
Also its direction is always constant (perpendicular to the plane of circular motion). whereas, direction of at changes continuously $\vec{a_t}$ is not constant.