Vectors fundamentals CBSE

8/10/2019 Vectors fundamentals CBSE

1/53

FZK 101DERS NOTLARI

VECTORS

Scalars:Quantites which have only magnitude ( mass, length , energy...)Vectors:Quantites which have both magnitude and direction ( velocity..)

Adding Vectors :Axiomatic Definition:

Null vector : A + 0 = A 0= null vectorInverse Vector : A + B= 0 A = -B Binverses vector of ACommutative Law : A + B = B + A

Associative Law : ( A+ B ) +C= A+ ( B +C )

1)

Graphical Method :

B C B A + (-B ) = C

A + B = C A

A C

2) Analytic Method : This method is useful for 3- dimensional vectors. The processs is

called revolving a vector into its components , draw perpendicular lines from A to the

axis , the quantities ax and ay so formed are called the components of the vector A.

ax = | A|. cos |A| = [ ax2

+ ay2

]ay = |A| . sin tan = ay / ax

ay A

ax

Unit Vector: It is a vector of unit length in a given direciton. These unit vectors are

symbolized by i , j and k for x, y and z respectively.

| UA| = 1 | UA| = A/ | A | A = Ax. i + Ay. j +Az.kor ( A = ( Ax, Ay)

3-D:

ax= A.sin . cos az ay= A. sin .sin

A az= A . cos ay

ax


2/53

Changing Coordinate Systems:a) By moving the origin to another location in the xy-plane

y y

ay

ayax x

x

ax

b) By pivoting the xy axes about the fixed origin which is a rotation of the coordinate

systems.

y

ax x

ay

ax3 ) Component Method :

A = ( Ax, Ay, Az) A + B = C C = ( Ax+Bx, Ay+By, Az+Bz)B = ( Bx, By, Bz)

Multiplication of Vectors :

1)

Multiplication of a vector by a scalar :

k .A= B A // B

2) Multiplication of Two Vectors to Yield a Scalar : ( Dot Product )

The scalar product of two vectors a and b written as a . b is defined to be ;

A. B = | A | . | B | .cos

A ( Bu tr bir arpm bir vektrle zerine den bir

izdmn arpm. )

b. cos ( Bu tr arpma dot product yada scalar arpm

da denir )

a. cos B

A = ( Ax, Ay, Az) A. B = Ax.Bx+ Ay.By + Az. Bz

B = ( Bx, By, Bz) | A |. | B | .cos = Ax.Bx+Ay.By+ Az.Bz

Properties :

a) A.B = B. A

b) A .( B + C ) = A. B + A .C

c)

(a+ b).A = a.A +b. Ad)

d (A. B ) / dt = (d.A / dt) .B +(d.B / dt) .A


3/53

3 ) Multiplication of Two Vectors to Yield Another Vector : ( Cross Product )

The vector product of two vectors a and b is written as ax band is another vector c ,

where c = a x b. The magnitude of c is defined as ,

| A x B | = | A |. | B| . sin

( Bu tr arpmlar 3 boyuta evrilmeli. Product olan c vektrnn yn , a ve b vektrlerininyzeylerine diktir ve sa el kuralyla bulunur. )

c = a x b b Sa el kural :4 parmak arpmda nce gelen

vektt ynndedir. Avu ii dier

b vektr gsterir.Ba parmak c vektrnn

a ynn gsterir.

a c =b x a

|c| = |c|= |a|.|b|.sinBu tr arpmlara vector product yada cross product denir.

sayfa iine doru sayfa dna doru

i x i = j x j = k x k = 0

k i x j = k j x i = -k

j x k = i k x j = -i

j k x i = j i x k = -j

i

A = ( Ax,Ay,Az) A x B = i j k

B = ( Bx, By, Bz) Ax Ay Az

Bx By Bz

i j k

Ax Ay Az = i.Ay. Bz+ AxBy.k + Bx.j.AzBx.Ay.kiBy. AzAx.j.Bz

Bx By Bzi j k

Ax Ay Az

Properties:

a) A xB = -( A x B )

b) A x ( B + C ) = ( A x B ) + ( A x C )

c) d (A x B )/ dt = ( dA/ dt x B )+ ( dB/ dt x A )


4/53

Direction of a Vector :

Ax= A. cos xAz cosx= Ax/ A

z cosy= Ay/ Ay Ay cosz= Az / A

Ax x cosx2+ cosy

2+ cosz2= 1

B

A B = |A|. |B|. cos

cos = A. B /|A|. |B| = ( AxBx+ AyB + AzBz) / [ Ax2+ Ay

2+ Az2].[ Bx

2+ By2+ Bz

2]

A

Exercises1) Prove that two vectors must have equal magnitudes if their sum is perpendicular to their

difference.

a (x1, y1) | a | = ( x12+ y1

2) ( x12+ y1

2) = ( x22+ y2

2) olduunu

b ( x2, y2) | b| = ( x22+ y2

2) ispatlamamz gerekli

a + b ab

( x1+ x2, y1+ y2) ( x1x2, y1y2)( x1+x2)(x1-x2) + (y1+y2) (y1-y2) = 0

x12x2

2+ y12y2

2= 0

x12+ y1

2= x22+ y2

2 ( x12+ y1

2) = ( x22+ y2

2)

2) a = + 3j2k Find a.(b x c) and a x ( b+c)

b = -i4j +2k

c = 2i + 2j +k

a ) a. ( -8i, 6k , 5j ) = ( 3i + 3j2k ) . ( -8i + 5j + 6k ) = -24 + 1512 = -21

b ) ( 3i + 3j2k ) x ( i- 2j + 3k ) = ( 5i11j8k )

3) a = 5i + 4j6k r = a- b + c

b = -2i +2j +3k a) Calculate angle between r and +z axis .

c = 4i + 3j + 2k b) Find the angle between a and b .

a) z = ( 11i + 5j7k ) ( 0, 0, 1) . ( 11i + 5j7k )

-7 = 1. | (121 + 25 +49 ) | . coscos = -7 / 185 = arccos ( -7/185 )

b) 10 + 818 = 45.17.cos

-20 / 4517 = cos


5/53

4) Show that the magnitude of a vector product gives numarically the area of the

parallelogram formed with the two component vectors. Does this suggest how an element

of area oriented in space could be represented by a vector ?

A D

b | a x b | = |a|. |b| .sin A( ABCD ) = 2(ABC)

= 2 |b|.sin.|a| / 2 = | a x b |B C

a

5) Show that a( b x c ) is equal in magnitude to the volume of parallelepiped formed on the

three vectors a, b nad c as shown in the figure.

a = ( x1,y1,z1) | a.( b x c) | = |a|.| b x c |

b = ( x2, y ,z2) = |a|.|b|.|c|.sinc = ( x3, y3,z3)

ekildeki gibi bir aktarm yaplrsa paralelkenar bir

dikdrtgenler prizmasna dnr.

v = |a|.|b|.|c| 0 |a|.|b|.|c|.sin 90

6 ) |a| = 3

|b| = 4 a) Calculate the x and y components

|c| = 10 c b) c = pa +qb p = ? q = ?ba

30a)

a = ( 3,0 )

b = ( 23 , 2)c= ( -5, 53 )

b) ( -5, 53 ) = ( 3p + 23q , 2q )53 = 2q -5. 3p + 23. 53 / 2 q = 53 /2 p = -20 / 3

MOTIONPosition Vector ( r ) :

(t) The particle is located by x , y and z

( r ) which are components of the vector r

that gives the position of the particle.

r = xi + yj+ zk


6/53

Displacement Vector : ( r )

A

r r = rB- rA

rA At r, origin and path are independent.rB B This means that what they are is not important.

Trajectory :trajectory

equation of trajectory y = y(x)

Worldline :

xB B

xA A equation of worldline x = x(t)

tA tB

Average Velocity :

Vav= r / t = ( rBrA) / t BA

tan = Vav Vavis the slope of the worldline.

Ex: You drive your car down a straight road for 5,2 km. at 43 km/ h. , at which

point you run out of gas? You walk 1,2 km. farther to the nearest gas station, in 27 min.

What is the average velocity from the time that you started to drive your car to the time that

you arrived at the gas station ?

V = x / t = ( 5,2 + 1,2 ) / ( t1+ 27/60 h ) = 6,4/ (0,12 + 0,45) = 6,4 / 0,57 = 11,2 km.

Average Speed: This is a scalar quantity.

Vav= total distance traveled / time interval

Ex:A runner runs 100 m. in 10 sec. and then walks back 100 m. in 80 sec.. Vav= ? Vav= ?

Vav= 0 / 30 = 0 Vav= 200 / 30 m/s

Vav mi Vav mi olduu ok nemli !


7/53

Instantaneous Velocity:

V = limt0r / t = limt0[r(t+t)- r(t) ] / tV = dr / dt

r(t) Vx= dr/dt Vy= dy/dt Vz= dz/dtr(t+t)

Average Acceleration:

aav= V / tInstantaneous Acceleration : a = limt0V / t = dV / dt = d

2r / dt2

ax= dVx/ dt ay= dVy/ dt az= dVz/ dt

Properties:1) r = V V = a

2) V t

dV = a dt V = a dt + V0V0 t0

r = r0+ V(t) dt

Ex: r(t) = ( 2t35t ).i +( 67t4) .j

Calculate r, V and a when t = 2 sec.

r(2) = 6i + ( 6- 7.16 ) .j V = dr/ dt = ( 6t25)i128t3.j

V (2) = 19i + 28.8j

a = dV / dt = 12t.i + 28.3t2.j

Ex: r(t) = i+ 4t2.j + t.k a) V(t) = ? a(t) = ? b) What is the shape of the trajectory?

a) V(t) = dr/ dt = 8t.j + k a(t) = dV/ dt = 8.j

b) x = 1 y = 4t2 z = t y = 4z2 ( this is a parabola )

Ex: V0= 30.i a = ( 5+ 3t3).i + (45t2).j

After 4 sec. what is the particles position and V ? ( x0= 0 )

V = a(t) dt = ( 5t + 3 /4t4+ 30.i ) + ( 45t - t3/3).jV(4) = ( 20 + 3.43+ 30 ).i + ( 18043/3 ).j

x = V(t) dt = ( 5/ 2.t2+ 3/ 20.t5+ 30.t ).i + ( 45/ 2.t2t4/ 12 ).jx(4) = ( 5/ 2. 16 + 3/ 20.45+ 120 ).i + ( 45/ 2.1644/12).j

Motion With Constant Acceleration

V(t) = a.t.dt + V0V(t) = a.t + V0 r(t) = 1/ 2.a.t + V0.t + r0

Ex: t = 0 V0= 3.6.i r0= 0 a = (-1.2.i , -1.4.j )

t = ? V = ? r = ? when x = xmaxx(t) = 1/ 2.(-1.2).t2+ 3.6.t = -0.6.t2+ 3.6.t

a) To maximize x(t) we differentiate it.

x(t) = -1.2t + 3.6 = 0 1.2t = 3.6 t = 3

b) V = at + V0 V = at + V0 V = ( -3.6i ,-4.2j) + 3.6i = -4.2j


8/53

c)r(s) = 1/ 2.(-1.2, -1.4 ).9 + ( 3,6. 30,6.9 ; -0,7.9 )

t = ( VV0) / a xx0= 1/ 2.( V +V0).t = 1/ 2.(V+V0).( (V-V) / a ) V2V0

2= 2a.(xx0)

Ex: You break your car from a velocity of 35 km/h to 45 km/h over a distance of 105 m.

a)

What is the acceleration?b) What is t ?

c) If you were continuing to break your car with same a, how much longer would it

take for you to stop and how much distance would you cover additional ?

a) 85000/ 3600 = 23.6 m/s 45000/ 3600 = 12.5 m/s

( 23.6)2( 12.5)2= 2a.105 560156 = 210.a a 2 m/s

b) 12.5 = 23.62t 2t 11 t 5.5 sec.c) - ( 23.6)2= -2.2 x 557/ 4 = x x 140 m 140105 = 35 m.

2t = 23.6 t = 11.8s5.5 s = 6.3 sec.

Ex: An particle travels along the inside of a straight hollow tube 2 m. long which formspart of a particle accelerator.

a) What is a if it enters at a speed of 10 4m/s and leaves at 5.106m/s ?

b) How long is it in the tube ?

a) V2V02= 2ax 25.108= 2a.2 108.( 25.1041) / 4 = a

b) V = at +V0 5.106= 108.( 25.1041).t / 4 - 104

4.104( 500 + 1) / 108.( 25.1041 ) = t

Free Falling Bodies

a = +g

V = V0+ g.t

y = 1/ 2.g.t2+ V0.t + y0

V2V02= 2.g.( yy0)

Ex:

h

12 m.

55km/h

Vk

55000/ 3600 = 275 /18 m/sec

275/ 18.t = 12 t = 0,78 sec.

Vk= g.t 0 10 .0,78 = 7,8 m/s

( 7.8 )2= 2. 10 .h h 3 m.


9/53

Ex: A ball is thrown vertically upward from the ground with a speed of 25.2 m/s.

a) How long does it take to reach its highest point ?

b) How long does it rise ?

c)

At what time will it be 27 m. above the ground ? ( g = 9,8 m/s2)

a) V = g.t + V0 0 = -9.8 t + 25.2 t = 2.57 sec.b) 0( 25.2)2= 2.9,8.h h = 635 / 19.6 = 32.39 m.

10,3 = 9,8t t = 1sec. 2,57 + 1 3,6 sec.c) V2(25.2)2= -2.9,8.27 V = [ 635529] = 10.3 m/s 10,3 = -5,8t + 25.2

t = 14,8 / 9,8 = 1,5 sec.

Projectile Motion

V g

V0 H

Vy Vx Vx

R(range) Vy V0

Vxo = V0.cos Vyo= V0.sin Vx= Vxo= V0.cos Vy= Vyog.t V = Vx+ Vy

X = X0+ Vxo.t h = y = y0+ Vyo.t + 1/ 2.g.t2 V2Vyo

2= 2.g.(yy0)

R = V02

.sin2/ g H = V02

.sin2

/ 2g Tuu= 2Vyo/ g

Circular Motion

r

r = x.i + y.j

r x = r.cos y = r.sinr and are unit vectors and r

tan= y/ x

r = r/ r = ( r.cos.i + r.sin.j ) / r r = cos.i +sin.j = cos(+/2).i + sin( + / 2).j = -sin.i + cos.jHence r r.= 0Then the position vector r= r0. r

The velocity vector V = dr/ dt V = d(r0.r )/ dt = r0.dr / dtdr /dt = d/dt . dr/ d= .( cos.j + sin.i ) = .V =r0. . V= r0.w. V = r0. w

The acceleration :

a = dV / dt = d( r0.w. ) / dt = r0.( dw/dt. + d/dt. w )Let dw/dt = d2/dt2= ( angular acceleration )

d/dt = d/dt. d/d= ( -cos.i - sin.j) = -.r = -w.ra = r0.. r0.w2.r a = -ar.r + at. ar= r0.w2= V2/ r0

at= r0.= dV/ dt ar= radial acceleration at= tangential acceleration


10/53

Uniform Circular MotionParticle moves with a constant speed in a circular path. So both the velocity and

acceleration are constant in magnitude but they change their directions continuously. The

rotation of the earth is an example of uniform circular motion.

r= r0.r V= V. a= -ar.r ar= V2/ r0= r0.w

2

Frequency is the number of the tours that are traveled in 1 seconds. (f)

w = 2.f f = V / 2r0 w is constantw = d/dt w = d/dt wt + = = integral constant

r= r0.os( wt + ).i + r0.sin( wt + ).j

Ex: P travels with a constant speed r = 3m. and T = 20 sec when t = 0 particle is on the

3 point 0. With respect to the origin 0 , find

a) R(t) = ? ( position vector )

P b) V(t) = ?

2 c) a(t) = ?

d) R1,2= ? , R2,3= ?

10

a) R = r.j +r = w.t + R = r.j + r.cos( wt + ).i + r.sin(wt + ).jR(t) = r.cos( wt+).i + r.(1+ sin(wt+) ).j

then = ? when t = 0 and R = 0 = ?R(0) = r.cos .i + r.( 1+sin).j = 0

cos = 0 ( 1+sin) = 0 = 3/2cos( wt + 3/ 2 ) = sinwt sin ( wt+3/2) = - coswt

R = r.sin(wt) .i + r.( 1-cos(wt) ).j

b) V(t) = R(t) /dt = wr.cos(wt).i + r.w.sin(wt).j

c) a(t) = V(t) /dt = -w2r.sin(wt).i + rw2.cos(wt).j

d) At the point 1 R1= 0

At the point 2 R2= r.i + r.j ( = 90)At the point 3 R3= 2rj ( = 180)

R1,2= R2R1R2,3= R3R2

Ex: At a certain instant V = 17,4 m/s. And its acceleration is an angle aof 22from the direction to the center of the circle.

a) At what rate is the speed of the particle is increasing ?

b) What is the magnitude of the acceleration ?

( r = 3,64 m. )

a) this means what is the a ? ( dV/dt)

ar= V2/r = a.cos at= a.sin= V

2/r .tana = ar/cos= V

2/ r.cos= ( 17,4)2 / (3,64. cos22)

d) a = V2/ r.cos= ( 17,4)2/ ( 3,64.cos 22)


11/53

Ex: A particle moves in a plane according to

x = R.sin(wt) + wRt y = R.cos(wt) + R ( w and R are constants )

This curve is called a cycloid is the path traced out by a point on the rim of a well which rolls

without slipping along the x-axis.

a) Sketch the path.

b)

Calculate the instantaneous velocity and acceleration when the particle is at its max andmin value of y.

a)

R 3Rb) Vx= dx/dt = R.w.cos(wt) +wR Vy= dy/dt = -R.wsin(wt)

V = ( Vx , Vy)

ax= dVx/dt = Rw2.sin(wt) ay= dVy/dt = -Rw

2cos(wt) a = ( ax, ay)

Relative Motiony

y P S and S are observers and they move

with the speed of V

r1 r= S ye gre P nin yer vektr.

r r1= S ye gre P nin yer vektr

Rss s x R= S nn S ye gre yer vektr

s x

r = R + r1 V = U + V1 a = a1dr/ dt = dR/ dt + dr1/dt

Ex: P

VC= 2mi/h VC= 2 mi/h VB= 4mi/h

4 mi a) Boatun P ye ulamas iin ban

hangi yne evirmesi gerekir?

VB b) P ye gelmesi ne kadar srer?

c) 2 mi aa inp yukar gelmesi ne kadar srer ?

d) En ksa srede karya ulamas iin hangi ynde gitmeli? Bu ne kadar srer ?

a) sin = 2/ 4 = 30 120from the axis.b) | VB + VC| = [ 164 ] = 12 = 23

x = V.t 4 = 23 .t t = 2/ 3 h.c) VC + VB= 6 mi/h 2 = 6.t1 t1= 1/ 3 h.

VBVC= 2 mi/h 2 = 2.t2 t2= 1 h

d) t = t() = L/ VB.cos dL/ d= L/ VB ( -sin/ cos) = 0 =0then tmin= t(0) = L/ VB= h /4 = 1 h.


12/53

Exercises1)

A ball is thrown upwards vertically and

the graph of its place and time is drawn.

height tu According to this graph prove that:H g = 8H / ( tL2- tU2)

tL time

V0= g.tL/ 2 H = -1/ 2.g.( tL/2tU/2 )2+ V0( tL/2t0/2 )

H = [-1/2.g (tLtU)2/4 ] + [ 2.g.tL/2 . (tLtU) /2 ]

8H = 2.g.tL.( tLtU)g.( tLtU)2

8H = g.( tLtU) ( 2tLtL+ tU) g = 8L/ ( tL- tU)2

2)a Show that if d > ( V1V2)

2/ 2a ,

there will be no collision.

V1 V2 Car #1 slows down with the acceleration a.

d

V1> V2 ( V1V2)2= 2a.d ( they collide )

So it must be ( V1V2)2< 2a.d

( V1V2)2/ 2a < d

3) V0= 3,6i a 01,2i1,4j

a) At what time does the particle reach its max x coordinate ?

b)

What is the velocity at that time ?c) Where is the particle at that time ?

a) V = ( -1,2i1,4j ) dt + V0 x = [ (( -1,2t + 3,6 )dt).i , -1,4j ]V = (-1,2t.i , -1,4t.j) + 3,6.i x = ( -0,6t2+ 3,6t).i , -0,7t2.j

V = (-1,2t + 3,6).i , -1,4t.j max x -1,2t + 3,6 = 0 t = 3 sec.

b) V = (-1,2t +3,6 ).i , -1,4t.j V(3) = ( 0 , -4,2j )

c) x = ( -0,6t2+ 3,6t).i , -0,7t2.j x(3) = 5,4.i , -3,6.j

4) a particle A moves along the line y = d( 30 m) with a constant velocity 3 m/s directedparallel to the positive x-axis. A particle B starts at the origin with 0 speed and constant

acceleration a = 0,40 m/s2 at the same instant the particle A passes the y-axis. What angle between a and the positive y-axis would result in a collision between these two particles ?

30 = 1/ 2.0,4.cos.t2 3t = 1/ 2.0,4. sin.t23m/s 150 /15 = cos.t2/ sin.t 15 = sin.t

30m 10.sin= cos.t 15 = sin.10.sincos0,4 m/s2 3 /2.cos= sin 3/ 2.cos= 1cos2

x2+ 3/ 2.x1 = 0 ( x- 1/ 2 ) ( x + 2) = 0

cos= 1/ 2 = 60


13/53

5) A ball is dropped from a height of 39 m. The wind is blowing horizontally and imparts a

constant acceleration of 1,2 m/s2.

a) Show that the path of the ball is a straight line and find the values of R and .b) How long does it take for the ball to reach the ground ?

c) With what speed does the ball hit the ground ?

a) Xy= 1/ 2.g.t2 Xx= 1,2.t

2

Xy= 4,9 t2 Xx= 0,6t

2

Xy= 8,16.Xx this shows a straight line

b) 39 = 1 /2.g.t2 t2= 8 t = 22 sec.

c) V2= 2ax V2= 2. 9,8. 39 Vy= 27,7 m/s

Vx= 1,2. 22 Vx= 3,34 m/s V = Vx+ Vy

6) Prove that for a projectile fired from the surface of the level ground at an angle abovethe horizontal.

a) The ratio H/ R = 1/ 4.tanb) Find the angle of projection of which the max H and R are equal.

a) ( V0.sin)2= 2gH H = V0

2.sin2/ 2gR = V0.cos.2t R = V0.cos.V0.sin.2 / g R = V0

2.cos.sin.2 / gV0.sin= g.t 2t = 2.V0.sin/ g

H / R = V02.sin2.g / 2.2.g.V0

2.cos.sin H / R = sin/ 4.cos= 1/ 4.tanb) V0

2.sin2/ 2g = 2V02.cos.sin/ g tan= 4 = 76

7)

A projectile is fired from the surface of level ground at an angle above the horizontal.a) Show that the elevation angle of the highest point as seen from the figure is relatedto by; tan= 1/ 2.tan.

b)Calculate for = 45.

a) tan= H / R/2 = 2H /RH = V0

2.sin2/ 2g R = V02.sin2/ g

2H /R = 1/ 2.tan tan= 1/ 2.tanb) tan= 1/ 2. tan 45 tan= 1/ 2

V H = Arctan 1/ 2

R

Force and Newtons Law

F1 F2 Fnet = F1+ F2+....+ Fn

Equilibrium: If Fnet= 0 then the situation is called equilibrium.


14/53

Contact Force:

F Ff F and F1are contact forces.

F1

Non-contact Forces: Gravitational force , electromagnetic forces etc...Newtons Law:1) Consider a body on which no net force acts. If the body is at rest , it will remain at rest. If

the body is moving with constant velocity , it will continue to do so.( the law of inertia ).

And a system whose Fnet= 0 is called inertial systems.

2)

F = m.a and a has the same direction with F.

3) When one body exerts a force on a second body , the second body always exerts a force

on the first. Furthermore these forces are always equal in magnitude. To every action

there is an equal and an opposite reaction. If our goal were to study the dynamics of one

body only one force of the action-reaction pair would be considered; the other is felt by a

different body and would be considered only if we were studying the dynamics of that

body.

Ex: a) The ox is pushed from A to B on a

130N frictionless surface. What is its velocity

at B ?

240kg F = m.a 130 = 240.a a = 13/24 kg.m/s2

A B V2= 2ax V2= 2.13/24. 2,3 V = 1,6m/s

2,3 m

d)

If we want to convert its velocity to the opposite direction when the box is at B and

the time is 4,5 sec., What amount of force we have to apply ?

V = at + V0 -1,6 = a.4,51,6 a = 32 /45 m/s2

F = 32/ 45. 240 = 170 N

Ex:m V0 m = 360 kg, V0= 120km/h

The car slows down to 62km/h within 17 sec..

What is the force on m ?

V = V0+ a.t (VV0)/ 17 = a a = -0,95 m/s2

F = m.a F = 360.0,95 = -340 N

Frictional Forcesf is a contact force which acts as a resistance to the motion. It always occurs in the

opposite direction of the motion.

V V

F F

f f


15/53

Normal Force : The force that is opposite to the surface. It is always perpendicular to the

surface.

N

N mg

mg

Force of Static Friction:The frictional forces acting between surfaces at rest with respect to

each other are called forces of static friction . The max. force of static friction will be the

same as the smallest applied force necessary to start motion.

fss.N ( fs)max= s.N fs= static frictions= the coefficient of static friction

Force of Kinetic Friction:Once motion is started ,the frictional forces acting between the surfaces usually decrease so

that a smaller force is necessary to maintain uniform motion .The forces in relative motionare called the forces of kinetic friction.

fk= k.N k= kinetic frictionIn general s> k

fg

(fs)max

fs=s.N

Fstatic region kinetic region

Applications of Newtons Laws

1) Clearly identify the body that will be analyzed. Sometimes there will be two or more such

bodies, each is usually treated independently.

2) Select a suitable ( non-accelerating ) coordinate system.

3) Draw all applied force vectors. ( draw free body diagram)

4) Define a direction for acceleration.

Ex: A sled of mass m = 7,5 kg. is pulled along a frictionless horizontal surface by a cord.A constant force of P = 21 n is applied to the cord. Analyze the motion if

a)

the cord is horizontal

b) the cord makes an angle of = 15with the horizontal .

a)

F = m.a 21 = 7,5a a = 2,8m/s221N b)

21 N F = m.a

15 21.cos15= 7,5.a75N a = 2,7m/s2

75N


16/53

Ex: A block of mass m = 18 kg. is held in place by a string on a frictionless plane inclined

at an angle of 27a) Find the tension in the string and the normal force exerted on the block by the plane.

b) Analyze the subsequent motion after the string has cut.

18kg a) N T T = 18.9,8.sin27= 80NN = 18.9,8.cos27= 153N

27 mg.sin mg.cos

b) If the string has cut t = 0 50 block starts to move along the direction ofmg.sin

F = m.a 80 = 18.a a = -4,45m/s2 ( along the x-axis )

Ex: A passenger of mass 72,2 kg. is riding in an elevator while standing on a platformscale. What does the scale read the elevator cab is ;

a)

descending with constant velocity.

b) ascending with acceleration 3,2 m/s2

a) m.g = 72,2. 9,8 = 708 N

b) f = Nm.g = ma N = m.( g +a ) N = 72,2( 9,8 + 3,2) = 939 N

Ex:m1 m2 m3 a) Find the force exerted on m3 by m2 ( there is

F an acceleration )

b) Find the force exerted on m2by m1.

a) N1 N2 N

F1 F F2 F1 F2

m1g m2g m3g

F2= m3.a

b) F1F2= m .a or FF1= m1.a

Ex:

C =0 mA= 4,4kgA mB= 206kg

s= 0,18 Nk= 0,15 N

0 a) mC= ? to keep the system from sliding.b) If c is lifted , what is a ?

B

a) For A N For B T N = (mA+mB)g

T = Ff = mBg = sN= s(mA+mB)g

Ff T 2,6g = 0,18(4,4+x).g( mA+mC)g mBg x = 10kg.


17/53

b) for A N N = mAg

Fnet= mBg - k.mA.g = ( mA+ mB).afK mBg g ( 2,60,66) = 7a

a = 2,7 m/s2

mAg

Ex: F = 12 N s= 0,6 k= 0,4What is the force exerted on the block by the wall?

F Ff

Fs= s.N = 0,6.12 = 7,2 NN 12N Fs7,2 N

( system doesnt move) and Fs= 5N

5 N Force exerted by the wall 12+5= 13 N

5N

Ex: sWhat is the Fmin? ( Required to hold m against M )

F s.N1= m.gm

M

no friction

For M N2 F / (M +m) = a Fmin/ M+m = N1/MFmin= ( M+m ).mg / s.M

N1

Mg

Ex:Find the Fmaxthat m1moves with m2

m1 s N2For m1 N1 For m2

F m2 Fs Fs F

no friction m1g m2g +N1

Fs= s.m1gF / ( m1+ m2) = a

F / ( m1+ m2) = Fs/m1

F = s.m1g( m1+ m2) / m1 F = ( m1+ m2).g.s

Ex: What is the Fmax so that they move together ?

F m2 s N2 N1Fs F

m1 Fs

m2g

no friction m1g + N2

F / (m1+ m2) = a

F /( m1+ m2) = Fs/ m1 F = s.m2.g( m1+ m2) / m1


18/53

In this type of questions first write F = m.a for the whole system. Then instead of a writethe acceleration of the object moved by fs. Write the total force which is counted in the free

body diagram, instead of fs

Relative Acceleration

A

Ais the acceleration of the car

ais the acceleration of m according to the

observer that is outside

ais the acceleration of m according to theobserver that is inside

a = a + Am

Ex:6m A = 8m/s2 How long later the object m

arrives the point A?

A m

k= 0,5

N Ff = m.a = .mga = 0,5.9,8 = 4,9 m/s2

Ff a = a +A 4,9 = a + 8 a = -3,1 m/s2

x = 1/ 2.a.t2 6 = 1/ 2.3,1.t2 t 2 sec.

mg

The acceleration in the relative acceleration questions is always according to the outsideobserver.

Ex: Surfaces are frictionless.

a) What horizontal a must have relative to them table to keep m with M ?

b) What horizontal force must be applied to

M achieve this result ?

c) Describe resulting motion if there is no

force .

a) N.cos= mg N.sin= m.aa = N.sin/ m = mg.sin/ cos.m = tan.g

b) F = ( M+ m) a F = ( M+ m).tan.gc) N

mgN.cos= m.ayN.sin= m.axN.sin= M.A

mg


19/53

Nm

N.sin= M.AA = ( A,0) a = a +A

a ( ax, ay) tan= ay/ | ax+A |a = ( ax- A, ay)

NMg

Dynamics of Circular Motion

Fnet = m.a radial component Fr= -m.arr tangential component Ft= m.ata r =r.r ( r is constant )

V = V.

a= -a.r+ at.Fnet= Fr.r+Ft.

Dynamics of Uniform Circular MotionWe have to apply a continuous force to an object to keep it on its orbit. This kind of

forces are called centripetal forces .

1)

Horizontal Surfaces :

V V is constant. T = mV2/r

Surface is frictionless

Fr= centripetal force T = Fr

Ft= tangential forceFr= mv

2/ r is always the combination of the forces

that are in the direction of the center.

2)

Conical Pendulum :

V is constantL m.V2/r = T.sin

T m.V2/ L.sin= T.sin

R m.V

2

/ l.sin= mg.sin/ cos tan = V2/r.g = V2/ L.sin.ga

m.g

3) The Rotar :

N = m.V2/r fs= N.sfs m.g/ s= m.V

2/ r

R V2= r.g / sN

mg


20/53

4)Curve Problems :

m rotates with the turn table ;

What must sbe to keep the object away fromescaping ?

0mV2/r = fs mV

2/ r = N.smV2/r = m.g.s s= V

2/r.g

V2= s.r.g

5)Banked Curve :

N N.sin= mV2/rN.cos= m.g

mg.sin/ cos= mV2

/r= 0 mg tan= V2/grR

If 0 s= V2/r.g s= tan

Ex:

s0 find max and minto net move withrespect to the cone.

N m.V2/ r = N.sin- fs.cos

N.cos+ fs.sin= m.gN.cos+ s.N.sin= m.g

m.g

mg.sin/ (cos+ s.sin)mg.s.cos/( cos+ s.sin) = mV2/r

( sin- s.cos) / ( cos+ s.sin) = V2/g.r

V = 2.r. ( sin- s.cos) / ( cos+ s.sin) = ( 2min)2.r / g

max ;mV2/ r = N.sin+ N.s.cos

N.cos= Ns.sin+ m.g

N ( cos- s.sin) = m.g ( 2.max)2.r / g = ( sin+ s.cos) / ( cos- s.sin)

Time Dependent ForcesIf an acceleration or a force equation is given, then we can find the velocity equation

by taking the integral of the acceleration equation.

t

F(t) = m.a(t) V(t) = V0+ a(t).dtt0 t

X(t) = X0+ V(t) dtt0


21/53

Ex: A particle of mass m is subjected to a net force F(t) given by ; F(t) = F0( 1- t / T)i

that is F(t) = F0 at t = 0 and decreases linearly to 0 in time T . The particle passes the origin

with velocity V0i. Show that at the instant t = T that F(T) vanishes , the speed and distance

traveled are given by ; V(t) = V0+ 1/ 2.a0.T

X(t) = V0.T + 1/ 3.a0.T2 where a0= F0/ m

a(t) = F0/m . ( 1-t/T ).i V(T) = [ F0/m .( 1- t/T ) ]dt + VoV(T) = F0.t/mF0t

2/ 2mT + V0= F0.T /mF0.T /2m +V0

V(T) = 1/ 2.( F0T/m ) + V0= 1/ 2.a0T + V0

X(T) =

F0/m ( t1/ 2.t2/T ) +X0= X0+ ( F0t

2/2mF0t3/6mT + V0t )

X(t) = 1/ 3. a0T2+ V0T

Drag Forces ( Velocity Dependent Forces )When an object makes free falling , it is affected by the air friction force. This force is

called drag force .

F= m.a a(V) = F(V) / m a(V) = dV / dt dt = dV / a(V)v t v dV/ a(V) = dt = t t = dV / a(V)

v0 0 v0

EXERCISES:1)

R If the object is turned with a string on the horizontal

surface and it has a velocity V on A and B , then

m what is the tension on A and B ?

TA mV2/ r = T - mg

A TA= mg + mV2/ r

mg

B mg T + m = mV2/ r T = mV2/r - mg

T

2)

T1= ? T2= ? ( m1= m2)

If they are on the same rope, their frequenciesm2 must be equal.

rr V1= 2r V2= 2r2r2V1= V2

T2= m.( 2V)2/ 2r T2= m.2V

2/ r

T1T2= mV2/r T1= mV

2/ r +2mV2/r

T1= 3mV2/ r

T1 T2


22/53

3)

K On the vertical plane, at the point K, the force between

the object and the rail is 2mg. What is the force that is

applied on the object by the rail at L ?

r Kmg 3mg = mV2/r

2mg

L

T

L Tmg = mV2/rT = mV

2/r + mg = 3mg + mg = 4mg

mg

WORK AND ENERGYWork Done by a Constant Force :

m

F W = F.x W is scalar.

x

F

W =F.cos.x W = work done by F

x

W = 1 N.m = 1 joule 1 eV = 1,6.10-19joule

Work is path dependent.

Ex: m

4,65m m = 11,7 kg. Calculate how much work you would do if

m 2,86m you applied a force parallel to the incline to push the block

up at constant speed.

F F.cos= N.sin N.cos+ F.sin= mgN N = F.cos/ sin F.cos2/ sin+ F.sin= mg

F( cos2+ sin2) / sin= mg F = mg.sinW = mg.sin.s= 11,7. 9,8. 2,86. 4,6 / 4,65 = 328 J

mg

Space Dimension Forcesa) One Dimensional Space :

F = m.a F(x) = a(x) dV/ dt = F(x) / mdV/ dx . dx/ dt = F(x) / m

v. dV/dx = F(x) / m m.dV = F(x) dx

m.dV = F(x)dx F(x) dx = 1/ 2.m.V2

W = 1 /2.m.V21/ 2.mV0

2 W = Ek


23/53

b) Two Dimensional Space :

F(x)

Area = F(x) dxArea = W = Ek

A

x0 x

c)Generalization to Higher Dimensions:

F L

ds = the infinitesimal line element

ds ds( dx, dy , dz )

F = ( Fx, Fy, Fz)

dW = F.d.s

i L x

W = F(s) ds W = F(x) dxi x0

If F is constant W = F.s W > 0 , F sW < 0 , F s W = | F|.|s|.cos

If F is constant W = F.s

F is not constant W = F(x) dx

Ex: WF= ? Wfk= ? Wmg= ?F WF= F.cos.x

Wfk= N.k.cos.x = 0 ( N s)Wmg= m.g.s = 0

fk

0 Forces that are perpendicular to the direction of the motion

dont do any work.

Work Done by Gravitation1)

s = ( 0, h) Wgrav= -mgh

h Fgrav= ( 0,-mg ) Wext= +mgh

2)

s = ( 0,-h ) Wgrav= mgh g sWext= -mgh Fexts

h


24/53

3)

x h Wgrav = -m.gh

Work Done by a Spring Force

m F = 0 ( relaxed )

x

Fs< 0 ( streched)

x Fs> 0 ( compressed )

fext Fs Fs= -kx ( Hooks Law )F is the restoring force and always tends to restore the particle to its position x = 0

xg xf

Ws= Fs(x) dx = ( -kx) dx = 1/ 2.k.xi21/ 2.kxg2xi xi

if xi= x0 Ws= -1/ 2.kx2

WFext= 1/ 2.kx2

Ex:Fs=-kx Fext=kx

W = F(x) ( if there is a force that isdependent on x , you have to take its

Ws=(-) Wext=(+) integral )

Wext=(+) Ws=(-)

Ex:A 263 gr. block is dropped onto a vertical

spring with force constant k = 2,52 N/cm. The block

sticks to the spring and the spring compresses 11,8 cm.

before coming to rest. While the spring is being

compressed, how much work done

a) by the gravity force

b) by the spring

c) What is the speed of the block just before it hits the

spring?

d) if this initial velocity is doubled , what is the max. compression of the spring ?

a) Wgrav = +mgx b) Ws= -1/ 2.kx2

c) 1/ 2.mV2+ mgh = 1/ 2.kx2 1/ 2mV2= 1/ 2kx2-mgh

d) 1/ 2mV

2

+ mgx = 1/ 2.kx

2

4.1/ 2.mV

2

= 1/ 2kx

2

mgx4(1/ 2.2,52.(11,8)2263.9,8.11,8 ) = 1/ 2.2,52.(x)2263.9,8.x x = 22,5 cm


25/53

POWERPower: Work done within the unit time.

Pav= W / t Pinstantaneous= dW / dtdim [P] ) [ mL2/ T ] unit of P = watt = joule / sec

1 horse power = 746 watt

P = W /t P = F. x/t P = F.V

Ex: m = 2,8 kg. X(t) = 3t4t2+ t3

a) Find W during first 4 sec.

b) t = 3 sec what is Pins?

a) W = Ek V = dx/ dt = 38t +3t2

W = 1/ 2.mV421/ 2.mV0

2 W = Ek(4)Ek(0)

b) P = W/ t = F.x / t = F.V = ma.V

P = m.dV/ dt. dx/ dt = m.( -8 +6t ) ( 38t + 3t2)

P(3) = m.10.6 = 60 m.

Ex: If P is constant show that V = ( 3Px /m)1/3 where x is the distance traveled from rest.

P = W/ t = Ek/t P.t = 1/ 2.mV2

V = [2Pt/ m] dx/ dt = [2Pt/m ]dx = [2Pt/m] dt x = [2P/m] . 2/ 3.t3V3= ( [2P/m] )3.t3 V = ( 3Px /m )1/3

Exercises1) A swimmer moves through the water at a speed of 0,22 m/s. The drag force opposing this

motion is 110 N. How much power is developed by this swimmer ?

W = P.t F.x = P.t

P = F. x/t = 110.V = 110.0,22 = 24,2 watt2) A ball loses 15 % of its kinetic energy when it bounces back from a concrete wall , with

what speed must you throw it vertically down from a height of 12,4m. to have it bounce

back to that same height ?

mgh + Ek= mgh + 15/100.( mgh + Ek)

Ek= 15/100.mgh + 15/100.Ek

85/100.1/ 2mV2= 15/100mgh

V2= gh/2,8 = 9,8. 12,4 / 2,8 = 43,4 V 6,5 m/s3) A running man has half of the energy that a boy of half its mass has. The man speeds up

by 1 m/s and then has the same kinetic energy as the boy . What are the original speeds of the

man and the boy ?

2. 1/ 2.mVM2= 1/ 2.m/2 .VB

2 4VM2= VB

2 2.VM= VB1 /2.m(VM+1 )

2= 1/ 2.m/2.VB2 2.(VM+1) = VB

2.( VM+1) = 2.VM VM= 1/ ( 21) VB= 2/ (21)

3) An object of mass m accelerates uniformly from rest to a speed Vfin time tf.

a) Show that the work done on the object as a function of time t.

W = 1/ 2.m .Vf2 / tf

2.t2

b) What is the instantaneous power as afunction of time ?

a) W = F.x W = F.1/ 2.a.t2 W = Vf. m /tf .1/ 2. Vf/tf. t2

W = 1/ 2.m.Vf2

/ tf2

. t2

Vf= V0+atf a = Vf/ tf F /m = Vf/tfb) P.t = W P = dW /dt = m. Vf

2/tf2.t


26/53

Conservation of EnergyThe constant forces that are path independent are called conservative forces.

i.e. spring force , gravity force...

W1= W2= W3 W = F.dr12 3 Non-conservative forces are the forces that changes according to time.

Ex: F( x, y) = ( -k1yi, -k2yj)

a) Show that the work done by this force is path independent.

b) k1= k2 is force conservative ?A

W1= F.ds ds= ( dx,dy )0 D A

B 4 A(a,b) W1= F.ds + F.ds F.ds = (-k1ydx , -k2xdy)0 D1 3 W1= -k2xdy + -k1adx = -k1a.b

C A A

W2= + = -k2.a.dy = -k2.a.b2 C 0 C C

A

W3 = F.dr y = m.x y = b/a.x dy = b/a dx0

W3= k1. b/a dxk2. b/a. dx = -ab /2.( k1+ k2)

W1W2W3 so they are non-conservative.b) k1= k2 -k1.a.b = -k2a.b = -ka.b

so F is a conservative force.

Potential EnergyIt is shown as U . It is the energy that the object possesses because of its position.

W = -U F(x) = -dU(x) / dxW = K = -U K + U = 0 ( K + U ) = 0

Let total mechanical energy E = K+ U

E = 0 Energy is conserved.

Ex: F(x) = -3x5x

2

m = 1,18 kg.a) U(x) = ? when U(0) = 0

b) Vi= 4,13m/s at xi= 4,31m what can be Vfat xg= 1,77 mxf x

U(x) = -F(x) dx + U(xi) U(x) = -(-3x-5x2)dxxi 0

U(x) = 3/2 x2+ 5/3 x3

1/ 2.mVi2+ Ui= 1/ 2.mVf

2+ Uf


27/53

1) Gravitational Potential Energy:

f

h U = mgh

i

2) Mass-Spring Systems Potential Energy :

i f Ws= 1/ 2.k ( xi2xf

2)

F(x) = -kx

U = 1/ 2.kx2 U(x) = 1/ 2.kx2

One Dimensional Conservative Systems

U(x) + 1/ 2.mV

2

= EV = [ 2/m. ( E-U(x) ) ]E = F.x Slope of the E-x graph gives the force

E3

E2

EkE1 F(x) = - dU(x) / dx

E0U(x)

The slope is zero at the points E0, E1, E2and E3. So the force changes its direction.

Ex:U(x) = a/x12b/x6 The potential energy of the attraction between two atoms

a) Find the force between the atoms.

b)

Find the equilibrium position between the atoms.

c) Find Eminto break the molecules apart.

a) F(x) =- dU(x) / dx F(x) = -( -12a/x13+ 6b/x7) = 12a/x136b/x7

b) F(x) = 0 dU(x) / dx = 0 -12a /x13+ 6b/x7= 0 x = ( 2a/b)1/6

c) W = U W = - (UiUf)W = - ( a/x0

12b/x06( a/xf

12b/xf6) ) W = -( ab2/4ab2/2a )

W = b2/4a

Higher Dimensional Conservativexf yf zf

U = U( x, y, z ) U = -Fxdx - Fydy - Fzdzx0 y0 z0

-Fx= dU /dx -Fy= dU /dy -Fz= dU/ dz


28/53

Ex: In a certain system of particles confined to x-y plane, the force has the form

F( x,y ) = -kyikxj. If the particle is initially at the origin and finally at a point

( a,b ) , find if the force is conservative or not. If it is conservative find U(x) .

0 b a

b W1= Fxdx + Fydy + -kydx0 0 0

1 W1= -kba

0

2 a a a a b

W2= -kydx + -kxdy + -kydx + -kxdy0 0 a 0

W2= -kab 0

W1= W2 so F(x) is conservative

W = -U W = -( UfUi) W = -Uf Uf= kab U(x) = kxy

Ex:

l = 2,13m = 35 Find the velocity at A.l cos35= (2,13h) / 2,13

1,74 = 2,13h h 0,4mmg.0,4 = 1/ 2.mV2 0,8.9,8 = V2 V = 2,8m/s

m

A

Ex:3L/4 F Length of the chain is L , mass is m

What is the work done to pull the L/4 part of the

L/4 chain ?

= 0 W = mgh W = m/4 .g. L/8W = mgL/ 32

Ex:P

a) What is FNthat affects P at A.

B b) What is the height that the box has to be dropped

5r to make Fnet= 0 at B.

r A

a) 5mgr = 1/ 2.m.V2+ mgr mV2/r = m.8gr / r

V2= 8gr

8mg

mg Fnet= 65 mgb) mgN = mV2/ r gr = V2 kmgr = 2mgr + 1/ 2.mgr

k = 5/2 h = 5/2. r


29/53

Work- Energy TheoremThe energy is not lost in the universe. But it is changed into the heat energy with the

non-conservative forces .If we show the non-conservative energies as WNC

WNC= E the change of the mechanical energyInternal Energy : The energy that is appeared by the objects that is affecting each other.

W = K + U + Eint

Ex: A baseball is dropped ( m= 0,143 kg. ) from a height of 443m. Max speed of the ball is

42 m/s. Find the change in the internal energy of the ball and the surrounding air during the

fall to the surface of the ground.

W = U + Ek+ Eint0 = -mg.443 + 1/ 2.m.422+ Eint0 = -621 + 126 +Eint Eint= 495 J

Ex:

m = 4,5kg. V = 5m/s1,5m a) What is the energy that is consumed by the friction?

5m b) What is the velocity at the initial point when the

V block slides back from the top?

30 a) K + U + Eint= Wf-1 /2.mV2+ mgh = Wf- Eint Wf- Eint= 3356 = -23 J

b) K + V + Eint= Wf1/ 2.mV233 = Wf- Eint 1/ 2.mV

2= -23 + 33 1/ 2.mV2= 10 V = 2,1 m/s

Exercises1) m = 2,5 kg h = 5 m k = 2 N/m

Find the max compression. ( g = 10 m/s2)

mg (h+x ) = 1/ 2.kx2 125 + 20x = x2

x220x = 125 x( x20 ) = 125

x 0 25m

2) F(x) = G.m1.m2 / x2 a) U(x) = ? ( U(x) 0 as x)

b) W = ? to make x1to x + d .a) W = -U(x) F.x = - U(x)

G.m1.m2/ x2 . x = -U(x) U(x) = G.m1.m2/ x

b) W = -U W = [G.m1.m2/ x1]-[G.m1.m2/ (x1+d)] = G.m1.m2.d /x1(x1+d)


30/53

3)

P Show that d > 32/5 for P object to complete

the circle with radius r.

d mgL = mg2r + 1/ 2.mV2

mgL = 2mgr + 1/ 2.mgr L = 5/2r

r L = d+ r L = d + 2/5L d = 3L/ 5

Systems of Particles

Center of Mass:1) Two Particle Systems:

Xcm= ( m1x1+ m2x2) / ( m1+m2)

m1.d1= m2.d2

d1 d2

m1 cm m2

Motion of CM : Vcm= dRcm/ dt = (m1V1+ m2V2) / (m1+ m2)

If a 0 acm= dVcm/ dt = (m1.a1+ m2.a2) / (m1+m2) ( m1+m).acm= m1.a1+ m2.a2

( m1+ m2)am= F1+F2 Fnet= ( m1+ m2).acmif Fnet= 0 a = 0 ( so Vcmis constant )

2) Many Particle Systems :

m1

R1 Rcm= [m1R1+ m2R2+ m3R3+...] / (m1+m2...)

R2 m2 Rcm= (x,y,z) xcm= (m1R1x+ m2R2x+...) / ( m1+m2...)ycm= ( m1R1y+m2R2y+...) / ( m1+m2...)

zcm= ( m1R1z+ m2R2z+...) / ( m1+ m2...)

Ex: m1= 3kg

m2= 4kg Rcm= ?

m3 m3= 8kg

m2xcm= ( m3+ 2m2) /15 = 16/ 15

m1 ycm= ( m2+2m3) /15 = 20/ 15 = 4/ 3

Rm= ( 16/15 , 4/3 )

Motion of Vcm : Vcm= dR/ dt = 1/ ( m1+m2+...). (m1.dr1/dt + m2.dr2/ dt +...)

Vcm= ( m1V1+m2V2+...) /(m1+m2+...) acm= ( m1a1+ m2a2+...) / (m1+ m2+...) ( m1+ m2+...).acm= F1+ F2+....


31/53

Ex:F1 m1 m1= 4,1kg m2= 8,2kg

F2 m3= 4,1kg F1= 6N

F2= 12N F3= 14 N

m2 a) Rcm= ? b) acm= ?

m3 F3

a) Xcm= (-2. 4,1+ 4. 8,2 + 1. 4,1) / 16,4 = 1,75

Ycm= ( 4,1. 3 +8,2 .24,1.2) / 16,4 = 1,25 Rcm = (1,75, 1,25 )

b) 16,4.axcm= -6 + 62 +14 axcm= 116,4.aycm= 0+ 62 + 0 aycm= 0,5 acm= (1 , 0.5 )

Ex:

What is the acm?m1

L-y

=0 m2cm y

( m1+ m2) acmx= m.g acmy( m1+m2) =m2g

acmx= m2g /(m1+m2) acmy= -m2g / (m1+m2)acm= ( m2g/(m1+m2) , (-m2g/(m1+m2) )

3) Solid Objects :

cm Rcm= 1/ (m1+m2....). r.dmRcm xcm= 1/( m1+m2+..) . x.dm

dm ycm= 1/ (m1+m2+...). ydmR zcm= 1/ ( m1+m2+...). z.dm

If the object is 1D , dm = .dx = linear mass density

= m /LIf the object is 2D , dm = .dA = surface mass density

= m/ AIf the object is 3D , dm = .dV = volume mass density

= m/VEx: du

xcm= ?

x L

xcm= 1/m .x.dm dm = .dx dm = m/ L.dx0

xcm= 1/m.m/ L.x.dx xcm= 1/ L.x.dx xcm= x2/2L ] xcm =L/ 2


32/53

Ex:

(x) = ax ( bar isnt uniform)L dm xcm= ? m = ?

xcm= 1/m.x.dm xcm= 1/m..x.dx0

xcm= 1/ m.a.x2dx xcm= ax3/3m xcm = a.L3/3mL

m = dm = dx = ax.dx = ax2/2 m = a.L2/20

xcm= a.L3.2 / 3.a.L2= 2L/ 3

Ex:In the figure the part with radius R was separated. Find Xcm.

2R

x( 4r2

- r ) = R.r2

x.3R2= R.r2

x = R / 3

R

m1 x cm c.m.

Ex:

dm

y

R x

Find Xcm?

Xcm= 0 ( it can be seen )

= M / r dm / M = d/ y = R.sin

Ycm= 1/ M. dm.y = 1/ M / . Rsin

0Ycm= R /. -cos] = R /- ( -R/ ) = 2R / P Xcm= ( 0, 2R /)

P.S : In this kind of questions make x and y dependent to angles and dont forget that

dm / M = d/ . This means that the unit angle is equal to the unit area.


33/53

Ex:A ball is places in a shell which has the

same weight with the ball .Ball starts moving

when dropped freely and stops at the point O.

During this time shell moves towards left.

Find the place displacement.

At first position CM = ( -R/ 2,0 )

O

R 2R

At the second position :

d CM = ( d, -R/2 )

cm ( 0,0 )

There is not any horizontal external force that affects the object. There is only a vertical force.

Because of this the horizontal CM of the object must not change.

So -R /2 = d d = R /2

Ex: L

If cannon throws all of the balls to the opposite

wall , then what is the place displacement of the car ?

What is the velocity of the car after the cannon was shot?

x2 cm x1 x1 x1 cm x2

CM moved 2x1. If Mballs>>> Mcar

x1= L/ 2 2x10 Lb) Vcm= 0 ( Fnet= 0 )


34/53

Ex: dMdog, d , L and Mboatis given. Find x = ?

L x2 x1

x = dL + 2x1

x Md. x2= Mb.x1 cmx1+ x2= L/ 2

( L/2x1).Md= Mb.x1 x2 x1x2

L.Md/ 2x1.Md= Mb.x1

x1= L.Md/ 2(Mb+ Md)

x = dL + 2x1 x = dL + L.Md/ ( Md+ Mb)

P.S : In this kind of questions , the center of mass of the boat and the dog together, does not

change.

Linear MomentumMomentum ; D = m.V

Fext= dP / dt ( if m is constant ) Fext= m. V/dt Fext = m.a( if m isnt constant ) Fext= dP / dt

P of systems of particles ; If there is more than one particle total P is ;

P = m1V + m2V2+ ....

P = Mtotal.VcmConservation of Linear Momentum:

If Fnet= 0 Momentum is conserved. This means thatdP / dt = 0 and Pi= Pf

Ex: mV VE ( wrt earth )

V

(wrt cannon )

Mass of the cannon is M = 1300 kg. Mass of the ball m = 72 kg.

V = 55 m/s Find V and VEPi= 0 Pf= m.VE+MV m.VE= MV| V| = | V | + | VE|

VE/ V = 1300 / 72 = 18 VE= 18 V 55 = V + 18 V 19 V = 55

VE= 52 m/s V = 2,9 m/s

Ex: What fraction of the total Ek of the system will

V1 V2 each block have at any later time ?

Pi= Pf 0 = m1V1mV2

fraction i = K1/ (K2+K1)

m1 m2 = m1V12/ [m1V1

2+m1V1(m1V1/m2) ]

= 1 / ( 1+ m1/ m2)

= 0 fraction i = m2/(m1+m2)( Blocks are pulled and then left free fraction f = m1/ (m1+ m2)

towards each other )


35/53

Ex:M V M = 4,88 kg

V = 31,4 m/s

Find the change of the linear momentum.

42 42

P = PfPi = ( PfxPix) + ( PfyPiy)= ( MVcos42- MVcos42) + ( MVsin42+ MVsin42)P = ( 0,2sin42) P = ( 0,205 )

Ex:m V = wrt train

V What is the change in V0just before the man

M V jumps from the car with V ?

( m + M ).V0= ( m+ M ). VfmV

m.V = ( VfVo) ( m+ M )

VfV0= m.V /( m+M )

CollisionsImpulsive Force : Forces that act for a short time compared with the time of observation

of the system are called impulsive forces.

Impulse and Momentum :

F Pf tf

f(t) F = dP / dt dP = F dtPi tiPfPi= F.tfF.ti

P = F.t

ti t tfIf F is a variable , like in the graph , we use the average force ( F )

P = F.t Area under f(t) in the graph gives p.

Conversation of P During Collision

F12= - F21 tf

P1= F21dt P2= F12.dtF21 F12 ti

1 2 P1= -P2 P = P1+ P2= 0 If Fext= 0 then P = 0


36/53

Ex:m h1= 2m h2= 1,5m m = 1 kg

time to hit the ground = 10-2sec.

a) What is the impulse on ball ?

h1 h2 b) What is the momentum of ball after collision ?

c) What is the average impulse force ?

a) I = F .t = P P = PfPi mg.h1= 1/ 2.mV2

Pi= m. 2gh2 Pf= m.2gh1P = m.( 2gh2+ 2gh1)

b) Pf= m.2gh2c) P = F.t F.10-2= 11,6 F = 1160

Collisions in One Dimension

1) Elastic Collision : The objects are separated from each other completely and no energyappears. For this reason energy is conserved , too ; besides momentum.

m1 m2

V1 V2

1/ 2.m1V12+ 1/ 2.m2V2

2= 1 /2.m1V1f2+ 1 /2.m2V2f

2

m1V1 + m2V2= m1V1f+ m2V2f

V1f0 ( m1m2) / ( m1+ m ). V1+ 2.m2 / ( m1+ m2)V2

If m1= m2 V1f= V2 ,V2f = V1If m1>>m2 V1f= V1 , V2f= 2.V1V2If m1


37/53

Ex:V m2 m1 What is the movement of m1

m2 m1 V = ? and the angle of its movement ?

V = 0 V/2

m2V = m1Vf.cos m2.V/2 = m1Vf.sincos= m2V / mVf sin= m2V / 2m1Vftan= 1 /2 = 22.5

V1

Ex: V0 V = 0 m 1m 2

m m

V2

It is an elastic collision. Then 1+

2= ?

mV0= mV1cos1+ mV2cos2mV1sin1= mVsin2 V0

2= V12+ V2

2

V02= V1

2cos21+ V22.cos22

V02= V1

2+ V22+ 2V1V2( cos1.cos2- sin1.sin2)

1+ 2= /2

Ex:m1 V1 m2 V2

m1= 1,88kg V1= 10,3m/s m2= 4,92kg V2= 3,27m/s k = 11,2N/cm

Find the max compression . ( Hint : At max compression blocks move as one complete object)

( complete inelastic )

m1V1+ m2V2= (m1+ m2) .Vf

Vf= 5,2m/s

1/ 2.m1V12+ 1/ 2.m2V2

2= 1/ 2.( m1+ m2)Vf2+ 1 /2.kx2

253 0 184 + kx2 kx2= 68

11,2 / 10-2. x2= 69 x2= 0,69 /11,2 x 0 0,25 cm

Ex:Find hmax.

m V = 0

V0 V2 hmax

m.V0= (m+M ).V2 V2= m.V0/ (m+M)

1/ 2.(m+M) V22= ( m+M).ghmax V2

2= 2ghmax

( m.V0/(m+M) )2. 1/2g =hmax


38/53

Ex:m V0 V1 V0 V3

m

3m

spring is relaxed Find xmax.4mV0= 2mV1+ 3mV0 mV0= 2mV1 V1= V0/2

2mV1+ 3mV0= 5mV3 V3= 4V0/5

1 /2.2m. (V0/ 2)2+ 1 /2.3m.V0

2= 1/ 2.5m.(4V0/5)2+ 1/ 2kx2

7mV02/2 = 16mV0

2/5 + kx2

3mV02/ 10.k = x2 x = V0. 3m /10k

Rotational KinematicsRotational Motion :Pure Rotation : If all the points over the objects follows a circular path and if the rotation

occurs on an axis that passes through the center of these points , then this movement is a pure

rotation.

z

y

P

x Rotational Variablesy

s = length of the arc s = r. (=radian)V ds / dt = r.d/ dt

r V = r.W ( W = rad/s , = rad/s2 ) s

z x

angular displacement = 2- 1average angular velocity W = d/ dtaverage angular acceleration = d/ dt = d2/ dt2

Rotation with ConstantIf is constant

W = t + W0 W = (W0+ W) / 2= 1/ 2.t2+ W0.t + 0 W

2= 2+ W02


39/53

Ex: = 3,2rad/s2. At t 0 0 reference line AB ishorizontal. a) Find the angular displacement of the

line AB

b) Find angular speed of the grindstone after 2,7 sec.

B a) = 1/ 2.t2+ W0t + 0 = 11,7 radb) W = t = 3,2. 2,7 = 8,6 rad/s

A

In that example W0= 8,6 rad/s. Stone stops at 192 sec because of the friction on the wheel.

a)

What is ?b) What is the total angle turned through during the slowing down ?

a) W = t +W0 0 = 192+ 8,6 = - 0,045rad/s2

b) W2W02= -2. 74 = 2. 0,045. = 822 rad = 131 revolution

Rotation with Constant :

= 0 = W.t + 0 W = 2/ TFrom the rotational quantities is scalar W and are vectoral quantities.

Ex:W2= 43 rad/s describe the rotation of the

W1 = 84 rad/s disk as seen by an observer in the room.

W1+ W2= 94,4

tan= 43 /84 = 27Disk makes pure rotation with 94,4 rad/s

W2

Relationship Between Angular and Linear QuantitiesLinear quantities are similar to the radial components of the angular quantities.

V = W.r dV / dt = dW /dt .r

ar= r.W2= V2/ r at= r. = dV /dt

a = ar+ at

Ex: An object rotating about the z-axis slowing down at 2,66 rad/s2. Consider a particle

located at r = 1,83j + 1,26k ( in meters ) . When W = 14,3 k ( rad/s) find

a) V = ?

b)

Its acceleration

c) the radius of the path

a) V = W.r V = (0,14.3 ) x ( 1.83 , 1.26 ) = -( 14,3) ( 1,83) i

b) ar= W x V = ( -14,3)2. ( 1,83).( k x i )= -( 14,3)2

at= x R = -( 2,66k ) x ( 1,83j + 1,26k ) = ( 2,66) (1,83)ic) r = 1,83

Vectoral Relationship : V = W x r ar= W x V

at= x r a = ar+ at


40/53

Exercises1)

A flywheel engine is rotating 25,2 rad/s. When the engine is turned off , the flywheel

decelerates at a constant rate and comes to rest after 19,7 sec.

a) = ? ( in rad/s2)b)

the angle = ? ( in rad )

c)

the number of the revolutions ?

a) W = .t 25,5 = .19,7 = 1,28 rad/s2b) = -1 /2.t2+ Wt = 248 radc) 2rad 1 rev 248 /2= 39,47 rev

2) A flywheel completes 42,3 rev as it slows from an angular speed of 1,44rad/s to a

complete stop.

a)

If is uniform , find time for slowing

b)

= ?c) How much time is required for it to complete the first one half of the 42,3 revolutions ?

a) 42,3 rev = 42,3. 2rad = 266 radW = t 266 = -1/ 2.t2+ 1,44t

266 = -1/ 2.1,44 + 1,44t = 0,72t t = 369 sec.b) 1,44 = .369 = 0,0038 rad/s2c) Vs

2 + ( 1,44)2= -2. 0,0035. 133 Vs2= 1,0362 Vs= 1,02rad/s

Vs= -.t + V 1,02 = - 0,0035t + 1,440,0035t = 0,42 t = 108 sec.

3) What is the time when

wheel A reaches = 1,6 rad/s2 andwheel B reaches V = 100rev/min ?

100.2/ 60 = 10,5rad/sr1= 0,1m r2= 0,35m

# rev . r = # rev.r

W1. r1= W2. r W.0,1 = 10,5. 0,25 W = 26 rad/s

W = .t 26 = 1,6.t t = 16,25 sec.

4) a) ar= ?

b) at= ?

c) If a makes an angle of 57with at, what is the totalr angle that the object traveled ?

a) ar0 V2/ r = W2.r = 2.t2.r

b) at= .rc) = 1/ 2.t2 a.cos57= r a.sin57= 2t2r

= 1 /2.1,54 t2= 1,54= 0,77rad


41/53

Rotational DynamicsKinetic Energy of Rotation

Rotational Inertia ( I ) :( Moment of inertia ) : The sum of the products of the mass of each

particle by the square of perpendicular distance from the axis of the body with respect to the

axis of the rotation.

I = m1r12+ m2r22+ ... + mnrn2K = 1/ 2.mV2 ( V = W.r )

K = 1/ 2m.W2r2 K = 1/ 2.I W2 ( W in radian )CALCULATING I

a) The System of Particles :

I = miri2= m1r1

2+ ....+ mnrn2

Ex: m1= 2,3kg

m2= 3,2 kg m2

m3= 1,5kg

3

m14 m3

a)Find I perpendicular to xy plane and passing through each of the three particles.

b) Find I that is perpendicular to xy plane and passing through the center of mass.

a) I1= 2,3.0 +3,2. 32+ 1,5. 42= 52,8 kgm2

I2= 2,3. 32+ 3,2. 0 + 1,5. 520 58,2 kgm2

I3= 2,3. 4

2

+ 3,2 .5

2

+ 1,5. 0 = 116,8 kgm

2

b) Xcm= ( 2,3.0 + 3,2.0 + 1,5.4 ) / 7 = 0,86m

Ycm= 2,3. 0+ 3,2.3 /7 = 1,37 m

r12= 2,62

r22 = 3,4

r32= 11,74 Icm= 2,3. 2,62 + 3,2. 3,4 + 1,5. 11,74 = 34,5 kgm

2

It is easy to rotate around the axis whose I is smaller. It is the easiest one to rotate around the

axis of Icm

Parallel Axis Theorem : If I is the inertia of any axis that is perpendicular to the Icm.I = Icm+ M.h

2 M = total mass

h = distance between the two axes.

Proof : y

I = miri2= mi( xi

2+ yi2)

yi P yi I = mi[ ( xi +xcm)2+ ( yi +ycm)

2]

I = mi( xi2+ yi

2) +2xcmmixiycm xi cm x +2ycmmiyi + (xcm

2+ ycm

2)mi

I = Icm+ Mh2

xi xcm


42/53

Ex: a) Find I

b) Show that I is minimum when x = xcm

x

m1 m2

La) I = m1x

2+ m2(L- x)2

b) dI / dx = 0 2m1x2m2L2m2x = 0x( m1+m2) = m2L

b) Solid Objects : We can divide the object into small parts.

I = ( 0,1 m ) ( 0,45L )2+ (0,1 m ) ( 0,35L )2+ .....

When the number of the parts approaches to infinity,

I = limmi0 ri2.mi

I = r2dmdm = dx dm = d dm = dV

Ex:L/2 L/2 I = ? L/2

I = x2dm = x2.dx = x2m/L dxuniform solid bar -L/2

I = m/L. x3/3 ] = mL2/24 + mL2/24 = mL2/12

Ex:

I = ?x I = x2m/L dx = x3m /3L ] = 1/ 3.mL2

L

uniform bar

Ex: dm I = ?

I = R2dm dm = m /2d2

I = R2

m/ 2d= R2

.m./ 2]0 I = mR2

Ex: Compute the I of the disk

I = r2dm R dm = m / R2= 2rdrI = r2m /A dA

h 0

I = 2r2m / R2 .r dr = 2m/ R2. r4/ 4 = 1/ 2mR2

R dr

uniform disk


43/53

Ex:

r I = 1/2 .dmr2 r = R2z2I = 1 /2 ( R2z2) .m / (4/3r3) dV

z dm = m / (4/3r3) .dV dV = r2dzR R

I = 3/8 ( R2z2).m /R3. (R2z2)dz-R

solid sphere I = 2/5.mR2

Ex:

I = 1/ 2.dm.R2 dm = m/ R2L.dVR I = 1 /2.R2.m/ R2L. R2dz dV = 1/ 2.mR2

I = 1/ 2.mR2/L. L I = 1/ 2.mR2

d

L

Ex: P

R Icm= 1/ 2.mR2 Ip= ?

Ip= Im+ mR2

Ip= 1/ 2mR2+ mR2= 3 /2mR2

Torque Acting on a ParticleTorque is defined as ;

= r x F ( vectoral product )= r.F.sin ( Unit of = N.m joule )

Ex:

a) = ? (wrt origin )(4,6) F(3,2) b) O = ? if wrt origin O is -0/2O

r a) = r x F= (4,6) x ( 3,2) = - 10k

b) 0 = ( r yj)x F 0 = r x F = -yjx F-3 /2. 0= -yjx F 15k = y.(3k ) y = 5


44/53

Ex: O

m = 0,17 kg L = 1,25 m = ? ( wrt O ) = 10

L

= L.mg.sin = 0,36 Nm

m ( through the inside of the screen )

mg

Rotational Equation of Motion1) Single Particle Case :

F.( -Fr,Ft)

Ft F = m.a Fr= m.ar

Ft= m.at

a.( -ar, at)

If Ft= 0 this is uniform rotational motion

Fr If Ft0 rotational motion= r x F = r x ( F r.r + Ft. )

= r. Ftk | | = |r|. |Ft|= r.m.at= r

2.m.= I.

2) Solid Object :dF = dm.a

d= r x dm.ad= r.d.Ftk

Ft = . I

Work Energy Theorem in Rotation

W dW = F.ds = F.ds.cos (/ 2- )F dW = F.r.d. sin

fW = .d W = Ek

iP = dW / d P = .W

Ex: V0=0

Find a) W = ? b) = ? c) acm= ?L d) Fp= ? e) W( ) and ( ) = ?

a) Ei= Ef mg L/2 = 1/ 3.mL2W2

W = 3g/LP

b) = -mg.L/2 -mg.L/2 = Ip -mg.L/2 = 1/ 3.mL2

= -3g /2L =


45/53

c) acm ax= V2/r = W2.r2/r = W2. L/2 = 3g /L. L/2 = 3g /2

ay = r.= L/2. = -L /2. 3g/2L = -3/4g acm( 3g /2 , -3/4g )

d) F = m.a Fn= m.ax Fn= 3/2.mgFr= mg = m.ay Fr= 1/ 4.mg

F = ( -3/2.mg , .mg )

Ex:R m1= 512gr m2= 463gr R = 4,9cm

m2takes 76,5 cm within 5,11 sec. What is the I of the pulley?

m1 T1m1g = -m1a T2m2g = m2a

( T1- T2) R = = I. a = R.0,765 = 1/ 2.a.(5,11)2 a = 0,06 m/s2

m2 0,06 = 0,049. =1,2 T1= 5N T2= 4,6N0,4. 0,049 = I. 1,2 I = -(m

1+ m

2)R2+ (m

1m

2)R2.g /9

Second Way : Ei= Ef

m1g.y1i+ m2g.y2i= 1/ 2(m1+ m2) V2+ 1 /2.IW2+ m1g.y2f+ m2g.y2f

V2= 2aL ( m1m2).gL = 1 /2.( m1+ m2+I/ R2). 2aL

( I can be found from this equation )

Ex:M = 2,5kg R = 20cm m = 1,2 kg

M Find a, T , a) mg = T = ma mgT = m.2T / M

T( 1+ 2m / M ) = mg T = 6NT.R = = I. T.R = 1 /2.M.R2

a m T = 1/ 2.MR a = R T = 1/ 2.M.a

b) mg6 = m.a a = 4,8

c) .R = a = 4,8 / 0,2 = 24 rad/s2

Combined Rotational and Translational MotionRolling Motion :

K = 1/ 2.MVcm2 + 1 /2.IcmW

2

Ex:M = 0,023kg R = 2,6cm R0= 0,3cm L = 0,84 cm

What is W needed for the yo-yo to climb up the string ?

1 /2.IW02= mgL + 1 /2.IW2+ 1/ 2.mV2

1 /2.IW02= mgL W0= 221 rad/ s.


46/53

Rolling Without Slipping :If Vp= 0 then it is rolling without slipping. There is

2V friction sbetween the ground and the object. But wecant say anything about k, because Vp= 0. Friction

V force does no work. Energy is conserved.

Vp= 0 sThere are two solution ways for rolling without slipping:

1)

2Vcm Vcm= R.W Ip= Icm+ M.d2

Vcm K = 1 /2.Ip.W2

K = 1 /2.IcmW2+ 1 /2.m.Vcm

2

P

2) Vcm Rw Vcm+Rw

Vcm Vcm

Vcm Rw

Vp=V

cm-R

w

Vp= 0 Vcm= R.WK = 1/ 2. m.Vcm

2+ 1/ 2.IcmW2

K = 1 /2.Ip.W2= 1 /2.IcmW

2+ 1/ 2.m.Vcm2

Ex: solid

cylinder Find the speed of its center of mass at

the bottom.

mgh = 1 /2.mVcm2+ 1/ 2.IcmW

2

h mgh = 1 /2.m.Vcm2+ 1/ 2. 1/ 2.mR2.Vcm

2/ R2

gh = 3 /4.Vcm2

Vcm= 4gh / 3W

Second Method : mg.sin- fs= m.acm Nmg.cos= 0fs.R = Icm Icm= 1/ 2.m.R

2 acm= R.fs.R = 1/ 2.mR

2.acm/ R

fs= m.acm/ 2 mg.sin- macm/2 = m.acmg.sin= 3 /2.acm acm= 2.gsin/3V2= 2a.L V2= 2.2g.sin/3. h/sinV = 4gh /3


47/53

Ex: (i) Find a) T in rope b) acm c) Vcm

h a) ( mgT ) = acm.m

(f) acm= R. Icm.=.Rm 1 /2.m.R2. acm/R = .R ( T forms )

1 /2.m.acm= T m.acm= 2TmgT = 2T T = mg /3

b) mg. mg/ 3 = m.acm 2g /3 = acm

c) mgh = 1/ 2.mVcm2+ 1/ 2. 1/ 2mR2.V2/R2

mgh = 3 /4.m.Vcm2 Vcm= 4gh /3

Ex:L It is released from the horizontal. What is W

L when it becomes vertical ?

L 3mgL = mgL /2 + mgh + 1/ 2.IpW2

3mgL = 1 /2.4/ 3.mL2W2

W = 3/2. g / L

Rolling Motion in Accelerated SystemsEx:

O A fs= ? a = ? a = ? = ?

rolls without a = a + A a = -r+AO slipping A = a + r

N fs.R = I fs= m.acm a A = R+ acm

mg wrt O mA fs= m.a

Ex:I =0

Find acceleration of masses.

T m1gT = m1a T.R = I.T m2gT = m2acm a = R.- acm

m2 a = R.- acm

a m1

Ex: If it rolls without slippingfind acm , fs

F Ffs= m.acm I.= fs.R + F.racm= .R

R acm= ( fsR + Fr) / I.R = 2F.(r +R ) / 3M.R

fs= F( R2r ) / 3R

0


48/53

Ex: I=0

m1 Find acm , a , , t ,fsm2gT = m2a

cylinder rolls T ( Tfs).R = I.without slipping a = .R + acm

m2 ( T- fs) = m1.acm acm= .R

a

acm= 4m2g / ( 3m1+ 8m2) a = 8.m2g / (3m1+ 8m2) T =3m1m2g / (3m1+8m2)

fs= m1m2g / (3m1+ 8m2)

In this kind of questions remember that the acceleration of the part that touches the ground

must be 0.

Angular Momentum

Angular momentum is represented with L.

l with respect to the origin

I L = r x P

L = r x mV = r x m( W x r )

dL/ dt = d( r x P )/dt = dr /dt x P + r x dP/ dt

r dL /dt = V x P + r x f

P dL/ dt = dL/ dt = I.

Net torque acting on a particle is equal to the time rate of change of its angular momentum.

If there are ore than one particle

L = L1+ L2+.....+Ln = 1+ 2+...+nEx:

3 Find L and at 1,3,42 L1= r x P = 0

1= r x F = 0V0 H L3= r3x mV

1 4 L3= ( R/2 ,H) x m.(Vcos, 0)R/2 R/2 L3= -m(H V0cos) k

L4= ( R,0 ) x m(V.cos, -Vsin) 3= ( R/2,H ) x ( 0, -mg) = -mgR/2 k

L4= -m.RV0sink4= ( R,0) x ( 0,-mg) = -Rmg kAngular Momentum and Angular Velocity

In generally W is not parallel to L. But in some cases it may be. For this reason :

L = r x (mV) L r , L VL// // W LW

L = L//+ L


49/53

P2 W L

P1 L = m( r x V ) = mrV k

W = Wk L // WV = r.W

r F L = mR2W k

Lz= I.WdLz/dt = d( IW )/ dt z= I.

This equations are seen in the systems that shows symmetry with the axis of rotation.

( L // W ) ( cylinder , sphere , disk , hoop....) Lz= I.W

Rotational Equation of Motion

LA= r x P + Lcm LA: wrt A

A= r x Fnet+ cm LA= Icm.WVcm cm= Icm.

R

A

Angular Momentum ConservationIf ext= dL / dt = 0 Angular momentum is conserved

Li= Lf Ii.Wi= I.Wf

A

Ex: If the cylinder rolls without slipping ,

= ? acm= ? LA= ?Vcm

a) Lp= R x m.Vcm+ Lcm Lp= -mrVm- ImW

dLp/ dt = -mram- Icm p= R x Fnet+ cmp= R.fsmg.sinR = R.fs p= -mgsin.R

b) acm= R. -mgR.sin= -( m.ram+ 1/ 2.m.r2

acm/ 2)acm= 2/3.g.sin

c) LA= m( R x Vcm) + Lcm LA= -mrVcm- ImW

A= R x Fnet+ cm cm= -r.fsR x Fnet= R x ( fs+N + mg ) = r.fs+ R//NR//.mg.cos- mg.sin

N = mg.cosR x Fnet= r.Fs- mg.sinTA= r.fs- mgr.sin- r.fs dLA/ dt = -( mracm+ Icm)


50/53

Ex: m1 If the upper disk falls over the lower disk, what is Wf?

Fext= 0 = 0 Li= LfI1.W0= ( I1 + I2) Wf

1 /2.m1R2. W0 = 1 /2.(m1+ m2) R

2.Wf

m2 W0 Wf= m1W0/ (m1+ m2)

Ex:W What is Wtrain if the wheel has M and R.

Train starts moving and reaches the velocity V.

Fext= 0 Li= LfW.I = m.R ( VR.W )

m (train) 0 = Lwheel+ Ltrain

W.M.R2+ mR2= m.RV mVR = IW

W = mV /( m+M )RW = mRV / t = mV /MR V = V + Vtrain V = V - RW

Rolling with Slipping Motion

Ex: W0 W

(disk) Vi=0 Vcm

k

a) Find W when rolling without slipping starts.

b) Find Ek c) Show that t = R.W0/ 3k.gd) Show that x = R2W0

2/ 18k.g = ?( In this kind of questions acm= R Vcm= RW X = R equations arent used )

a) First Method :

A = 0 LA1= LA2W0 Wf -IcmW0= mR x Vm+ Lm

-Icm.W0= -( mRVcm+ IcmWf)

R1 R2 Vcm 1 /2.mR2W0= mR2Wf+ 1/ 2.mR2WfWf= W0/3

A

Second Method : fk= m.acm = I. = R.fkR.macm= I R.m.acm= 1/ 2.mR

2= 2acm/ R

Vf= acm.t Wf= W0- t Vf= Wf.RVf/ acm= ( W0Wf).R Wf= W0/ 3


51/53

b) 1 /2.mVcm2+ 1/ 2.IWg21/ 2. 1/ 2.mR2Wf

21/ 2. 1/ 2.mR2W02= Ek

3 /4.m.W02.r2/ 91/ 4.mR2W0

2= Ek-2mW0

2R2/ 12 = -mW02.R2/ 6

c) Wf.R = Vf Wf.R = am.t Wf.R = fs.t / m

Wf.R = k.mg.t / m t = W0R / 3k.g

d) Wf2W0

2= -2B W02/ g = 2.(WfW0) /t

4.W02/ 9 = -2W0.3.k.g./ 3.W0R

= 2 RW02/ 9.k.g

Ex: Wi=0 How far does it move before rolling without

slipping ?

V0

m kFirst Method : fk.R = Icm. fk= m.acm fk0 k.mg

k.mg = m.acm m.kgR = 2/5.mR2 = 5k.g / 2R

Wf= .t t = Wf / = Vf /2Vf= V02/5R.Vf/ R Vf= 5/7.V0Vf

2V02=2acm.X 25V0

2/49V02= -2kg.X

24.V02/ 49 = 2kg.X X = 12.V0

2/ 49kg

Second Method: L1= L2 m.V0.R = m.V1R + Icm.Wf

m.V0.R = m.VfR + 2/5.m.R2.Vf/ R

V0= 7/5.Vf

Ex: The insect flies for sometime with the velocity

V. After it stops,

a) W = ? b) Ek= ?a) Li= Lf

Icm.W0m.VR = If.Wf

insect Icm.W0mVR = ( Icm+mR2)Wf

W ( If= Itable+ Iinsect )

( while the insect is moving L = mVR

otherwise L = IW ( I = mR2) )

Wf= ( I.W0mVR ) / ( I+ mR2)

b) 1 /2.mV2+ 1/ 2IW02= 1 /2.IWf

2

Angular ImpulseAngular Impulse = J

J = I = .dtIf is constant J = L = .t


52/53

Ex: According to the picture decide the motion of the object.

L/3 The given acceleration causes a circular motion and also

the object gains a horizontal velocity.

F.t P = F.t ( m.Vcm0) = F.tVcm= F.t /m

L = Jm Icm.W = F.t ( L/2L/3 )

W = F.t.L / 6.Icm W = F.t.L2/ 6.(1/12mR2)

W = 2.Vcm.m.L / mR2 Vcm= W.L/ 2

Ex: m V1 V2

Find V2in terms of V1, m , I , R

R J = LL = R.fk dt = L2L1R.fkdt = IW

initially P = -fk.dt = P2P1= m.( V2V19at rest m( V2V1) = I.V2/ R2

J = IW /R = I.V2/R2

V2= m.V1/ ( m+ J/R2)

Ex: m collides elastically

L What is m, so it remains at rest after collision

in terms of M, L , d.

d

m V0 Ek, P , L are conserved.

M

m.V0= M.Vcm L1= L2m.V0.d = IW

Ki= Kf

1/ 2.mV02= 1 /2.M.Vcm

2+ 1/ 2.IW2

1 /2.mV02= M:Vcm

2+ IW2

m.V02= m.V0

2/M + I.m.V02d2/ I2

V02= m.V0

2/M + m.V02d2/ (1/12.ML2)

m = M:L2/ ( L2+ 12d2)

Ex: W0 m Wf= ? after completely

1 inelastic collisions.

Vi2 Li= Lf

m m Lf= Ip.Wf

3

m

Icm.W0m.Vi= ( Icm+ md2)Wf ( path 1 )

Icm.W0= ( Icm+ md2) Wf ( path 2 )

Icm.W0+ m.Vid = ( Icm+ md2)Wf ( path 3 )

Icm= 2.( 1/12.2.m(2d)2) = 4/3.m.d2


53/53

Ex: W1 W2

Show that h = 4R /5

F Fdt = P2- Ph V0 9/7V0 Fdt = m.V0

h.Fdt = I.W1

h.mV0= IW1W2= 9V0/ 7R

L1= L2

mRV0+ IcmW1= m.R.V0.9/7 + Im.W2

mRV0+ 2/5. mR2W1= 9/7 mRV0+ 2/5. m.R

2. 9V0/ 7R

2/5.mR2W1= 2/7.mR.V0+ 18/35.mRV0

2mR2W1= 28mRV0/7 W1= 2V0/ R

h.mV0= 2/5.mR2. 2V0/ R h = 4R/ 5

Ex:

m a) Find L before collision.

V0 W b) Find L after collision.L ,M c) W = ?

d) K = ?

at rest

a) L = mVR L = m.V0.L/2

b) L1= L2 L2= m.V0L/2

c) m.V0L/2 = ( 1/12M.L

2

+ m.L

2

/4) . Wm.V0L/2 = ( ( M+3m) L2/ 12 ).W

W = 6mV0/ ( M+3m )

d) 1/ 2.IpW21/ 2.mV0

2

1/ 2. mV0L/2. 6mV0/(M+3m).L1/ 2.mV02

= 3/2. ( m2V0mV02) / ( M+3m )

Documents

Vectors fundamentals CBSE