## Circular motion,Vertical circular motion and Radius of curvature

If Any object moves such that it covers an angle q at the fixed point (center of the circle) and its gap from the fixed point remains constant (radius of the circle)

Here R is constants

Now finding linear speed of

(i) hour hand (30 cm)

(ii) Minute hand (90 cm)

(iii) Second hand (60 cm)

Solution:- (i) time period of hour hand T = 12 hr

= 12 x 3600 sec

v = r w

(ii) Minute hand:- time period = T = 60 min = 60 x 60 sec

(iii) Second hand:-

Time period= T = 60 sec

v = r w

**Types of circular motion: (on the basis of speed)**

(1) Uniform circular motion:** (U C M)**

Only direction of velocity is changing, magnitude remains unchanged.

Now acceleration due to change in direction of velocity

Using law of parallelogram

= 2v^{2} - 2v^{2} cosq

= 2v^{2 }(1- cosq)

Now

sinq » q if q <<<<

a = v w,

Direction of this acceleration is towards center

Centripetal acceleration or Radial acceleration

(2) Non uniform circular motion (Non U C M)

Here in this case direction as well as magnitude both are changing continuously therefore here two different named acceleration will be as,

(i) a_{c} = a_{R} (centripetal or radial acceleration)

(ii) a_{t} (Tangential acceleration)

Now,

**Tangential acceleration. **

**“**Rate of change of magnitude of velocity”

v = R w

So a_{t} ^ a_{c}

Net acceleration for non-uniform circular motion.

**Example:-** A particle moving in a circular path of radius 2 meter and its velocity varies as v = 10t^{2.} Then net acceleration of the particle at t = 2 sec.

**Solution:** Given v = 10t^{2}

= 20 x 2

= 40 m/s^{2}

**Centripetal force:-** the force require to move in a circular path for any object with respect to inertial frame is called centripetal force.

**Note:-** Centrifugal force will also have same magnitude but direction opposite to centripetal force.

Direction of this force is towards the center.

**HORIZONTAL CIRCULAR MOTION**

**(1) Only banking**

Here N sinq is C.P. provider

But N cosq = mg

Safe speed

**(2) Only friction **

Here now m N is C.P. provider

N = mg

Safe speed

**Friction and Banking both:- **To prevent inward sliding

cosq + m N sinq = mg

Divide

**TO PREVENT OUTWARD SLIDING**

N cosq - m N sinq = mg

v_{1} < V_{safe} < v_{2}

**CONICAL PENDULUM**

T sinq = mrw^{2}

T cosq = mg

Time period of conical pendulum

## Vertical CM and String Mass

**VERTICAL CIRCULAR MOTION **

For top point

For middle position

v^{2 }= u^{2} - 2gh

T_{min} ≥ 3 mg

**VERTICAL CIRCULAR MOTION **

For Lowest point

& Lowest to top

v^{2} = u^{2 }– 2gh

T_{min} ≥ 6 mg

**STRING MASS SYSTEM**

**RIM & MASS SYSTEM**

**RIGID ROD AND MASS SYSTEM**

To Complete the Circle if minimum velocity at top point = 0

Then Top to bottom

v^{2 }= u^{2} - 2gh

(o)^{2} = u^{2} – 2g x 2l

At mid point

Lowest to mid point

v^{2} = u^{2} – 2gh

**INCOMPLETE CIRCLE**

Case I velocity given at lowest position

Case II velocity given at lowest point

**Example:** As shown if figure a small mass m is put on the top of smooth hemisphere now this mass slightly hit to disturb it. At what angle it will leave the hemisphere.

Leaving the Circle so N = 0

Now from top to point where leave the Circle

v^{2} = u^{2} + 2gh

v^{2} = (0)^{2} + 2gR(1 - cosq)

gR cosq = 2gR (1 - cosq)

Cosq = 2 - 2 cosq

3 cosq = 2

**RADIUS OF CURVATURE **

Case I At top of projectile motion

if we assume war the top point of projectile is a small Circular part then

Case II At any point before the top of projectile

At a point where velocity makes angle q/2 with horizontal

V cosq/2 = u cosq

**Condition of toppling of a vehicle on circular tracks**

While moving in a circular track normal reaction on the outer wheels

(N_{1}) is more than the normal reaction on inner wheels (N_{2}).

Or N_{1} > N_{2}

This can be proved as below.

Distance between two wheels = 2a

Height of centre of gravity of car from road = h

For translational equilibrium of car

N_{1} + N_{2} = mg … (i)

and for rotational equilibrium of car, net torque about centre of gravity should be zero.

Or N_{1} (a) = N_{2} (a) + f (h) … (iii)

From Eq. (iii), we can see that

Or N_{2} < N_{1}

From Eq. (iv), we see that N_{2} decreases as v is increased.

In critical case, N_{2} = 0

and N_{1 }= mg [From Eq. (i)]

N_{1} (a) = f (h) [From Eq. (iii)]

Therefore, for a safe turn without toppling

From the above discussion, we can conclude while taking a turn on a level road there are two critical speeds, one is the maximum speed for sliding and another is maximum speed for toppling *.* One should keep ones car’s speed less than both for neither to slide nor to overturn.

**Case I**

The length of each light rod is *l*. The mass of each small ball B and D is m. Balls are connected with a light smooth sleeve at point A as shown in the figure below. The frame is rotating with constant angular velocity w. A force F_{0 }is

**Question.** If tension in AB wire is T_{1} and that in wire BC is T_{2}. Then the value of (T_{2} – T_{1}) is

(a) mg

(c) 2 mg dmg2

**Solution:** At the ball B,

T_{2} sin 45° = mg + T_{1} sin 45° (in vertical direction)

T2-T1=mgsin45°=2 mg …(i)

**Question.** The required force F_{0} on the sleeve is

(c) 2 mg

**Solution:** T_{1 }cos 45° + T_{2} cos 45° = mrw^{2} (in horizontal direction)

or T_{1} + T_{2} = mlw^{2} …(ii)

From Eqs. (i) and (ii), we get

For bead, 2T_{1} cos 45° = F_{0}

**Answer (a)**

**Case II**

A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (sec fig.) The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted byq. **(2002 JEE Advance)**

(a) Express the total normal reaction force exerted by the spheres on the ball as a function of angleq.

(b) Let N_{A} and N_{B }denote the magnitudes of the normal reaction forces on the ball exerted by the spheres A and B, respectively. Sketch the variations of N_{A} and N_{B} as function of cos q in the range 0 £ q £ p by drawing two separate graphs in your answer book, taking cos q on the horizontal axis.

**Solution:**

Velocity of ball at angle q is

Let N be the normal reaction (away from centre) at angle q.

Substituting value of v^{2} from Eq. (i), we get

mg cos q - N = 2 mg (1 – cos q)’

N = mg (3 cos q -2)

(b) The ball will lose contact with the linner space when N = 0

After this it makes contact with cover sphere and normal reaction starts acting towards the centre.

N_{B }= 0

and N_{A} = mg (3 cos q - 2)

N_{A} = 0 and N_{B} = mg (2 – 3 cosq)

The corresponding graphs are as follows.

**Case III**

A small smooth block of mass m is projected with speed from bottom of a fixed sphere of radius, R so that the block moves in a vertical circular path.

** Question.** The speed of the block as function of q, the angle of deflection from the lowest vertical.

**Solution: **From free body diagram of the block, - mg sin q = ma_{t}

or v^{2} = 2gR – 2gR (1 – cos q)/

**Answer (b)**

** Question.** Normal reaction at the moment when vertical component of the block’s velocity is maximum.

(a) mg

(d) 2 mg

**Solution:** From the figure, vertical component of velocity is

**Answer (c)**

** Question.** The angle q between the thread and the lowest vertically at the moment when the total acceleration vector of the block is directed horizontally.

**Solution: **From the figure, the centripetal acceleration is

and a tangential acceleration is a_{t} = a cos q

or g sin q = a cos q

a = g tan q

or v^{2} = gR tan q sin q

or 2gR cos q = gR tan q sin q

**Answer (c)**