Dynamics

Chapter 3

DYNAMICS

Dynamics can be divided into two main branches (a) Kinematics

(b) Kinetics

In kinematics, motion of particles or rigid bodies is studied without considering the forces that produce or change this motion.

In kinetics, motion of particles or rigid bodies is studied with the unbalanced force system that produces or changes this motion.

Kinematics of Rectilinear Motion

Where u=initial velocity, v=final velocity, s=distance of travel, t=time and a=acceleration Example 1

Velocity of a particle is defined as V=Kx3_-x2 _{6x , where “V” is in m/s and ‘x’ is in ‘m’ compute}

the acceleration when x=2 and k=1 Solution Given, V= Kx3_-x2_+6x 6 , ( 6 ) ( 6) = (123_-22₊₆₂₎ ₍₃₁₂2_-2₂₊₆₎ = (8-4+12)(12-4+16) =1614=224m/s2

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 67 Example 2

The brakes of a train reduce its speed from 60 to 20 km/h while it runs 200 m. Assuming that there exists constant retarding force, find

(a) How much further the train will run before coming to rest, and (b) How long will it take.

Solution 6 / , / v u as 400 – 6 α a = /

(a) For the train to come to rest , v u as

6

=225 m

Further distance moved = 225 -200 =25 m (b) v u at 0 =20 – 8000 t 6

Motion of Bodies Projected vertically upwards

When a body is projected vertically upwards, it is under the effect of the downward acceleration due to gravity, i.e., it moves with retardation. Its velocity, therefore, gradually decreases until it becomes zero the body is then for an instant at rest and immediately begins to fall with a velocity which increases numerically but is negative. Thus, we get

v u gs

Time to reach the highest point,

Maximum height reached,

Time for returning to the starting point time of flight = Example 3

A particle is dropped from the top of tower 200 m high and another particle is projected at the same time vertically upwards from the foot of the tower so as to meet the first particle at a height of 50 m. find the velocity of projection of the second particle.

Solution

Let the particle meet after seconds. The first particle falls from rest a distance = 200 -50 = 160m.

Hence 150

√

( )

If is the required velocity1 of the second particle, then ( )

Adding Equation (a) and (b), we get 200 =

6 / Example 4

Two balls are projected simultaneously with the same velocity from the top of a tower, one vertically upwards and the other vertically downwards. If they reach the ground in times t and t respectively, show that √t t is the time which each will take to reach the ground if simply let drop from top of the tower.

Solution

Let h = height of the tower u = velocity of projection

Taking downward direction positive, for the first ball, we have , , ,

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 69 ( )

For the second ball, , , ,

∴ ( ) Eliminating from equation (a) and (b), we get

If is the time to reach the ground when initial velocity is zero, we get

∴ √ Example 5

The speed of a train increases at a constant rate from zero to and then remains constant for an interval and finally decreases to zero at a constant rate β If d be the total distance described, prove that the total time occupied is

( )

Solution

Let t , t and t be the times for acceleration, constant speed and deceleration, then ,

Total distance travelled = Or t t t

Now total time And t ,

∴

t d ( )

( )

= ( )

Example 6

Two stopping points of an electric tram car are 450m apart. The maximum speed of the car is 20km/h and it covers the distance between the stops in 100seconds. If both acceleration and retardation are uniform and the latter is twice as great as the former, find the value of each of them and also calculated how far the car runs at the maximum speed.

Solution

Let t , t , t be the times for acceleration, uniform velocity and deceleration. Then

Maximum speed of car = 20 km/h m/s 450 = = ( ) ( ) ∴t t t 6 Also ∴t =2t 2t t t 6 t t 6 and t t 6 t 6 s, t 6 s, t s Distance covered during constant speed = 6 Example 7

A person going Eastwards with a velocity of 4 km/h finds that the wind appears to blow directly from the North. He doubles his speed and the wind seems to come from the North-east. In what direction and with what velocity is the wind blowing?

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 71 Solution

Let the actual velocity of wind be km/h along OA making an angle θ with OE, as shown in figure.

(a) When the velocity of the man along OE is 4 km/h the velocity of wind is along ON. Therefore, the relative velocity of wind will be along OS. Therefore, the relative velocity of wind will be along OS. The resultant of 4Km/h and the relative velocity of wind is . Resolving along OE, we get

4 =

(b) When the velocity of man is 8km/h the relative velocity of wind is along OB, whose resultant is again .

Resolving perpendicular to OB, we get cos( ) cos √ (cos sin ) √ (cos sin ) ∴ √ 6 6 √ / tan  = Example 8

The driver of a car travelling at 72 km/h observe the traffic light 300m ahead of him turning red. The traffic light is timed to remain red for 20s before it turns green. If the driver wishes to pass the light without stopping to wait for it to turn green, determine (a) the required uniform deceleration of the car and (b) the speed with which the driver crosses the traffic light.

S W B C A E N.E N O θ ( ) /

Solution 6 6 / , Now 300 = 20 ×20 + / Also / = 6 / Example 9

Two trains A and B of length 400 m each are moving on two parallel tracks with uniform speed 72km/h in the same direction, with A ahead of B. The driver of B decides to overtake A and acceleration by 1ms _{If after 50 s, the guard of B just brushes past the driver of A, what was the}

original distance between them? Solution

Relative velocity of B w.r.t A = 0

Let d orginaldistance between the trains in m.

Distance covered by B in 50 s = 400 + d +400 =800 +d Using

( )

Example 10

A stone projected vertically upwards from a point A passes a point B after 3 seconds. If it returns to A after a further interval of 4 seconds, find (i) the height of B above A (ii) the velocity with which the stone passes the point midway between A and B.

Train B Train A

400 m 400 m

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 73 Solution

Let projection maximum height reached

Time taken from A to C and back to a A

= 3 +4 =7 s

Or

Time taken from A to B = 3 = 34.3 × 3 - or AB = 58.8 m AD = ( ) v 6 6 6

Kinematics of Curvilinear motion

Motion of projectile

Maximum height (h) =

Time required to reach maximum height (t) =

∴

Range (R) =

∴

A

Acceleration Analysis

Background Material: Rate of Change of Vector

A vector quantity has magnitude and direction and hence, the vector can change in time in two ways-In magnitude or direction. Therefore, when we take the time-derivative of a vector ie

lim

( ) ( )

, the numerator is change in vector from time t to t + ∆t can have two

components.

• Rate due to change in Magnitude: This is got by differentiation of the scalar magnitude

| |

n̂ (n̂ | | is the unit vector in the direction of velocity vector)

Ex: For vector ̂ ̂,

̂

• Rate due to change in Direction: This is more involved, but it can be shown that a fixed vector V rotating in a rate ω has a rate of change due to direction change as

So, in general we have that for a generic vector changing in magnitude and rotating at rate ω, we have | | ̂

Acceleration Analysis

Given in the figure, is a fixed ground Coordinate System (CS) inside of which a rotating CS with origin at O rotates at ω, has angular acceleration of , translates at V0& accelerates at A0. Inside this rotating CS O, is a point P which translates at OVP & accelerates at OAPlocally.

OVP=Velocity of Point P as viewed from Origin of Rotating Cord. Sys. O

OAP=Acceleration of Point P as viewed from Origin of Rotating Cord. Sys. O

VO=Velocity of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G

AO=Acceleration of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G Figure : Diagram for Acceleration Analysis

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 75

O=Angular Velocity of Rotating CS O w.r.t Fixed Ground CS with origin at G

O =Angular Acceleration of Rotating CS O w.r.t Fixed Ground CS with origin at G

OP =Position vector of a point in w.r.t Rotating CS with origin at O

GP =Position vector of a point in w.r.t Ground CS with origin at G

GO =Position vector of Origin O of Rotating CS w.r.t Ground CS with origin at G

We are interested in acceleration of point P as seen from ground because as coordinate system O is accelerating and rotating system, it is a non-inertial coordinate system and hence Newton’s laws are not applicable for vectors in that frame. Hence, we need to write P in terms of ground coordinate system if we are to apply the same for subsequent force/torque analysis.

So, we start by writing down the position of point P as seen from ground coordinate system.

GP = GO + OP ⇒ ⃗ ⃗ ⃗

Now, differentiating w.r.t time, we get ⃗

⃗

⃗

But we know that ⃗

(velocity of point P as viewed from O) & ⃗

(velocity of point

P as viewed from ground). Now, ⃗ is a vector which has change in magnitude (motion in the

frame O with a local velocity of OVP, acceleration OAP) & also has change in direction (rotates at ω & α). So, we can write as follows.

⃗

⇒

We now have the velocity of point P w.r.t ground G. To get acceleration, further differentiate both the sides.

We now have the velocity of point P w.r.t ground G. To get acceleration, further differentiate both the sides.

Now, we know that is a rotating vector in the CS at O and hence has magnitude change( ) and direction change(rotates , ) ⇒

⇒

( )

Simplifying the above expression gives as follows ( )

So, we see that we have five terms in total here on the Right Hand Side which is why we call this above formula as the five point acceleration formula.

 ( ): This is the acceleration of the origin O of rotating CS w.r.t ground CS at G.

 Term 2( ): This is the acceleration of the point P local to/when viewed from within the rotating CS at O.

 Term 3( ) This component is the Coriolis acceleration component produced due to velocity inside a rotating frame.

 Term 4 ( ): This is the Tangential acceleration component which acts in the

instantaneous tangential direction of motion of point P.

 Term 5(ω (ω rop)) This is the Radial/Centripetal acceleration component and acts radially inward towards the point of center of rotation.

Thus, all the five terms of motion are clearly derived from this highly generic method. Newton's law can now be applied to the LHS term which is got by adding up the five acceleration components in the RHS.

Acceleration Analysis in Cylindrical or Path or Curvilinear Coordinates

In case of cylindrical coordinates, we have the coordinate directions as shown below. Clearly, for a motion from time , The direction of CS itself changes in & direction whereas remains constant. We proceed with derivation in the same manner as above.

⇒ ( )

Now, as shown in figure, when the CS rotates by in a time dt, we see that

( ) ( ) ∆ ( )

Now, note that in the limiting case, as , the direction of ∆ ( ) approaches tangential direction in . Also, as far as magnitude is concerned,

Trajectory of the particle P

∆ ( ) ( ) ( ) dθ ( ) O (Projecting Outward) ( ) ( ) P ( ) ∆ ( ) ( ) ( ) Rate of change Rate of change Rotation of CS from t to t+td

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 77 | | | ( ) | ( | | ) ⇒

Similarly, from the diagram above, using the same logic, we can prove that

However, as does not change direction in time

Using the above results, we proceed to get as follows. ̇

̇ ̇

Again differentiating, we get ̈ ̇ ̇ ( ̈ ) ( ̇ ) ̈

This is the final formulae for velocity and acceleration of a moving particle in cylindrical coordinates.

Examples 11

Acceleration of a point moving in a rotating frame. Consider the rotating tube shown in figure. It is given that the arm OAB rotates with counterclockwise angular acceleration / and at the instant shown the angular speed / . Also, at the same instant, the particle P is falling down with speed / / and acceleration / / w.r.t. tube. Find the

absolute acceleration of the particle at the given instant. Take in the figure. Solution

Fixing a ground coordinate system on ground and a rotating coordinate system with origin at O and moving along with the frame OAB, we see that for point P, position vector

, But clearly, & are all fixed vectors from geometry which does not change in

magnitude (length) with time. So, only change is change in direction. But changes in length

because point P has a local velocity when viewed from B & also it changes it direction like the other two vectors.

Figure: Problem 1 Figure ω, ̇ ℓ = 2 f t ℓ = 2 f t ℓ ft B O P A 2ℓ / /

So, ⇒

( ) / . Same way, on further differentiating, we

get acceleration

( ) ( ( ) / ) / / .

Now, we have to represent all vectors in same CS. Here, we can write

̂ ̂, ̂, ̂, ̂, / ̂ & / =- ̂ Substitute these values

and upon simplifying, we get the answer as ̂ 6 ̂ / & 66 ̂ ̂ / Example 12

A particle moves along the spiral as shown in figure, and defined by the equation r= 12t and o=2nt. Determine the velocity and the acceleration of the particle when t=0 < t=0.3 see

Solution Given, R= 12t and  = 2nt  We know, for r- System,

 = 12+12t 2 =12(1+2t) m/s Acceleration a= ar+a [ (  ) ] *   + =  -(12t2)2 _{+ 0+ 2}₁₂₂ a=48-5762_t2 _m/s2 V= 12(1+2t)m/s a=48-5762_t2 _m/s2 0 O Y X

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 79 Example 13

Find the maximum horizontal range of a cricket ball projected with a velocity of 15m/s. If the ball has a range of 20m, find the least angle of projection and the least time taken.

Solution

Horizontal range =

For maximum horizontal range, α ∴R m Now Sin α α 6 or 6 6

∴ Least angle of projection = 6 Least time of projection

=

The value of the maximum range r

( )

( )

Example 14

A particle is projected from a point on an inclined plane with a velocity of 30m/s. The angle of projection and the angle of plane are and to the horizontal respectively. Find the range and time of flight of the particle.

Solution

Range on an inclined plane

*sin( ) )

Here

For the range to be maximum α – β

Or 2 ×5

Thus, this corresponds to maximum range on the given inclined plane. Maximum range,

r

u

= _{( ) ( )} = 6 6 Time of flight, T = ( ) ( ) = Example 15

A projectile is launched as shown in the figure. (i) Compute the velocity of the projectile if R = 250m (ii) Find the flight time of the projectile

Solution (i) R =

Sin ] sin ]

Given, -

R = 250m (range of the projectile)

G = 9.81 m/s2 _{(Acceleration due to}

gravity)

Angle of the projectile = 600

Β Angle of inclination, 0

U = velocity of projectile = ? ∴

sin( 6 ) sin ]

∴ u m/s (ii) Flight time, - T = ( ) = ( ) = 10.09sec. u = 42.89m/s T = 10.09sec 600 300 R A u B

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 81 Example 16

A rope, to which a weight is attached, passes around a pulley 50cm in diameter. The angular acceleration of the pulley is 18 rad/s . If the pulley is initially at rest, find(a) the time required for the weight to attain a velocity of 15m/s, (b) the number of revolutions through which the pulley rotates during that period and (c) the total acceleration of a point on the rim of the pulley 0.5 second after it was at rest.

Solution (a) ω 6 / ω ω αt 60= 0+18t (b) θ ω t = ( ) Number of revolutions = (c) Tangential acceleration α rα cm /s Normal acceleration α rω After 0.5 s, we have ω ω αt rad/s ∴ cm/s Total acceleration, √ = √( ) ( ) cm/s Example 17

A ball is thrown from the top of a tower 60m high with velocity of ms at an elevation of above the horizontal distance from the foot of the tower to the point where it hits the ground. (g = 10 ms ₎

Solution

Refer the figure sin 6 sin 6 or 6 or t t √ √ √ 6 6

Hence, R = cos cos 6 = 20 × √ 6 = 79.67 m

Example 18

A shell is required to be projected so as just to pass over the wall of a fort horizontally. If the wall is 20m high and 60 m distant from the gun, find the angle and the velocity of projection.

Solution

The shell grazes the wall horizontally. Therefore, the height of the wall is the maximum height attained. / A B O R 60 m

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 83 With reference to figure

( ) Also 6 or R = 120m Now 120 = ( ) Dividing (1) by (2) 6 tan Or tan _{( ) 6} From (1), 20 = u 6 /

Kinetics of Rectilinear Motion

Figure: Diagram for Acceleration/Force Analysis

, G ,

=Velocity of Point P as viewed from Origin of Rotating Cord. Sys. O =Acceleration of Point P as viewed from Origin of Rotating Cord. Sys. O =Velocity of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G

=Acceleration of Origin of Rotating CS O w.r.t Fixed Ground CS with origin at G =Angular Velocity of Rotating CS O w.r.t Fixed Ground CS with origin at G =Angular Acceleration of Rotating CS O w.r.t Fixed Ground CS with origin at G OP =Position vector of a point in w.r.t Rotating CS with origin at O

GP =Position vector of a point in w.r.t Ground CS with origin at G

GO =Position vector of Origin O of Rotating CS w.r.t Ground CS with origin at G

As seen in acceleration analysis, the acceleration experienced by a point P moving in a rotating and accelerating coordinate system, as seen from the global frame of reference G, is given by(5-term acceleration formula)

( )

Now, we know that Newton's law is F = ma or Force F required to make an object of mass m move at acceleration a is ma.Therefore, standing in the frame G, we see acceleration given by previous formula & therefore, if we multiply both sides of the equation by mass of particle P, i.e., m, we get force as follows.

( ( ))

However, from the rotating frame of reference O, if we write the Newton's law, we see that it because an observer in O frame only sees the)AP acceleration for the point P. However, there can be only one force required to move the object from any frame. So, which one is correct value, or FG? Clearly, the observer in frame O is applying a lesser force as per Newton's law but that's because he is neglecting many acceleration components including Coriolis acceleration components & hence, FO is wrong. So, the conclusion is observer in frame O cannot employ Newton's law to estimate force whereas observer in frame G can. Frames where Newton's law can be applied (G) are called Inertial Frames of Reference & those frames where Newton's law cannot be applied(O) are called Non-Inertial Frames of Reference. Simply put, any frame accelerating or rotating w.r.t a stationary frame is a Non-Inertial Frame and any frame stationary or moving at a constant velocity w.r.t a stationary frame is a Inertial Frame. Clearly, between an Newton's Law equations for Inertial & Non-inertial frames, the difference comes in 4 terms. These 4terms are 4 forces not in strict physical sense as such forces don't physically exist but come up in the equations. Hence, these terms are called Fictitious Forces or Phantom Forces. They are as follows.

1. Inertial Force: The term & ( ), caused by translational & rotational acceleration

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 85 2. Coriolis force: The term ( )caused by velocity of particle P inside the

accelerating frame O is a fictitious force. A major physical manifestations of this force include cyclones/hurricanes.

3. Centripetal Force: The term ( ( ))due to rotation of the frame O is a fictitious

force.

Further, we see that ( ) ( ) ( ( )) So,

from the above equation, we can conclude that, if we indeed want to write force balance equation in frame O, we should take into consideration, the fictitious forces & draw them in opposite direction in the Free Body Diagram (FBD).

Finally, if we also consider the term as a fictitious inertial force and draw it in opposite direction, this is equivalent to writing an equation

( ) ( ) ( ( ))

Which is the form∑ which is the equation for equilibrium/statics. Thus, we see that considering each fictitious force in opposite direction of motion in the FBD, we can convert a dynamics problem to statics problem. This famous principle is called the D'Alembert's Principle. D’ Alembert’s Principle

It was pointed out first of all by D’Alembert that on the line of equation of static equilibrium, equation of dynamic equilibrium can also be established by introducing inertia force in the direction opposite to acceleration in addition to the real forces acting on the system.

According to Newton’s second law of motion, F = ma where

Or

Now is the inertia force Example represents the D’ Alembert’s principle, which may be stated as follows:

When different forces act on a system such that it is in motion in a particular direction, the algebraic sum of all the forces acting on the system in the direction of the motion, including the inertia force taken in opposite direction to motion is zero. Thus in general

Where , , or

F ma

∑ F ∑ ma ]

Example 19

An elevator weighs 25 kN and is moving vertically downwards with a constant acceleration. Write the equation for the elevator cable tension. Starting from rest it travels a distance in this time. Neglect all other resistance to motion. What are the limits of cable tension?

Solution

Here W = 25kN

For downwards motion of elevator, using D’Alembert’s principle, we get T = W – T = W ( ) Now , , 35 = 0 + / T = 2( ) 6 kN Maximum tension T = 25 =( ) 6 Example 20

A collar ‘P’ is free to slide along the smooth shaft ‘Q’ mounted in the vertical frame as shown in the figure below Determine the horizontal accretion ‘a’ of the frame necessary for maintain the collar in a fixed position on the shaft.

P

Q

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 87 Solution: F B D of the Collar ‘P’

Consider mass is “m’

Acceleration of ‘P’ & ‘Q’ same, ∴ ∑

N cos -mg = 0  1 &

∑

ma-Nsin = 0  2

From eq 1 & 2 we, can write 

a= gtan Example 21

A load W is used to raised by a rope from rest to rest through a height h and the greatest tension which the rope can safely wear is

W. Show that the least time in which the ascent can be made is √_{( )}

Solution

The time required will be least if the motion is first with maximum possible acceleration and then with maximum possible retardation. The maximum possible acceleration is obtained when the tension in the rope, is , and maximum retardation is when this tension is zero.

For the first part when the load is going up, we have by D’Alembert’s principle T –W -

( ) For the second part a =-g

Let t time for accelration

ma N

mg

a x

t time for retardation v maximum velocity

Then acceleration = ( ) Retardation =

Total distance moved

= ( ) Now t ( ) , t t ( ) _{( )} Or ( ) ( ) ( ) ( ) t t √ nh ( ) Example 22

A system of weights connected by spring, passing over pulleys A & B is shown in Fig below. Find the acceleration of the three weights assuming weightless springs and ideal conditions for pulleys.

Solution

Pulley A has two weights 150N and 60 + 40 = 100N acting on it. Let T1 be the tension in the

string for the downward motion of N weight using D Alembert’s principle, we have T1 – 150 + 150/g =0 T1 – 100 -10 a1 /g = 0 Solving these, a1= 1.962 N N 6 N

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 89 For pulley B, let T2 be the tension in the rope and a2 be the acceleration of the 60N downwards.

Again using D Alembert Principle, we get T2- 60 + 60a2/g =0

T2- 40 - 40a2/g=0

Solving, a2=1.962 m/s2

Hence the acceleration of 150N and 60 N weights is 1.962 m/s2 _{downward and the acceleration}

of 40 N weight is also 1.962 m/s2 _upwards.

Example 23

Two blocks A and B are connected to each other by a spring and a string. The string passes over a frictionless pulley as shown in Figure below. Block B slides along the vertical side of a stationary block C and the block A slides along the vertical side of C, both with same uniform speed. The coefficient of friction between the surfaces of the block is 0.2. Force constant of the spring is 1960N/m2_{. If the mass of block A is 2 Kg, calculate the mass of block B and the energy}

stored in the spring. Solution

Free body diagram is shown below

R= mB g T=F= R =  mB g ---(1) For Block A =T- mAg ---(2) From (1) & (2) mAg =  mB g mB = 10 kg T= mAg=19.6N Extension of spring = T/k = 19.6/1960= 1/100 m Energy stored in the spring = ½ kx2 _{= 0.098 J}

B A A B T R F C T

Example 24

Two weights 800 N and 200 N are connected by a thread and move along a rough horizontal plane under the action of a 400N applied to the first weight of 800N as shown in Figure below. The coefficient of the friction between the sliding surfaces of the weights and the plane is 0.3. Determine the acceleration of the weights and the tension in the threads D Alembert’s principle Solution

Using D Alembert’s principle, we have 400 - W2 a /g- R2 –T =0

T - W1 a/g – R1 =0

Solving, a = 0.981 m/s2 _{and T = 80N}

Kinetics of Curvilinear Motion

Centrifugal force = Where r = radius of the path = angular velocity

v = linear speed

g = acceleration due to gravity

Moment of momentum (angular momentum)of the whole body Iω Where I = mk , k being the radius of gyration.

Moment of Inertia

We know that Newton’s laws of motion define the equations that govern motion of bodies in macro scale. As far as translation is concerned,

• First Law: At its crux defines Inertia or the resistance offered by body to change in state(rest or motion). For translation, inertia is same as mass of body & is the resistance offered to change of translation state of the body.

• Second Law: This defines the relation between inertia and force required to induce an momentum change in the body as

( )

• Third Law: This proposes the action and reaction pairs of Forces.

Now, if we extend this to rotation, the concept analogous to these are generated as follows. • First Law: At its crux defines Moment of Inertia or the resistance offered by body to

change in state(rest or motion). For rotation, moment of inertia is the resistance offered to change of rotation state of the body.

• Second Law: This defines the relation between moment of inertia and the torque required to induce a change in angular momentum in the body as

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 91 • Third Law: This proposes the action and reaction pairs of Torques.

The moment of the inertia force on a particle around an axis multiplies the mass of the particle by the square of its distance to the axis and forms a parameter called the moment of inertia (denoted by I). The moment of inertia of an object is defined by the distribution of mass around an axis. It depends not only on the total mass of the object but also on the square of the perpendicular distance from the axis to each element of mass. This means the moment of inertia increases rapidly as masses are distributed more distant from the axis. For example, consider two wheels that have the same mass, one that is the size of a bicycle wheel and one that is half the size. The larger wheel has four times the moment of inertia even though it is only twice the diameter.

• For system of particles, moments of inertia of individual particles sum to define the moment of inertia of a body rotating about an axis ie ∑

• For rigid bodies moving in a plane, such as a compound pendulum, the moment of inertia is a scalar got by integrating over the body mass ie I ∫ r dm ∫ r dV

• For rigid bodies moving in three dimensions, such as a spinning top, the moment of inertia becomes a matrix, also called a tensor with 9 components (6 independent components). For easiness, moment of inertia is written in lumped form as where k is called radius of gyration.

Theorems for computing Moment of Inertia for compound bodies

The main theorems which will help in computing Moment of Inertias are as follows.

• Superposition: The moment of inertia of the body is additive. That is, if a body can be decomposed (either physically or conceptually) into several constituent parts, then the moment of inertia of the whole body about a given axis is equal to the sum of moments of inertia of each part around the same axis ie ∑

• Perpendicular Axis: If Ix, Iy, Iz are moments of inertia around three perpendicular axis passing through the bodies center of mass, then each of them cannot be greater than the sum of two others: For example . Here the equality holds only if the body is flat/planar body (negligible z values ie z ≈ 0)). When body is planar, ∫ ∫(

) ∫ ∫ ∫( ) ∫( ) ( ≈ ) ⇒ ,thereby proving the above identity.

• Parallel Axis: If the objects moment of inertia around a certain axis passing through the center of mass is known (ICM), then the parallel axis theorem or Huygens Steiner theorem provides a convenient formula to compute the moment of inertia of the same body around

a different axis (Id), which is parallel to the original and located at a distance d from it. The formula is only suitable when the initial and final axis are parallel & is given by Id = ICM + Md2

Examples 25

1. Given 2 masses - m@2r & 2m@r from axis of rotation, to find the moment of inertia, we see that this is a particle system and hence have Inet = £\ Ij = 2m(r)2 + m(2r)2 = 6mr2. 2. Given a disk of radius R & mass M, we can divide the continuum into discrete rigid bodies

and take integral as follows. I= L,r2_{dm = f„ 7rnfR nr}2_{(rd6dr^L) = ±MR}2_{. However, note} that in case of a loop/hoop it has same mass and radius,

The Moment of Inertia / = MR2_{because in that case the whole of the mass is distributed at} a larger distance from the rotation/central axis.

3. For a given right circular cone with radius R, mass M and height H about the vertical axis, we have as follows. We divide the cone into stacked discs of diminishing radius as shown and do a summation of individual I’s for getting the total moment of inertia of the

compound body. ∫ ( ) ∫ ( ) ⇒ ( ) On Simplifying, we get

Further, for any axis passing through the apex of cone, parallel to base (3rd _{figure in above}

diagram), we proceed as follows. Here, a common mistake people make is to employ the perpendicular axis theorem but the point to be noted is that it can be used only for planar bodies. So, we again break up the bodies into smaller bodies and do integration as follows. We see that the moment of inertia about an axis parallel to base but passing through the elemental disc shown is given by (obtained by applying perpendicular axis

Figure: Computation of Moment of Inertia for a cone Figure: Computation of Moment of Inertia for a disc

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 93 theorem for the disc which is a planar body). Now, applying the perpendicular axis theorem for getting the moment of inertia about the axis parallel to base passing through the apex, we get IdApex = Id + dm(H - h). Finally, integrating this, we get as follows.

∫ ( ( ) ) ∫ ( ( ) ) (

)which on solving

gives ( )

Similarly we can find the moment of inertias for any given shape. I for some common shapes are given below.

No. Shape I

1 Rod of length L and mass in (Axis of rotation at the end of the rod)

2 Rod of length L and mass m (Axis of rotation at the center of the rod)

3 Sphere (hollow) of radius r and mass m

4 Ball (solid) of radius r and mass m

5 Thin rectangular plate of height h and of width w and mass m

( ) 6 Ellipsoid (solid) of semi-axis a, b, and c with axis of rotation a and

mass m ( )

Example 26

The drum shown in Fig. (a) has a radius of gyration of 30 cm and weighs 1.8 kN. It is supported by means of small hubs which rest in bearings. A weight of 1 kN is attached to one end of a rope, the other end being wrapped around the drum. Neglecting friction in the bearings, determine the acceleration of the weight, the angular acceleration of the drum, and the tension in the rope. Solution

Moment of inertia of drum,

( ) 6

The free body diagram is shown in Fig. (b) Let T be the tension in the rope Using D’ Alembert’s principle, we have for the 1kN weight,

T 37.5 cm ω α (a) 1kN 1kN T (b) 1.8kN W=1.8kN 37.5 cm R T a

∑ ( )

1000 – T- (a) And for the drum

∑ ( )

∴ 6 (b) Now a = r = 0.375 .. (c) Substituting Eq. (c) in (a), we get 1000 – T – ∴ 6 Substituting in Eq. (b), we get 0.375 (1000 – 38.226 )- 16.5 375 – 14.335 - 16.5 = 0 6 / 6 6 / 6 6 Example 27

Two weights and of 8 kN and 5 kN are attached at the ends of a flexible cable. The cable passes over a pulley of 100 cm diameter. The weight of the pulley is 600 N with a radius of gyration of 50 cm about its axis of rotation. Find the torque which must be applied to the pulley to arise the 8 kN weight with an acceleration of 1.5 / . Neglect friction in the pulley bearing. Solution

Let T and T be the tensions in the cable on the weights W and W sides respectively and be the torque applied to the pulley to accelerate the 8000 N weight with 1.5 / as shown in Fig. Using D’ Alembert’s principle, we have for ,

T = (a) and for

(b) For the pulley, we have

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 95 ∑

Or M ( ) (c ) Also a = r (d) From Eq. (d), we get

/ ∴ ∴ ( ) 6 ( ) a 5kN a 8kN W2 W1 α 50 cm 600N

Example 28

A step pulley and mass system is shown in the figure along side. Consider pulley is mass less & frictionless, string is non extendable

Find the m2>m1 and m2 moving downward

Solution F.B.D is,

From the F.B.D, we can write,.. S1-m1a1-m1g & S2-m2a2-m2g 2 m1g S1 a1 m1a1 m2g S2 a1 m2a2

m

m

r

r

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 97 For the rotation of ‘O’ of the step pulley, we can write

x

=r

0

x

= r20

∴

Considering equilibrium at that instant S1r1=S2r2

From eg 2,3,4, we can write

From eq 1 & 5 we can write

( )

( )

Kinetic of system with variable mass

( )

Where m = mass of the system v = Velocity of the system F = Force Consider mR= mass of rocket 0 n2 n1

mf= mass of the fuel

vR = Velocity of rocket

vf = Velocity of the fuel

( ) Now vr=v

And vf= -(u-v) u=Velocity of burn fuel with respect to rocket

( ) m0 = Initial mass of the system

( ) ( ) ( ) ( ) ∫ ∫ ( ) ( )

Friction

Friction is the "evil" of all motion and always resist relative motion between two bodies. Friction is actually a force that appears whenever two things rub against each other. Although two objects might look smooth, microscopically, they're very rough and jagged, as this picture shows:

Figure: Microscopic Source of Friction

1 Mass 2 1 2 u Rocket Fuel (u-v)

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 99 As they slide against each other, their contact is anything but smooth. They both kind of grind and drag against each other. This is where friction comes from.

But friction is not all bad. In fact, it has a lot to do with life as we know it here on Earth. Without it, we wouldn't be able to walk, sit in a chair, climb stairs, or use a mouse to surf the web. Everything would just keep slipping and falling all over the place. When we walk, we try to push Earth backward, but friction opposes this relative motion by a forward reaction on the body which pushes us forward making us walk. Another important example of friction is rolling. Friction is usually distinguished as being either static friction (the frictional force opposing placing a body at rest into motion) and kinetic friction (the frictional force tending to slow a body in motion). In general, static friction is greater than kinetic friction.

The force due to kinetic friction is generally proportional to the applied force. However, there is no strong consistent theory for friction and so, based on experimental observations, an empirical result is developed whereby a coefficient of friction is defined as the ratio of frictional force to the normal force on the body.

We'll carry out a thought experiment. Take a mass (say 1Kg) and start applying a force on it horizontally. Plot the frictional force and force applied against each other as shown below.

Figure: Friction thought Experiment

 Apply force of say 0.1N. The body does not move. So, body is in equilibrium ⇒ ∑ ⇒ Same continues till some critical . Note that in this whole region till

, the frictional force is same as applied force and hence a line is traced as shown

in figure.

 Apply force F till the body just start moving. Then in this limiting case = F . Now,

in this limiting case, let us denote frictional force f N where N=normal reaction=mg in this case. So, maximum frictional force is in this limiting static condition & is f = N.  Further increasing F, we see that body now moves with some acceleration. So, by

Newton's Law, we have F f ma. But, now on, we see that remains constant. Let us denote f N. Note that there is a kink in plot which shows that (f Dynamic Friction Force) (f Static Friction force)

f a F F F m m f =friction force F= Applied Force , /

Hence, we conclude that static friction force is not constant but dynamic friction force is always constant.

The study of friction is called tribology.

Rolling Friction

Rolling is direct consequence of friction. However, the mechanism differs slightly. As shown below, the microscopic surface irregularities act like microscopic gear tooth which mesh with each other, there by the contact force appear as friction forces as shown below. So, clearly, when we take the free body diagram, we see that ∑ (also in pure rolling, , ) . frictional force induces rolling in Force/Torque free motion. But again, the difference in mechanism can easily also help conclude that magnitude of rolling friction is only 1 or 2 orders lesser than sliding friction & just as in sliding case, we have frictional force due to rolling Coefficient of rolling friction. Example: A truck in neutral rolling down the incline has friction force pointing up the incline/backwards.

Figure: Rolling Friction Mechanism in Force/Torque Free Case

However, in case of forced/torque motion the case changes, as shown below. Here, when the body is generating a Torque for motion, this causes the microscopic tooth to push back at ground just like walking. Consequently, by Newton's third law, ground pushes back with a reactive frictional force as shown above which causes friction.

Figure: Rolling Friction Mechanism in Forced/Non-zero Torque Case

Example: A truck moving up an incline in gear has friction force direction forward/up the incline T C , O T , C O

At point of contact, surface irregularities act like Gear teeth meshing each other

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 101

Impulse and Momentum

Or

For a finite period of time, integrating, we get ∫ ∫

If is constant, above equation may be integrated, giving

Where indicates vector difference of two momentum terms.

If the forces are variable, they must be given as a function of time and should be similarly integrated. Forces that cannot be expressed mathematically as a function of time may be plotted on a force-time curve, in which the area under the curve is equal to the left side of the equation. Linear impulse of a force is defined as Ft and linear momentum is defined as mv. Thus, it may be expressed as follows:

Ft = mv – mvo

The resultant impulse of the external forces acting upon a body is equal to the change of momentum of the body. Both impulse and momentum are vector quantities. The units of impulse and momentum are Ns.

Conservation of Linear Momentum

If the sum of the external forces acting on any system of mutually attracting and impinging bodies resolved in any direction is always zero, the total momentum of the system in that direction remains constant during the motion.

Suppose a system consists of only two bodies that are in contact for a period of time t. Let there be no external forces acting on this system According to Newton’s third law of motion, if one body exerts a force on a second body, the second body exerts an equal and opposite force on the first since only a pair of equal and opposite forces act on the system, the resultant impulse acting on the system must be zero. It follows, then, that the net change of momentum must also to be zero.

Let the two bodies have masses and with velocities and respectively, before coming into contact with each other, and velocities and at the end of the period of contact. Then according to the conservation of linear momentum, we have

Example 29

A gun weighing 80kN shoots a 250N projectile in a horizontal direction with a muzzle velocity of 500m/s. The bullet weighs 150N and is assumed to move with one-half of the muzzle velocity of the projectile. Find the initial recoil velocity of the gun. If the recoil is resisted by a constant force, find this force if the recoil distance is 80cm.

Solution

Let v be the recoil velocity in m/s Total momentum =

Since all portions of the system are at rest initially, using the principle of conservation of momentum, we have

8 6 /

The negative sign indicates that the gun moves in a direction opposite to that of the projectile. Now using the work-energy relation, ∆ , we have

( ) ] ( )

= 21kN

Example 30

A car of mass 750kg is moving East with a velocity of 30km/h and collides with a second car of mass 1200kg moving NW@ W with a velocity of 45km/h. Find the velocity of the two cars after impact, assuming that they remain constant.

Solution

Applying the principle of conservation of momentum, we have | ( )

| | 6

From the vector diagram shown in Fig. we have

= ( ) ( 6 ) 6 cos = 133.125 + 766.847 - 541.924 = 448.048 21.17 km/h tan = /√ /√

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 103 =

= 6 66 6 66 Or = West of North

Therefore, the direction of the motion of the cars after impact is N W. Example 31

A body of mass 6kg moving with a velocity of 3m/s meets a body of mass 4kg moving(a) in the same direction, (b) in opposite direction, with a velocity of 1.5m/s. If they coalesce into one body, find the velocity of the compound body and the loss of kinetic energy.

Solution

(a) Let be the velocity after coalescing, then 6 (6 ) 18 + 6 = 10 ∆ ( ) ( ) , = (6 ) (6 )( ) = ( 6) 2.7N.m (b) (6 )(6 ) 12 = = 1.2m/s ∆ (6 ) = 24.3N.m N E 11.538 27.692

Example 32

A shot of mass m penetrates a thickness s of a fixed plate of mass M. Prove that if M is free to move, the thickness it penetrated is

( )

Solution

Let be the velocity of the shot and F be the average resistance to penetration. When the plate is fixed, using ∆ we get

Let be the thickness penetrated when the plate is free to move, then using the principle of conservation of momentum, we have

( )

and using the principle of work and energy, we have

( ) , = = ( ) ( ) ( ) ( )

Collision of Elastic Bodies

If two bodies suddenly collide, an impulsive force, or impact, is set up between them. When the direction of each body is along the common normal at the point where they touch, the impact is said to be direct. When the direction of motion of either or both, is not along the common normal at the point of contact, the impact is said to be oblique. If the pressure exerted on the surfaces of contact coincides with the line joining the mass centers of the bodies, the impact is central. If such is not the case, it is eccentric.

For a very short period of time after the two bodies come in contact, the mass centers continue to approach each other. This is known as the period of deformation. During this period the intensity of the force between the surfaces increases. For an instant at the end of the period of deformation, the mass centers are moving with the same velocity. If the bodies are elastic, the impulsive forces causes centers to begin separating and, after a second short interval, the surfaces of the bodies are no longer in contact. This second short period is known as the period

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 105 is known as the period of restitution. Time of impact is the sum of the period of deformation and period of restitution. The time of impact is very small. For this reason, the resultant impulse of the external forces acting on the system during this time must be small and can be neglected. On the bodies of this assumption, the sum of the momentum before impact is equal to the sum of the momentum after impact, i.e, the conservation of momentum holds, thus for direct central impact, we have

Coefficient of Restitution

For direct central impact Newton verified experimentally that the relative velocity after impact is in a constant ratio to the relative velocity before impact. If the bodies collide obliquely, the same fact holds for their compound velocities along the common normal at the point of contact. This ratio is known as the coefficient of restitution, and is denoted by e. Thus

In which the proper sign of the four velocities must be included.

The value of e lies between zero and one. It is zero for perfectly inelastic bodies and one for perfectly elastic bodies.

Example 33

A ball of mass 4kg moving with a velocity of 1.5 m/s is overtaken by a ball of mass 6kg moving with a velocity of 3m/s. (a) in the same direction as the first, (b) in the opposite direction. If e = 1/5, find the velocity of the two balls after impact. Find also the loss of energy in the first case. Solution Here = 4kg, = 1.5m/s, = 6kg, = 3 m/s, e = 1/5 = 0.2

(a)

2.5 = 5.25 = 2.1 m/s = 2.85m/s

Loss in kinetic energy = ( ) ( ₎ = ( 6 ) ( 6 ) = 6 N.m

(b)

A ball weighing 1kg and moving with a velocity of 3m/s impinges on a smooth fixed plane in a direction making 6 with the plane. Find its velocity and direction of motion after impact, the coefficient of restitution being 0.5. Find also the loss in kinetic energy and the impulse on the plane due to the impact.

Solution:

Here v = 3m/s, m = 1kg α , e = 0.5

Now (sin cos ) = 9(sin cos ) = 9(0.25 + 0.25 ) = 3.9375 1.984m/s

Also cot cot cot = 0.866

Loss in kinetic energy ( , ₎

( ) = 2.583N.m Impulse = ( ) cos

= 1(1 + 0.5) cos

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 107 Example 35

A solid cylinder, as shown in Fig. is supported on frictionless bearings. Find the angular velocity of the cylinder and the tension T in the rope, 8s after the system is released from rest.

Solution

The impulse-momentum equation for the 600N load is ( )

(600 – T) 8 =

( ) ( )

The angular impulse-momentum equation for the cylinder is: ( ) ( 6) ( 6) ( ) 6 ( ) Now 6 ∴ Eq.(a) becomes (6 ) 6 6 6 ( ) Comparing Eqs. (b) and (c), we get

∴ T = 375N rad/s 6 = 29.43m/s 60cm 2kN 600N T

Conservation of Angular Momentum

According to this principle, if a system of two rotating bodies are brought into contact for a short time period, and no external torque is applied to the system during this time, the resultant angular impulse on the system is zero.

Suppose the two bodies have moments of inertia and and angular velocities and repectively, before coming, into contact, and angular velocities and at the end of the period of contact. Then the principle of conservation of angular momentum may be stated as

Example 36

The 2.5kN body shown in Fig. rolls freely on the inclined plane and has a centroidal radius of gyration of 0.6r. Determine the velocity of each body and tension T, 4s after the system has been released from rest.

Solution

= ( 6 ) 6 6 For the rolling body,

= ( )

∴ ( sin ) 6 6 ( )

6 6 . . . (a)

For the translating body, ( ) ( ) 6 . . . (b) 1.5kN R F=µR 2.5kN O T r

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 109 Now = 2r

Or =

∴ Eq. (a) becomes

Or = 86.65 ( ) Adding Eqs. (b) and (c), we get

3500 = 239.55 ∴ = 14.61m/s .305m/s 6 6 6 N Example 37

A pile hammer weighing 12kN drops from a height of 75 cm on a pile weighing 6N. If the pile penetrates a distance of 4cm for every blow, find the resistance of the ground assuming it to the uniform and the impact to be plastic.

Solution

Velocity of hammer on impact

√ √ 6m/s

Let be the velocity of the pile and hammer after impact, then applying the principle of conservation of linear momentum, we have

6 6 ( 6 ) = 2.557m/s

Let be the resistance of the ground. Using the principle of work end energy ∆ , we get ( )

( )

6 6 N

Example 38

A cricket ball of mass 150g is moving with a velocity of 12m/s and is hit by a bat so that the ball is turned back with a velocity of 20m/s. The force of blow acts for 0.01s on the ball. Find the average force exerted by the bat on the ball.

Solution

m/s, = 0.01s Change in momentum

( ) _{( ) = 4.8kg m/s}

Force = Rate of change of momentum = = 480N

Example 39

A body of mass 8kg is moving 2m/s without any external force on it. An internal explosion suddenly breaks the body into two equal parts and 16 joules of translational kinetic energy is imparted to the system by the explosion. Neither part leaves the line of original motion. Determine the speed and direction of each part after the explosion.

Solution

Let and be the velocities of the disintegrated parts (Fig.)

Applying the principle of conservation of momentum, ( )

or = 4 . . . (1)

Now applying law of conservation of energy. 6 ( )

Or = 16 . . . (2)

Substituting the value of from (1) in (2), ( ) = 16 16+ = 16 2 = 0 Or = 4m/s And = 0

Work and Energy

Work: If a force acts on a body and causes it to move some distance, work is said to be done by the force. Thus, work is a measure of accomplishment. Therefore, work done by a constant force is equal to the product of the force and the displacement of its point of application in the direction of the force. It is measured in Nm.

2m/s

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 111 Energy: The capacity to do work is called energy. It is measured in N.m.

Potential Energy: This is the energy which a body possesses because of its position. Kinetic Energy: This is the energy which a body possesses because of its velocity. Power: The rate of doing work is called power.

Work Done by a Force

The work done by a force is equal to the product of the force, F, and its displacement, s, provided the force is constant and the displacement, of the body is in the same direction as the force. Denoting work by , we have

, = F.s

Relation between work and change of kinetic energy Net work = change in kinetic energy

, ∆

Where represents kinetic energy. This equation represents the principle of work and energy.

Conservative/Non-Conservative Force Fields and Energy Balance

We know work done is the dot product between force F & displacement s ie W=F.s. Further, assume we have a body of mass m being acted upon by a spatially varying force F(s) which is a function of displacement (Ex: Spring force which is ks where s is the displacement from equilibrium). Then, assume that the body moves from position s to a small change in position s + ds, during which time we can assume that F(s) remains constant. Then, the work done to achieve this motion is

dW = F(s).ds

Now, assume that during this motion, the velocity changes from v to v+dv. By Newton’s law, if

a(s) denotes the acceleration of the body during this motion,

F(s) = ma(s)⇒ dW ma(s) ds

But we know ⇒ ( ) ⇒ Then, if we move from position 1 to 2, then the total work to be done for this is

∫ ∫ ∫ = Change in Kinetic Energy Further, suppose for a given force function F(s), we can find some V(s) such that F(s) = — ∆V(negative gradient of some function V), then we get as follows.

∫ ∫ =Change in Function V between points ⇒ ⇒ ⇒

The function V is usually called Potential Energy Function. So, if we are able to find a potential function V which satisfies the above condition, clearly, then the quantity (V + KE) ie Sum of Potential and Kinetic Energies is conserved. This brings us to 2 main classifications of forces. • Conservative Forces: Those force fields F(s) for which a function V can be found out such

that F(s) = –∆V In these cases, KE+PE=Constant. Examples: Gravitational force mg whose potential function is mgh(Gravitational Potential Energy), spring force kx whose potential energy function is

• Non-Conservative Forces: Those force fields F(s) for which there is no potential function which satisfies F(s) = —∆V exists Examples Friction force, Damping force etc. In this case,KE PE Constant

Therefore, in a generic scenario with both conservative and non-conservative forces, one has to apply energy work balance and not energy conservation which in turn comes out of the

following form.

Power = (F )

v is the velocity of the point where the force F is acting.

is the angle between the directions of the force and the velocity. If both are in the same direction then .

One metric horse power = 735.5 watts

Work of the Elastic force: If a prismatic bar of area of cross section A, length and elastic constant E is stretched then the work of elastic force can be calculated by treating it as a spring of

stiffness k.

Principle of work and Energy

Work energy principle: The work done by a force acting on a particle during its displacement is equal to the change in kinetic energy of the particle during that displacement.

U ( ) & are

Net W.D. by the force for displacing a body from (1) to (2) ( )

W.D. by a force for displacing a body from (1) (2) is positive (+ve) and from (1) ⟵ (2) is negative ( ve).

Principle of conservation of energy: means that the sum of the potential energy and the kinetic energy of a particle (or of a system of particles) remains constant during the motion under the action of conservative forces.

K E P E K E P E

THE GATE ACADEMY PVT.LTD. H.O.: #74, Keshava Krupa (third Floor), 30th Cross, 10th Main, Jayanagar 4th Block, Bangalore-11 : 080-65700750,  [email protected] © Copyright reserved. Web: www.thegateacademy.com Page 113 Work of the gravity force

( ) is positive upwards is negative upwards.

( )

Force exerted by the spring is not a constant force but it varies linearly with the displacement from the undeformed position.

U ( )U ∫ du ∫ F dx

sign indicates that Force and displacement are in opposite directions.

∫

If a particle of mass m is moving with velocity , it’s kinetic energy k E is given by

Let v and be the velocities of the particle at points 1 and 2 and the corresponding distance be and . U ( ) (K E ) (P E ) (K E ) (P E )

W.D. by the springs +ve ⟸ W.D. by the system (springs) W.D. on the system ve. ⟸ W.D. on the system (springs) {

When spring is stretched W D by force is –ve ie work is done on the spring When the spring returns towards undeformed position W D is ve(or)work is done by the spring Example 40

Block A in Fig. weights 500 N and block B weights 200 N. The coefficient of friction under B is 0.2. Find the velocity of each block after A has moved 6 m from rest. Also find the tension T in the cord attached to B.

Solution

By inspection of the weights, we observe that A will move downwards while B will move up the inclined plane. Also B will travel by half the amount of A.

R T T T T T A B W C