## 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

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

Ans: (C)

Solution:

By energy conservation between

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

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

Now, radius of curvature r

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

## 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) $\frac{2 a}{g\sqrt{3}}$

Ans: (B)

Solution: $cos\theta = \frac{a/2}{a} = \frac{1}{2}$

$\theta = 60^o$

$N sin60^o = m g$

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

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

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

## 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

Ans: (C)

Solution: 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.

## A student skates up a ramp that makes an angle 30° with horizontal . He/she starts (as shown in the figure) at the bottom ….

Q: A student skates up a ramp that makes an angle 30° with horizontal . He/she starts (as shown in the figure) at the bottom of the ramp with speed vo and wants to turn around over a semicircular path xyz of radius R during which he/she reaches a maximum height h (at point y) from the ground as shown in the figure . Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only . Then (g is the acceleration due to gravity)

(a) $\displaystyle v_o^2 – 2 g h = \frac{1}{2} g R$

(b) $\displaystyle v_o^2 – 2 g h = \frac{\sqrt{3}}{2} g R$

(c) the centripetal force required at points x and z is zero

(d) the centripetal force required is maximum at points x and z

Ans: (a,d)

Solution: Applying Energy Conservation ,

$\frac{1}{2} m v_o^2 = m g h + \frac{1}{2}m v^2$

$\frac{1}{2} m v^2 = \frac{1}{2}m v_o^2 – m g h$ ….(i)

At top point y ,

$mg sin\theta = \frac{m v^2}{R}$

Putting in (i)

$\frac{m g R sin\theta}{2} = \frac{1}{2}m v_o^2 – m g h$

$\frac{m g R}{2} sin30 = \frac{1}{2}m v_o^2 – m g h$

$\frac{m g R}{4} = \frac{1}{2}m v_o^2 – m g h$

$\frac{g R}{2} = v_o^2 – 2 g h$

## A simple pendulum mounted on a car. The car starts to accelerate on horizontal surface …

Q: A simple pendulum mounted on a car. The car starts to accelerate on horizontal surface with constant acceleration a0 = g/√3 . The maximum deflection of pendulum from vertical is

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Click to See Answer :
Ans: (a)

Sol: Let deflection of pendulum from vertical is θ

If T be the tension in the String

T cosθ = m g …(i)

T sinθ = m a0 …(ii)

On dividing (ii) by (i)

$\displaystyle tan\theta = \frac{a_0}{g}$

$\displaystyle tan\theta = \frac{g/\sqrt{3}}{g} = \frac{1}{\sqrt{3}}$

θ = 30°

## A small block is placed at the top of sphere. It slides on the smooth surface of the sphere. What will be the angle made by the radius vector of the block with the horizontal, when it leaves the surface?

Q: A small block is placed at the top of sphere. It slides on the smooth surface of the sphere. What will be the angle made by the radius vector of the block with the horizontal, when it leaves the surface?

(a) 52°

(b) 48°

(c) 42°

(d) 38°

Click to See Answer :
Ans: (c)

## The string of a pendulum of length l and bob of mass m is displaced through 90°….

Q:The string of a pendulum of length l and bob of mass m is displaced through 90°. Minimum strength of the string in order to withstand the tension should be.

(a) mg

(b) 2 mg

(c) 3 mg

(d) 4 mg

Click to See Answer :
Ans: (c)

At mean position body is in vertical equilibrium

$T – \frac{m v^2}{l} = m g$ …(i)

$v^2 = 2 g l$ …(ii)

from (i) & (ii)

$T – 2 m g = m g$

T = 3 m g

## A ball of mass 0.6 kg attached to a light inextensible string rotates in a vertical circle …

Q: A ball of mass 0.6 kg attached to a light inextensible string rotates in a vertical circle of radius 0.75 m such that it has a speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is (g = 10 m s-2)

(a) 30 N

(b) 26 N

(c) 20 N

(d) 6 N

Click to See Answer :
Ans: (c)

Force at equilibrium ,

$T = \frac{m v^2}{r}$

$T = \frac{0.6 \times 5^2}{0.75}$

T = 20 N

## A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m….

Q: A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m. What is his inclination to the vertical?

(a) 30°

(b) 45°

(c) 10.5°

(d) 26.5°

Click to See Answer :
Ans: (d)
$tan\theta = \frac{v^2}{r g}$

$tan\theta = \frac{9.8^2}{19.6 \times 9.8}$

$tan\theta = \frac{1}{2}$

$\theta = 26.5^o$