DYNAMICS

KINEMATICS We study the geometry the motion For different time t, establish

position vector r→, velocity v→, acceleration a→, path taken by the body

The above quantities involve dimensions of time (t) and length (L) only (kinematics quantities)

We don’t inquire about the force system acting on the body.

Neither cause & effect relation studied.

KINETICS(Dynamics) Cause & effect relation studied

force/moment is the cause and linear acceleration and angular acceleration is the effec

Newtons 2nd law is studied P = ma

All concepts of statics are useful e.g. F.B.D. /equivalence.

Unit of mass is involved.

IDEALIZATION OF BODY

1. PARTICLE - The body is a point mass - The geometry of body is neglected

- Situation -motion in pure TRANSLATION

2. RIGID BODY - The body is considered in its true geometry (shape & size)

- But the body is infinitely rigid - i.e. the body does not deform under the action

of the applied forces - Situation - when motion is either ROTATION or TRANSLATION & ROTATION

Types of motion

1. TRANSLATIONAL 2. ROTATIONAL 3. TRANSLATIONAL + ROTATIONAL

(GENERAL PLANE MOTION) 4. 3D TRANSLATIONAL + ROTATIONAL

Types of translational

1) Rectilinear translation

2) Curvilinear translational

KINEMATICS OF A PARTICLE Or

KINEMATICS OF TRANSLATIONAL

1) Rectilinear translational

v v + ∆ v

X, S

O S S (t) (t+ t) Defination (1) Velocity at any time t is V =lim∆ → [ ( ∆ ) ( )

∆] - lim Δt→0[ ∆

∆]= lim Δt→0∆

∆

Defination (2)Acceleration of particle at any time t, is a= lim Δt→0 V(t+∆ t) – v(t) -lim Δt→0 v +∆ v = lim Δt→0∆v ∆ t ∆t ∆t

(1)& (2)

V = 풅푺풅풕

a =dV/dt

EQUATIONS OF KINEMATICS Entire motion is described by one coordinate

V =dS/dt

a =dV/dt = ( )= d2/dt2

=dV/dt.ds/dt = VdV/dt

(1) Different form is given S=S(t)

Then V=dS/dt =V(t)

a=dV/dt =a(t)

(2) Kinematics in integral form

Given a=a(t)

Then ∫ 푑푣=∫ 푎푑푡 ≫V-VO=∫ 푎푑푡

Or ∫ 푑푣 = ∫ 푎푑푡 ≫ V2-V1=A1→2

V2=V1+A1→2

∫ 푑푠 = ∫ 푑푡 ≫ 푆 − 푆O=∫ 푣푑푡

∫ 푑푠 = ∫ 푣푑푡 ≫S2-S1=A1→2

S2=S1+A11→2

Example: Given motion id rectilinear translation and at any time (t); x = - 3.6t3+10.8t2 (m, s)

Find v, a, at t?

Solution: x=-3.6t310.8t2

Example: Given-at any time t, V=4.5t2-18t-15(1)

when t = 6, x = - 75

Find x, a when t=4(s)

V=dx/dt= -10.8t2+21.6t

a=-21.6+21.6

Solution: Differentiate w. r. t ‘t'

a= =9t-18 when t=4s

a=9(4) -18=18m/s2

We know V=dx/dt∫ 푑푥 = ∫ 푣푑푡

X x-75 =∫ (4.5푡2-18t-15)dt = 1.5t3-9t2-15t 6t

x-(-75) =1.5t3-9t2-15t–(1.5(6) 3-9(6) 2-15(6) )

x+75=1.5t3-9t2-15t-(324 – 324 -90)

→v0=0 →a=-Kx-2

Example: a = - Kx-2

At t = 0 ,x0 = 800 mm ; vo=0

─ at x = 500 mm ,v=6m/s

Find K and v ,when x = 250 mm

Solution: a=-Kx-2 =vdv/dt

∫ 푣푑푣 = ∫ −퐾푥-2

1/2v20

v

= v2 = - = ( ) (1)

(1)

At x=5 ,v=6

Put in (1) 62 =k(8-5)/4(5); we get 36(2) = 3k

K = × or, K=24

(1) V2 = ( )

At x=25.. v2=60(8-25)/25 =240(.55)=132

V=11.49m/s

X=1.5t3-9t2-15t+15

V2= ( )

(2) solution by factorisation

3t2-12t-15 =0

3t2-15t+3t-15 =0

t(3t-15)+(3t-15)=0 ,(3t-15)(t+1)=0

t=-1,5

Special case of UNIFORMLY ACCELERATED RECT.TRANSILLATION

In this case a=ao =constant

dv/dt =a =aot ∫ 푑푣=a0∫ 푑푡

v-vo=a0t v =v0+a0t

dx/dt =v = vo+a0t ∫ 푑푥 = ∫ (푣표+ 푎표푡)푑푡

x-x0 =vot+ aot2

v =a=a0 ∫ 푣푑푣=a0∫ 푑푥

v2- vo2=a0(x-x0)

V2 = v02+2a0(x-x0)

RELATIVE VELOCITY AND RELATIVE ACCELERATION(rectilinear transillation)

aA aB

VA VB

SA S

SB/A

SB

SA=Absolute coordinates of A

SB=Absolute coordinates of B

SB/A=coordinate of B w.r.t. A

We know that SB/A = SB - SA or SB = SA + SB/A

VB/A = VB - VA or VB = VA + VB/A

aB/A = aB - aA or aB = aA + aB/A

DEPENDENT MOTION

(Blocks attached to INEXTENSIBLE ROPE passing over frictionless pulleys)

(1) loFixed pulley h L=total length of rope=constant

SA SB =SA-h+lo=constant

=SB-h=constant

SA+SB=constant B

A

(2)Sl1 L=SA-h1+l1

SA SB +SB-h1-h2+l2

l2 +SB-h2

h2∆SA+2∆SB=0

Snatch pulley

(3)

SA SC SA+SB=constant

SB VA+VB=0

SD aA+aB=0

VA+VB=0 aA+aB=0

A SA+2SB=constant

B

VA+2VB=0 aA+2aB=0

A

B

C

SB/DSC/D SB/D+SC/D=constant

‘ SB-SD+SC-SD=constant

, SB+SC-2SD=constant

‘ VB+VC-2VD=0

aB+aC-2aD=0

Eliminate vD,aD

VB+VC+2VA=0

aB+aC+2aA=0

C

SA SB

Given VB=450mm/s→

Find (a) VA (b) VB (c)VA/B (d)VC/D

Soln;- 3SB-2SA=constant

Differentiate w.r.t. ‘t’

3VB-2VA=0 (1)

Given that VB=+450 , put in (1)

3(450)=2VA VA=675mm/s→

Va=VB=VC

SC/A C SC/A+SD/A=constant

D SC-SA+SD-SA=constant

SC+SD-2SA=constant SD/A differentiate ; VC+VD-2VA=0

+450+VD-2(+675)=0

VD=900mm/s→

A

A

VA/B=VA-VB=675-450=225mm/s→

VC/D=VC-VD=450-900 =-450

VC/D=450mm/s←

Given a=-Kv3 (m/s2)

----When x=0 , v=25m/s

-----When x=40 ,v=20m/s

Find (i) distance travelled since velocity drops from 25 to 10 m/s

(ii)distance travelled at v=0

Soln;- a=v =-Kv3

-∫ . =K∫ 푑푥

= +1/v v25=Kx=+ -(+1/25)=Kx

-1/25 +1/v=kx =

= +kx=

V=

Now put x=40,v=20

20=×

=

20+20,000k=25

K=,

=,

Finally v= = × = ,

Question:- At what x , is v=0

(1) 0= , ≫≫ x=∞ When v=25 , x=0 V=10= , ≫≫1,600+10x=4,000 10x=2,400

X=240m

V=25 v=10

O X

240

Given X distance be A,B,C are

Equal(equal to d)

VC/B=200mm/s d

Kinematics SA+SB=constant

VA+VD=0 equation 1

SB/D+SC/D=0

VB-VD+VC-VD=0 or VB+VC-2VD=0 equation 2

From 1 and 2 eliminate VD

VC/B=VC-VB=200

For alignment

=

dSA dSB dSC - = -

VB-VA=VB

4)VC=200+VB

5)2VA-VB+(200+VB)= or 2VA-2VB=-200

3)2VB-VA-(200+VB)=0 [-VA+1VB=200]2

4VB=200

2VA+VB+VC=0

2VB-VA-VC=0

VB=50mm/s

VA=100 150

VA=150 mm/s VC=250mm

Given VB/A=450 after 6s

VB-VA=450 equation 1

3VA-2VB=0 equation 2

2[VB-VA=450] eqn1

-2VB+3VA=0 eqn2

VA=900

VB=1350

aA= 150푚푚/푠2

aB= = 225푚푚/푠2

at t=0

(VA)0=600

At t=6s

VB=(VB)=720

VB=720

at ant t

aA=constant

aB=0

to time ∆t lim∆

∆∆

= ∆∆

-∆∆

At any t

4VC=VB-VA

Equation 1

4aC=aB-aA

Equation 2

Motion of A

VA=VA0+aAt

720=600+aA (6)

aA= =20mm/s2

(2) = 4aC=ab-aA

=0-20 =aC=5 mm/s2

At t=0 -4VC=VB-VA

-4VC=720 -600=120

VC=-30 =VC0

Motion of c

VC=VC0+act

At t=4

VC=-30+5(4)=-10

VC= 10mm/s

SC=VC0t + act2

At t=6 SC = -30(6)+ (5) (6)2

= -180 + 90 = -90mm or SC = mm

CURVILINEAR TRANSLATION

(2D or 3D [rectangular component]

---at any time t ,particle is at P(x,y,z)

---at any time t ,position vector of a vector of particle is given by →=X→ + 푌 →푍 →

Note ; here x=x(t)

Y=y(t)

Z=z(t)

Def 1; velocity of particle at any time t is → =→

here →=→(t)

Vector which is a function of scelar.

→= LIMIT T=0 ∆→

∆=lim∆ → [

→(푡 + ∆푡)−→ (푡)/∆t

Geometrical or physical significance

→ = (lim∆ →∆∆

) ∑ = ( )

V= │→│ = speed at p =

Def; acceleration of particle at any function t is

(2)

= lim∆ →∆→

∆ = lim∆ → [

( ∆ ) → ( )⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

∆]

In Cartesian form at any function t , positive vector in

→ = 푥 → + 푦 →푘 → or x=x(t) , y=y(t) ,z=z(t)

Velocity → = →

= → +푥→

+ → +푦 + → + 푧→

= 0

→ =lim∆ →

∆→

∆= lim∆ →

→( ∆ ) ( )

∆=0

→ =푑 →

푑푡 = (푑푥푑푡) →+

푑푦푑푡 →+(

푑푧푑푡) →

=VX→+Vy→VZ→

Acceleration is

→ =→

=(d2→/dt2)= → + → + →

=(d2x/dt2)→ +(d2y/dt2)→(d2t/dt2)→

=aX→+ay→+az→

integration of vector quantities

푓(푡)( )

→푑푡 =→ 푓(푡)푑푡 −푑 →푑푡 푓(푡)푑푡

=→∫푓(푡)푑푡

See SHAMES for v.good problems.

→relative velocity and relative acceleration or kinematics with respect to A MOVING FRAME OF REFERENCE.

→ =푑 →

푑푡

→A=ab.s positive vector of A

=position vector of A with respect to fixed from of reference

→b=abs position vector of B

→=position vector of B with respect to A

=relative position vector of B with respect to A

= +/⎯

= +/⎯⎯

= +/⎯

/⎯ = +

/⎯⎯ = −

/⎯⎯ = −

VB/A=1125

=VA+VB/A equation (1)

Algebra ; = 700→

= (푉B/A)x→+(VB/A)Y→

Put in (1) =[700+(VB/A)x]→[(VB/A)Y]→

VBX=700+(VB/A)x eqn 2

VBY=(VB/A)Y eqn 3

(2)2+(3)2

5002=VB2=7002+11252+1400(VB/A)X

(VB/A)x=5002-7002-11252/1400=-1075

(VB/A)Y= 푣( )2- (푣( )X2=±√1125 − 1075in square

(2)VBX =700+(-1075)=-375

VBy=332

Tanα=

α=41.50s of W

alternative in graphical solution

cosu formula

11252=7002+5002-2(700)(500)cosβ

Cosβ=180-138.7=41.30 s of W

푘푖푛푎푚푎푡푖푐푠푖푛푛표푟푚푎푙푎푛푑푡푎푛푔푒푛푡푖푎푙푐표푚푝표푛푒푛푡푓표푟푚표푟푘푖푛푎푚푎푡푖푐푠푖푛푡푒푟푚푠표푓푝푎푡ℎ푝푎푟푎푚푒푡푒푟(푠)

We know at any time t particle is at p(s)

→ 푉 ∑ 푡 ;v= ∑푡 + 푉 ∑ eqn 1

∑ = ∑ = ∑ . . =V. ∑ 푡

푑 ∑푛푑휃 = − 푡

1) ← = ( )∑푡 + ∑푛

at∑ 푡 + 푎n+∑푛

at=tangential component of acceleration

= = an= normal component of acceleration =v2/p MIXED (HYBRID) PROBLEMS Curvature of path in cartesain coordinates

(eqn of path) 1/p =│d2y/dx2│/[1+( )2

Kinematics in Polar Coordinates

At any time t, particle is at P(r,θ). Here r=r(ε), θ=θ(t)

At any time position vector of vector of particle is

r=r εr.

Velocity of particle at any time t is

V= dr/dt- drεr/dt +rdε r /dt

dεr/ dt= ( dεr/dθ)( .dθ/ dt)=( dθ/dt)εθ

dεθ/dθ = dε r/d t .dθ/ dt =-(dθ/dt)εr

V=(dr/dt)εr+(rdθ/dt)εθ=Vr ε r +Vθ ε θ

Here V r = dr/dt=radial component of velocity

V θ=rdθ/dt=transverse component of velocity

Differentiate => acceleration at any time t , is

a= dV/dt=( d2r/dt2)εr+ (dr/dr)(dε r/ dt)+(dr/dt)(εθdθ/dt)+(rd 2 θ/dt)ε θ +(rdθ/dt)(dεθ/dt)

a=[d2 r/ dt 2 -r(dθ/dt)2 ] ε r+ [rd 2θ/dt 2 + 2(dr/dt)(dθ/dt)]ε θ

= ar- ε r+ aθ. εθ

Here a r=d 2 r/dt 2 -r(dθ/dt) 2 = radial component of acceleration

aθ=rd2θ/dt 2+2(dr/dt)(dθ/dt)=transverse component of acceleration

VelocityStudy

Vx=

Vy=

V=

Vø=푟

Acceleration Study

ax=dvx/dt=

ay=dvy/dt=

af=dv/dt=

ar= − 푟

aø= + 2( )( )

A model rocket is launched from Point A with an initial velocity v0 of 75 m/s. If the rocket’s descent parachute does not deploy and the rocket lands a distance d = 100 m from A, determine (a) the angle α that v0 forms with the vertical, (b) the maximum height above Point A reached by the rocket, and (c) the duration of the flight.

Set the origin at Point A. Xo=0, yo=0

Horizontal motion x=vt sin 푠푖푛훼 =x/vot

Vertical motion y=voy=푐표푠훼 − 푔t2

푐표푠훼 = 1/vot(y+ gt2)

sin 훼 + cos 훼 =1/(vot)2[푥 + 푦 + 푔푡 =1

x2+y2=gyt2+ g2t4=vo2t2

푔 푡 −(vo2-gy)t2+(x2+y2)=0

At point B, (x2+y2)1/2=100m, x=100 cos 30om

Y= −100푠푖푛30 = −50m

14

(9.81) 푡 − [75 − (9.81)(−50)]푡 + 100 = 0

=

t2=252.54 s2 and 1.648s2

t=15.8916 and 1.2829 s

Restriction on 훼: 0<훼<120

푡푎푛훼 =푋

푦 + 푔푡=

100푐표푠30−50 + (4.905)(15.891) = 0.0729

훼 = 4.1669

and ( . )( . )

= −2.0665

훼 = 115.8331

Use훼 =4.1669 corresponding to the steeper possible trajectory .

(a) Angle 훼 = 4.17 (b) Maximum height , vy=0 at y=ymax

Vy=vot 푐표푠훼-gt=0

t=vo푐표푠훼/g

ymax=vot cos훼 − 푔푡(푣 cos 훼)/ 2푔

=(75)2cos24.1669/2*9.81

ymax=285 m

(c) duration of the flite t=15.89

Given DE is arc centre A radius b OC is a straight rod rotating at constant speed 풅휽풅풕

=흎Pin B remaining on DE (which is fixed ) and OC which .is rotating. Find acceleration on pin B.

SOLUTION

1. In term of path parameter S in normal & tangential form. Let pinB be at O’ at t=0 Let if it at position B at time t, Let S be collinear chord Velocity of pin at point B 푉 = but=푏(2휃) = 2푏휃

So, = 2푏 = 2푏휔 푽 = ퟐ풃흎 acceleration at pin B at any time t, at=tangential component= =0 as V is constant an= normal component of acceleration =4bw2

2.in Cartesian coordinates At any time t coordinate of pin B are , X=OG=OA=AG=b+bcos2휃 Y=BG=bsin2휃 Velocity of pin B at any time t Vx= = −2휔푏푠푖푛2 = −푏푠푖푛2휃. 2.

Vy= = 2휔푏푠푖푛2휃

V=(vx2+vy

2)1/2-2bw Tan훼 =vx/vy=tan 2휃 훼 = 2휃 acceleration of pinB at any point ax=dvx/dt=d2x/dt2=-4w2bcos2휃 ay=dvy/dt=d2y/dt2=-4휔 푏푠푖푛2휃 a=(ax

2+ay2)1/2=−4휔 푏

tan 훽 = 푡푎푛2휃 = tan훽 = 푡푎푛2휃 훽 = 2휃

2. In polar coordinate /in term of radial and transvers form At any time ,point at B(r,휽)

Velocity of pin B

Vr= = −2푏휔 sin휃

Vø= = 2푏휔푐표푠휃

V=(Vx2+VY

2)1/2=2b휔

푡푎푛훼 =Vr/Vø=tan휃 = 훼 = 휃

acceleration of pin b is

ar=d2r/dt2−푟 = −4푏휔 푐표푠휃

aø푟=푟 + . =-4휔 푏푠푖푛2휃 a=(ar

2+aø2)1/2=4b휔

RECTILINEAR MOTION OF PARTICLE

11.2] The motion of a particle is defined by the relation x=t3-9t2+24t-6 where x is expressed in meters and t in seconds . Determine the position ,velocity and acceleration when t=5s?

Sol; x= (t3-9t2+24t-6) m

At t=5sec, x=(5)3-9(5)2+24(5)-6 =14m

V = dx/dt = 3t2-18t+24

=3(5)2-18(5)+24

=9 m/s

A=dv/dt =6t-18

=6(5)-18

12m/s2

11.4] The motion of a particle is defined by the relation x=2t3-6t2+10 where x is expressed in m and t in seconds. Determine the time , position and acceleration when v=0 ?

Sol; GIVEN v=0

dx/dt=0 6t2-12t=0

t(6t-12)=0

At v=0, t=0 and t=2sec

CASE 1; At t=0 ,v=0

X=2(0)3 -6(0)2+10 =10 m

A=dv/dt =12t-12 = -12m/s2

=12m/s2

CASE 2; AT t=2s ,v=0

X=2(2)3 -6(2)2 +10

=16-24+10

=2 m

A=dv/dt =12(2)-12 =24-12 =12m/s2

11.6] The motion of a particle is defined by the relation x=t3-6t2+9t+5 where x is expressed in meters and in seconds . Determine time

(a) When the velocity is zero

(b) The position , acceleration and total distance travelled when t=5s

SOL; (a) v=dx/dt =3t2-12t+9=0

3t2-3t-9t+9=0 3t(t-1)-9(t-1) =0 t=1 or 3 sec

(b) at t=5s X= (5)3 -6(5)2+9(5)+5 =125-150+45+5 =25 At t=5s A =dv/dt =6t-12 = 6(5)-12 =18m/s2

At t=5s S=St=(0=>1)s +St=(1=>2)s+ St=(2=>3)s+ St=(3=>4)s+ St=(4=>5)s

St=(0=>1)s =Xt=1sec -Xt=0 sec

= 13 -6(1)2 +9(1) +5 =1-6+9+5 =4 m

St=(1=>2)s =- 2m

St=(2=>3)s =-2m

St=(3=>4)s=4m

St=(4=>5)s=16m

TOTAL DISTANCE TRAVELLED = 4+2+4+2+16

=28 m

11.8] The acceleration of a particle is defined by the relation a= -4 m/s 2. If v= +24m/s and x=0, when t=0, determine the velocity, position and total distance travelled when t=8sec.

Sol; 1]. A = -4 m/s 2 , at t=0,v=24 m/s find at t=8 sec ,v=?

A = dv/dt dv= a dt

V-24= a(t) 8 = a× 8 = -4 ×8= -32

V-24= -32 v =-32 +24

= -8 m/s

2]. V= ds /dt ds=. V dt

S- 0 = v[8 -0]

S= -8×8 = -64m

= 64m

3]. dv/dt =a = -4 m/s 2

dv= a dt

V-24 = -4t

V= 24 -4t

dx/dt = v= 24-4t

dx= (24- 4t) dt

x=. 24t – 4t 2. = 24t -2t 2

x = 24t - 2t 2

S t=(0-1 sec) = x t=1 – x t=0

= 24(1)-2(1) 2 -0 = 24-2 = 22m

S t(1-2 sec) = x t=2 – x t=1

= 24(2)-2(2) 2 – 22= 48-8-22 = 40-22=18m

S t(2-3 sec) = x t=3 – x t=2

= 24(3)-2(3) 2 -40= 72-18-40=54-40= 14m

St(3 -4 sec) = x t=4 - x t=3

= 24(4)-2(4)2-54 = 96-32-54 =64-54= 10m

S t(4-5 sec) = x t=5 – x t=4

= 24(5)-2(5) 2 -64= 120-50-64=70-64= 8m

S t( 5-6 sec) = x t=6 – x t=5

= 24(6) – 2(6) 2 -70 = 144-72-70 = 72-70=2m

S t(6-7 sec) = x t=7 – x t=6

= 24(7) – 2(7) 2 -72= 168-98-72= 70-72=-2m

S t( 7-8 sec) = x t= 8 – x t= 7

= 24(8)-2(8) 2- 70 = 192-128-70= 64-70= -6m

TOTAL DISTANCE TRAVELLED

= 22+18+14+10+8+2+2+6

= 82 m

11.10] The acceleration of a particle is defined by the relation a = 9-3t 2. The particle starts at t=0 with v=0 and x= 5m. Determine

A) The time when the velocity is again zero

B) The position and velocity when t=4 sec

C) The total distance travelled by the particle from t=0 to t= 4 sec

Sol; a) a = dv /dt.

= dv = a dt

= V= 0 = (9 -3t 2 ) dt = [ 9t – t 3]

= 9t – t 3 =0

= t ( 9 – t 2) =0

= t = 3 sec

b) a = dv/ dt

= dv = a dt = V = ( 9 – 3 t 2) dt = [ 9t – t 3 ]

V = 9(4) – 4 3 = 36 – 64 = - 28 m/s

V= dx / dt

= dx = v dt = ( 9t – t 3 ) dt

= x – 5 = [ 9 t 2/ 2 – t 4/ 4 ]

= x - 5 = 9(4) 2/ 2 – (4) 2 / 4

= x = 72 – 64 +5 = 13m.

C). x = 4.5 t 2 – 0.25 t 4 +5

S t( o- 1) = X 1 - X 0

= 4.5(1) 2 - 0.25 (1) 4 +5 -5 = 4.25m

S t(1 – 2) =. X 2 - X 1

=. 4.5(2) 2 – 0.25 (2) 4 +5 -9.25 = 18 – 4 +5 – 9.25 = 9.75 m

S t(2 – 3) =. X 3 – X 2

=. 4.5 (3) 2 – 0.25 (3) 4 +5 – 19 = 40.5 – 20.25 +5 – 19 = 6.25m

S t( 3 – 4) =. X 4 – X 3

=. 4.25 ( 4) 2 – 0.25 (4) 4 -5 – 25.25 = 72 – 64 + 5 – 25.25 = - 12.25 m

TOTAL DISTANCE TRAVELLED = 12.5 + 6.25 + 9.25 + 4.25

= 32.5m

11.12] The acceleration of a particle is defined by the relation a= - kx -2 . The particle starts with

no intial velocity at X = 800mm , and it is observed that it’s velocity is 6 m/ s when X = 500mm.

Determine.

A) The value of k

B) The velocity of the particle when X = 250 mm

Sol; GIVEN a = - k x -2

At t= 0 ,. X 0 = 0.8 m ,. V 0 = 0

At X = 0.5 m ,. V = 6 m/ s

a = - k x -2 =. V dv / dx

v dv = a dx =. – kx -2 dx

[ v 2/ 2 ] = [ - kx -1/ - 1] = [ k/ x ]

½ v 2 = k/ x – k/ 0.8 = k ( 0.8 – x ) / 0.8 x

V 2 = k ( 0.8 – x) / 0.4x

A) At x= 0.5 , v= 6 put in V 2 6 2 = k ( 0.8 – 0.5 ) / 0.4(0.5) K= 24

B) At x = 0.25

V 2 = 60 ( 0.8 – 0.25) / 0.25

= 132

V = 11.49 m/s

11.14] The acceleration of an oscillating particle is defined by the relation a = - kx . Determine

A) The value of k such that v = 15 m/ s when x= 0 and x= 3m when v = 0

B) The speed of particle when x = 2 m

Sol; A) a= v dv / ds

V dv = a ds

V dv = - k x dX

= [ V 2 /2] = - k [ x 2/ 2]

= 0 – 15 2/2. = - k [ 3 2 /2 – 0]

- 15 2/2 = - k 3 2 /2. = K = 15×15×2/3×3×2. = 25 S -2

B). V dv = - k x dx

[ V 2 / 2 ] = - k [ x 2 / 2 ] =. – k [ 4/2]

V 2 /2 – 15 2 /2 =. – 25 ×4/2

V 2 = [62.5] 2. = 125

V = √125 = 11.18 m/s

11.16]. The acceleration of a particle is defined by the relation a = - 4x ( 1+ k x 2), where a is expressed in m/s 2 and x in meters . knowing that v= 17 m /s when x= 0 , determine the velocity when x = 4 m for

A) k=0.

B) k= 0.015

C) k= - 0.015

Sol; a = v dv /ds.

V dv = a ds

V dv = - 4x ( 1 + kx 2) dx

[ V 2 / 2] = (- 4x – 4kx 3 )dx

V 2 / 12 – 17 2 /2 = [ - 4x 2 /2 – 4 kx 4/4 ]

V 2 /12 – 289/2 = - 2(4) 2 – k(4) 4

V 2/2 = - 32 -256k + 144.5

V 2 = 225 – 512 k

V = √ 225 – 512k

At , k = 0 , V = √ 225 – 512(0) = 15 m/s

At ,. K = 0.015 , V = √ 225 – 512 ( 0.015) = 14.74 m/s

At ,. K= - 0.015 , V = √ 225 – 512 (-0.015) = 15.25 m/s.

11.18 . The acceleration of a particle is defined by the relation a=-4V,where a ,and

V .

Knowing that at t=0 the velocity 30 mm/s , determine

(a) The distance the particle will travel before coming to rest .

(b)The time required for the particle to come to rest .

(c)The time required for the velocity of the particle to be reduced to 1%of its initial value

Sol . (a) a=V =>

=>

=>

(b)

(c)

0.

=>

=>

=> .

11.20 . The acceleration of a particle is defined by the relation . Where

The particle starts at with velocity of 25 . and when the velocity is

found to be . Determine the distance the particle will travel (a) before its velocity drops to .

(b) before it comes to rest .

Sol. Given

When

When

Find (1) distance travelled since velocity drops from 25to 20m/s

(2)distance travelled till v=0

Sol.

=>

-----------(1) Put at

Or putting value of k in (1)

(1) At

V=25 v=10

O 240m

11.22 . The acceleration of a particle is defined by the relation , where a is expressed in m/ . Knowing that the initial velocity of the particle is 15m/s at x=0 , determine

(a) the position where the velocity is 14m/s .

(b) the velocityb of the particle when x=100m .

Sol . (a) given

Find x= at V=14m/s

0.02

or,

=>

=>

(a) at V= 14m/s

(b) now

11.24 . The velocity of a particle is defined by the relation is

expressed in m/s and x in meters . Knowing that x=0 at t=0 ,

determine

(a) The distance travelled before the particle comes to rest. (b) The acceleration at t=0 . (c) The time when x=100m.

Sol. (a) given

At t=0 , at x=0

0.

.

11.30 . A Stone is thrown vertically upward from a point on a bridge located 40m above the water.

( c ) x=0 x=100

t=0 t= -----

푎 = 푣 푑푣 푑푥 푣 = 8 − 0 . 02 푥

Knowing that it stikes the water 4sec after release , determine (a)The speed withb which the

Stone was thrown upward 9b0 the speed with which the stone strikes the water

Sol. 푡 푣 = 0(2)푣 = 0

푦

푎 = −9.81푚푠 푎 = −9.81푚/푠

(1)

(3) 푣 푡

given 푡 + 푡 =4s

푦 = 40푚

Ball is thrown upward at constant acceleration from (1) to (2)

푣 = (푣 ) + 푎 푡 2(−9.81)푦 = 0 − (푣 )

0 = (푣 ) − 9.81푡 =>푦 = (푣 ) /19.62 ------(2) 푡 = ( )

. ----------------(1)

Ball will come down from (2) to (3) at constant acceleration

푣 = 푣 + 9.81푡 2(9.81)(푦 + 40) = 푣 − 0

푡 = 푣 /9.81 ---------------(3) 19.62푦 + 784.8 = 푣

푦 = 푣 /19.62− 784.8/19.62 --------------(4)

From (1) and (3)

푡 + 푡 = 4 =>(푣 )9.81

+푣

9.81

(푣 ) + 푣 = 39.24 --------------(A)

From (2) and (4)

=>(푣 )19.62 =

푣19.62−

784.819.62

=> (푣 ) = 푣 − 784.8

=>39.24 =V2

2 -784.8------------------(A) V0 =29.62 m/s and (v1)0=9.62 m/s 11.32 . An Automobile travels 360m in 30s while being accelerated at a constant rate of 0. .

Determine (a) its initial velocity. (b) its final velocity . (c) the distance travelled

during first 10sec.

Sol. Given S=360m ,t=30sec

.

(b) . (c) .5*10+ ½ (0.5) *

=70m.

11.34 . Automobile A starts from O and accelerates at the constant rate of 0. . A short time later it is passed by bus B which is travelling in the opposite direction at a constant speed of 6 m/s . Knowing that bus B passes point O 20 sec after automobile A started from there , determine when and where the vehicle passed each other .

Sol. V0 =27 km/h=7.5m/sec. a= 0.75m/s2 B A 150m

(a) S = ut +1/2 at2 150=7.5(t)+1/2 (0.75)t2 t2+20t-400=0

t=12.36 sec. (b) v=u+at =7.5+0.75(12.36)=16.77m/s. = 60.4 km/h

11.36 . Automobile A starts from o and accelerates at the constant value of 0.75m/s2 .A short time later it is passed by bus b passes point b which is travelling in the opposite direction at a constant speed of 6m/s . knowing that bus b passes point 20sec after a started from there . determin e when and whre the vehicle passed each other . Sol . let bus a travels x m from o where it meets bus b 2as =v2 –u2

2(0.75 )x =va2

Va =1.22 (x)1/2 m/s V =u+at

1.27 (x)1/2 =0.75t

T =1.22(x)1/2

From b reaches point o at 20 sec with constant speed 6m/s

So, total distance moved in 20 sec = 20 (6) =120

So , after t sec travels (120-x)

S =ut+1/2 at2

120-x = 6t

T = 120-x/6

1.22 (x)1/2/0.75 = 120-x /6

X = 50.59m

T = 120-x /6 =120 -50.59/6 =11.60

11.40 . GIVEN ; VE 5m/s

(a) 2SE+SC =0

2VE + VC =0

2(5) + VC =0

VC =-10 M/S

(B) SE + SW =0

DIFF , VE + V W =0

VW = - V E =-5m/s

VC/E = VC –VE =-10 -5

-15 M/S

(d) V W/E = VW –VE = -5-5

-10

11.42 .

The shells block b moveds to right with a constant velocity 0f 450mm/s , find

(a) VA (b) VB (c) V A/B (d) V C/D

GIVEN

VB =450mm/s

(11.46) Under normal operating conditions tape is transfered between the reels shown at a speed of 720 mm/s.at.t=0 pertionA of the tape is moving to the right at a speed of 600 mm/sand has a constant acceleration.knowing that for then B of the tape has a constant speed of 720 mm/s and that the speed of for then A reached 720 mm/s at t=6s determines the acceleration and velocity of compensator c at t=4s the distance through which of c will have at t=6s

At t=0

( VA )0=600mm/s

VB=(VB)0=720mm/s→ constant

At t=6s

VA=720mm/s

V B=720mm/s

At any t at time ∆t

aA=constant t-4∆sC/∆t=∆sB/∆t-∆sA/∆t

aB=0

∆t→0 at any t -4VC= VB-VA→1

-4aC=aB-aA→2

Motion of A from (2)

VA=(VA)o+aAt -4oc=aB-aA

720=600+aA(6) =0-20

aA=20mm/s→ ac=5mm/s2↓

motion of C at t=4

VC=VC˳+aCt

VC=-30+5(4)=-10mm/s

VC =10mm/s↑

Now t=0

-4VC=VB-VA

-4VC=720-600=120

VC=-30mm/s=vC˳

SC=VC˳t+1/2at2

At t=6s

SC=-30(6)+1/2(5)(6)2=-or SC=90 mm↑90mm

(11.44) Collar A starts from rest and move(3)to the lift with a constant acceleration knowing that

after 6s vel of collar B with respect 16 collar A is 450mm/s,find (a) aA and aB (b)SB and VB after 8 sec Given motion of A

aA=constant

at t=6s

VB/A=450mm/s

Find (a) aA&aB (b)SB&VB after 8sec

Sol] 2SA+SB/A+SB=constant

2SA+SB-SA+SB=constant

SA+2SB=constant

Diff , VA+2VB=0

aA+2aB =0 at t=0

VA+ =0=VB

AA = constant =aB

Motion of a , at t=6

Va=va0+aA t =0|+6 aA

VA=6aA VB=-3aA

VB/A =VB -VA =-450

-3aA -6aA AA= 50mm/s AB =-aA =-25 mm/s

AB = 25 mm/s

Motion of b at t=8s ,

V B =VB0 +aB t

V B =0-25(8) = -200 mm/s

S B =VB0 +1/2aB t2

=0+1/2 (-25)(8)2 = -800 mm

11.48]collar a starts from rest at t=0 and moves upward wuth a constant acceleration of 90mm/s knowing that collar b moves perpendicular with a constant velocity of 400mm/s Determine

(a) the time at wich the velocity of block I s zero

(b)the corresponding position of block c

At t=0 , VA =0 aA = 90mm/s2

V b =400mm/s

At consider movement of c is zero

SA+ SA/B+ SB =constant

Diff , VA+ VA/B+ VB

2VA + VB =0

2VA =400mm/s

VA= 200mm/s

(a)Consider the motion of a at vC =0

At ,VC= 0 VA =200mm/s

200=0+aA t

T=200/90

T=2.22sec

11.50] the three blocks shown are equally spaced horizontally and move vertically with constant velocities . knowing that initially thay are at same level and that the relative velocity of c w.r.t b is 200mm/s . downward determine the velocity of each block so that the three blocks will remain aligned during their motion

GIVEN

-X distance between A,B,C are equal

VC/B =200 mm/s

Kinematics

SA + SD =CONSTANT

VA + VD =0

SB/D + SC/D

VB+VC -2VD =0

FROM 2VA+VB+VC =0

VC/B =VC–VB =200

For alignment

dSB - dsa/d = dsc -dsb /d

VB–VA = VC-–VB

2 VB–VA - VC- =0

From 2VA+VB + (200 +VB) =0

2VB-VA - (200 +VB) =0

VA= 150mm/s

VC =250mm/s

4VB =200

VB=50 mm/s

AREA UNDER THE CURVE A-T,V-T,X-T, INTEGRATION ETC

(11.52) a particle moves in a straight line with the acceleration shown in the fig. knowing that it starts from the origin with v0 =-18m/s .(a)plot the v-t and x-t covers from o<t<20. (b)Determine its velocity ,position and the total distance travelled after 12s.

Soln;-Change in V=Ar(under a-T curve)

At , o ≤t≤4

V4-V0=4*3

V4=12+V0=12-18=-6m/s

From 4≤t≤10

V10-V4=6x6

V10=36+V4=36-6=30m/s

From 10≤t≤12

V12-V10=-10

V12=-10+V10=-10+30=20m/s

From 12≤t≤20

V20-V12=-40

V20=-40+V12=-40+20=-20m/s

Change in X=Ar(under V-T curve)

At 0≤t≤4

X4-X0=-6x4+1/2x-12x4=-48

X4=-24-24=-48+0=-48

X5-X4=1/2x1x-6=-3

X5=-3+X4=-3-48=-51

X10-X5=1/2x5x30=75

X10=75+X5=75-51=24

X12- X10=1/2x10x2+2x20

=10+40=50

Position X12 at t=12

50+X10=50+24=74

Velocity at t=12=20m/s

Total distance moved =(X4-X0)+(X5-X4)+(X10-X5)+(X12-X10)

=48+3+75+50=176m

(11.54) a particle moves in a straight line with the velocity shown in the fig .knowing that X=-16m and t=0 ,draw that and X-t curve for the 0<t<40sec and determine (a) the maximum value of the position coordination of the particle (b) the value of t for which the particle is at a distance of 36m from the origin

Solution:

Given ,X0=-16m, at t=0

Change in X=Ar(curve under V-t)

X10-X0=10X2=20

X10=20+X0=20-16=4m

X18-X10=8x2+1/2x8x4

=16+16=32

X18=32+X10=32+4=36

X24-X18=1/2x6x6=18

X24=18+X18=18+36=54m

X30-X24=-1/2x6x6=-18

X30=-18+X24=-18+54=36m

X40-X30=-6x10=-60

X40=-60+X30=-60+36=-24

(a) Maximum position of value co-ordinat =54m

(b) At X=36m, t=18,36 sec

From t(10-18)

V=6, U=2m/s t=8sec

V=U+at

6=2+a(8)

a=4/8=0.5m/s2 from

t(18-24)

V=0, U=6, t=6

V=U+at

0=6+a(6)

a=-1m/s2

from t(24-30)

V=-6, U=0, t=6

V=U+at

-6=0+a(6)

a=-1m/s2

(11.60)A motorist is travelling at 54km/hr when she observes that a traffic signal 280in ahead of her form red.She knows that signal is timed to at any red for 28 sec.What would she do to pass the signal at 54km/hr just has it tenrns green again ? Draw V-t curve selecting the solution which calls for the smallest possible acceleration and deceleration and determine (a) the determination and acceleration in m/s2 (b) the minimum speed reached in km/hr

Solution: consider the velocity between (1)to(2)

V=V0+at

V1=15-at1→(1)

V1=15-14a

S=Ut+1/2at2

X=15t1-1/2at12→(2)

X=210-98a→(2)

Consider the motion between (2)and(3)

V=V0+at

15=V1+a(28-t1)

15=V1+a(14)→(3)

S=Ut+1/2at2→X=15t1-1/2at12→(2)

X=210-98a→(2)

Consider the motion between (2)and(3)

V=V0+at

15=V1+a(28-t1)

15=V1+a(14)→(3)

S=Ut+1/2at2→ 280-x=V114+98a→(4)

x =210-98a→(2)

V1=70/14=5m/s

(3)→15=5+a(14)

a=10/14=0.714m/s2

a=±0.714m/s2

( maximum speed ) V1=5m/s=18km/hr

(11.62) an automobile at rest is passed by a track travelling at a constant speed of 54km/hr.the automobile start and acceleration for10 sec.at a constant rate until it reacher a speed of 90km/hr.if the automobile then massition a constant speed 90km/hr,determine when the automobile start

(a) just as the track passes it (b) 3sec after the track has passed it

54km/hr=15m/s

90km/hr=25m/s

(a)Motion of automobile for 10sec t1=10s u=0 v=25m/s

V=U+at

25=0+a(10)=2.5

a1=2.5m/s2

S1=Ut1+1/2a1t12

S1=0+1/2x2.5x102

S1=125m

25+S=25t2

S =15t2→(1)

25=10t2→(2)

t2=2.5sec

total time =t1+t2=10+2.5=12.5sec

S=15x12.5=187.5m

15xt=S

25xt=70±S

-10t=70, t=7sec

Total time=7+10+3=20sec

S=125+70+7x15

=195+105=300m

S=300m

CONSTANT ACCELERATION

(11.56) A bus start from rest at point A and accelerations at the rate of 0.8m/s2 until if reached a speed of 12m/s .it then proceed at 12m/s.it then proceeds,at 12m/s until the break are applies,if comes to rest at point B ,42m beyond the point where the breaker were the breakes are applies.assumining uniform decelaration and knowing that the distance bet A and B is 300m.determine the time required for the bus to travel from A toB.

Solution: V1=V0+at1, motion of bus from V0=0 to V1=12m/s

12=0+0.8t,→ t1=12/0.8=15sec

2as1=v12-v0

2

2x0.8xs1=122-02

S1=90m

Motion of bus at constant speed of 12m/s upto 42m before B where breake

applids S2=300-90-42=168m

t2=S2/t2=S2/v1=168/12=14sec

motion of bus from breake applied to point B

S3=42m

(V3)0=12m/sec

V3=0

S=ut+1/2at2

42=12xt3+1/2a3t32→(1)

V=U+at

0=12+a3t3→a3=-12/t3→(2)

Put a3 in (1)

42=12t3+1/2x-12/t3xt32

42=12t3-6t3

6t3=42

t3=7sec

total time=t1+t2+t3=15+14+7=36sec

11.68: The acceleration was obtained for a small airplane travelling along a straight curve knowing that

X=0 and v=60m/s when t=0, determine

A) the velocity and position of the plane at t=20 sec

B) its average velocity during the interval 6s<t<14s

a(m/s^2)

0.75

O 6 10 12 14 t(sec)

-0.75

Vo=60m/s V

t=0 sec ts

60

58.5

O 6 10 14 20

Velocities:

푉 =푉 +

=60+0=60m\s

푆 =푆 +

=0+[6*60+8*58.5+8*1.5*1/3+6*60]

=360+468+4+360=1192m

= =1192/20=59.6m\s

11.74: the maximum possible deceleration of a passanger train under emergency conditions

Was dwtermined experimentally . the results are shown in the figure.if the brakes are applied when the travelling at 90km\hr ,determine by approximate means

A)the time required for the train to come to rest

B)the distance travelled in that time

Solution:

At t=0,푉 =25m\s

Find t=?when v=0

And s=? till v=0

a=Vdv/ds

Vdv=ads

A) A=dv/dt

Dt=dv/a=1/a*dv

∫ 푑푡 = ∫ 1/푎푑푣t= (25-20) +

.(20-10)+ (10-0)

B) Vdv=ads When v=25 to 20 ,a=- 2

∫ 푉푑푣 = ∫ −2푑푠 - =-2S1

푆1 = 56.25푚

When v=20 to v=10 a=-2.5

∫ 푉푑푣 = ∫ −2.5푑푠S2=116.25m

When v=10 to v=0 a=-3

∫ 푉푑푣 = ∫ −3푑푠 S3=132.92m

Rectangular components of velocity and acceleration

+projectile,relative motion

11.80:the motion of a particle is defined by the equations x=2t^3-8t^2 and y=3t^2-12t

Where x and y are expressed in meters and t in seconds bdetermine the velocity and acceleration when t=1s,t=2s,t=3s

Solution :

dx/dt=6t^2-16t,d^2t/dt^2=12t-16

dy/dt=6t-12,d^2y/dt^2=6

푎⃗=(d^2x/dt^2)횤⃗+(d^2y/dt^2)횥⃗

푎⃗ =(12t-16)횤⃗+6횥⃗

푎⃗=-4횤⃗+6횥⃗ ,at t=1sec a=7.21m/s^2

푎⃗ =8횤⃗+6횥⃗ ,at t=2 sec a=10m/s^2

푎⃗=20횤⃗+6횥⃗ ,at t=3sec a=20.9m/s^2

푉⃗=(6t^2-16t)횤⃗+(6t-12)횥⃗ m/s

At t=isec 푉⃗=-10횤⃗-6횥⃗ V=11.66m/s

At t=2sec 푉⃗=-8횤⃗ V=8m/s

At t= 3sec 푉⃗=6횤⃗+6횥⃗ V=8.48m/s

11.82:the motion of a particle is deined by the equation x=t^2-8t+7 and y=0.5t^2+2t-4 where x and y ae expressed in metres and sec

A)determine the magnitude of the smallest velocity reached by the particle

B)the corresponding time,position and direction of the velocity

Solution:

V=(dx/dt)횤⃗+(dy/dt)횥⃗ = (2t-8)횤⃗+(t+2)횥⃗

At,t=0, 푉 = (−8) + 2 =8.25m/s

t=1, 푉 = (−6) + 3 =6.7m/s

t=2, 푉 = (−4) + 4 =5.66m/s

t=3, 푉 = (−2) + 5 =5.38m/s

t=4, 푉 =√0 + 6 =6m/s

t=5, 푉 =√2 + 7 =7.28m/s

At t=3sec, velocity is 5.38m/s which is minimum

푉 . = (−3) + 4.5 =5.4 m/s

Position: S=(3 -(8*3)+7)횤⃗+(0.5*3 +(2*3)-4)횥⃗

=-8횤⃗+6.5횥⃗

Direction is along horizontal

11.92:After rolling down a 20 incline a sphere has a velocity 푉 at point A.Determine the range of values of V0 for which the sphere will enter the horizontal pipe shown.

Solution:

Consider the vertical motion of the sphere (uniform acceleration):

Now,

(푉 ) =푉 sin 20 = 0.34푉

y= 푉 푡+ a푡

=0.34푉 푡 + (9.8)푡

1 0r 1.5=0.34푉 푡 + 4.9푡 (1)

Horizontal motion of ball:

Now , (푉 ) = 푉 cos 20=0.94푉

x= (푉 )t= 0.94푉 푡

2=0.94푉 푡푉 푡=2.13 (2)

Putting the values of (2) in (1)

1 or 1.5 = 0.34(2.13) + 4.9푡

Or, 1 or 1.5 =0.7242 + 4.9푡

Or, t= ..

or . ..

=0.237246sec or 0.3979sec

So 푉 = ..

or ..

from (2)

8.98m/s or 5.353 m/s

So,5.35≤푉 ≤8.98

11.94: An oscillating water sprinkle is operated at point A on an incline which forms an angle α with the horizontal the sprinkler discharges with an initial velocity 푉 at an angle φ with the vertical which varies from -휙 to +휙 knowing that 푉 =8m/s ,휙 = 40 and α=10 ,determine the horizontal distance between the sprinkle and point B and C which define the watered area.

Solution:

For +휙 (vertical motion uniform acceleration): 푦 =푉 cos 40 t- (9.8)푡

푦 =6.13t-4.9푡 (1)

for horizontal motion(uniform motion): 푉 sin 40 t=푥 = (2)

putting the value of 푦 in (2)

푉 sin 40 t= . .

5.14t= . ..

4.9푡 -5.223t=0

t= . ±√ ..

=1.0659 sec

So,푥 = 8 sin 40 (1.0659)=5.48m

Horizontal motion if -휙 is considered:

Uniform motion: 푥 =푉 sin 40 t=5.14t (1)

Vertical motion (uniformly accelerated):

-y=푉 cos 40 t- (9.8)푡

-y=6.13t-4.9푡 (2)

y=푥 tan 10 (3)

put (3) in (1)

-푥 tan 10 = 6.13t-4.9푡

-5.14t*(0.176) = 6.13t-4.9푡

4.9푡 -7.034t=0

t= . ±√ ..

=1.436 sec

So,푥 =5.14*(1.436) =7.38m

11.96:a nozzle at A discharges water with an initial velocity of 푉 at an angle of 60 with the horizontal. Determine the stream of water strikes the roof ,The range of values of initial velocity

For which the water will fall on the roof.

Solution:

Consider the vertical motion (uniformly accelerated):

4.8=푉 sin 60 t-4.9푡 (1)

Consider the horizontal motion which is uniform:

1)푥 =4.5=푉 cos 60 t (2)

2)푥 =12=푉 cos 60 t (3)

Putting the value of t of (2) and (3) in (1) to take the range of 푉

1st case :4.8= 푉 sin 60 * . -4.9( . )

4.8=7.794- .

-2.994=- . 푉 =11.51m/s

2nd case: 4.8=푉 sin 60 * -4.9( )

4.8=20.7846- . 푉 =13.28m/s

So, 11.51≤푉 ≤13.28

11.98: sand is discharged at A from a conveyer belt and falls onto the top of a stock pile at B. knowing that the conveyer belt forms an angle α=20 with the horizontal , Determine the speed 푉 of the belt.

Solution:

Consider the horizontal motion which is uniform: 푉 cos 20 t=10 (1)

Consider vertical motion which is uniformly accelerated: -6=푉 sin 20 t-4.9푡

-6= ∗ -4.9*( )

-6=3.64- 푉 =7.7m/s

11.100: knowing that the conveyer belt moves at the constant speed 푉 =8m/s , Determine the angle α for which the sand is on the stock pole B

Solution: consider horizontal motion (uniform):

8cosαt =10 (1)

Consider vertical motion (uniform acceleration):

-6=8sinα 푡-4.9푡

Putting the value of t of (1) in (2) -6=8sinα ∗ – 4.9 *( )

-6=10tanα - 7.656 (secα)

-6=10tanα - 7.656(1+(tanα) )

-6= 10tanα - 7.656-7.656(tanα) tanα =1.1 or 0.195

α = 47.72 or 11

11.104 A player throws a ball with an initial velocity v0=16m/s from a point A located 1.5m above the floor knowing that h=3.5m,determine the angle ( ) for which the ball will strike the wall at point B.

Solution:

Considering horizontal motion (uniform)

18=16cos( )t -----(1)

Considering vertical motion (uniform acceleration

ℎ − 1.5 = 16 sin(푎) 푡 − 4.9푡2

3.5 − 1.5 = 16 sin(푎) ×18

16 cos(푎) − 4.9 ×18

16 푐표푠 (푎)

2 = 18 tan(푎)− 6.2푠푒푐 (푎)

2 = 18 tan(푎)− 6.2(1 + 푡푎푛 (푎)

2 = 18 tan(푎)− 6.2− 6.2푡푎푛 (푎)

6.2푡푎푛 (푎)− 18 tan(푎) + 8.2 = 0

푡푎푛 (푎)− 2.9 tan(푎) + 1.32 = 0

tan(푎) =2.9 ± 2.9 − 4(1.32)

2

tan(푎) =2.9 ± 1.766

2 2.33,0.566

Ans : ° °

11.106 The velocity of Boats A and C are as shown in figure and the relative velocity of boat B w.r.t A is VB/A=1.2푚 푠⁄ angle =50°.Determine a)VA/C b)VC/B C)The change in position of B and C during a 10-sec interval. Also show that for any motion VB/A+VC/B +VC/B=0.

30°N

1.5m/S C A 1.8m/s

W E

S

a)VA/C=VA-VC

VC =1.5m/s

VA/C

120°

VA=1.8m/s

V

sin120° =1.8

sin(훼) =1.5

sin{180°− (120° + 훼) =1.5

sin(60° −훼)

1.8sin(훼) =

1.5푠푖푛60° cos(훼) − 푐표푠60°sin(훼)

1.8sin(훼)

=1.5

0.866 cos(훼)− 0.5sin(훼)=> 15.588 cos(훼)− 2.4 sin(훼) = 0

15.588− 2.4 tan(훼) = 0

tan(훼) = 0.6495 => 훼 = 33°

푉 =1.8 × 푠푖푛120°

푠푖푛33° = 2.86푚/푠

ANS:…… 27°

VA/C

b)

VB VB/A=1.2m/s

α 130°

VA

Given

VB/A=1.2m/s

50°

푉푠푖푛130° =

푉 /

sin(훼) =푉

sin{180° − (130° + 훼)}

1.2sin(훼)

=1.8

sin(50° −훼)

0.92 cos(훼)− 2.568 sin(훼) = 0 => tan(훼) = 0.358,훼 = 19.7°

푉 =1.2푠푖푛130°푠푖푛19.7° = 2.73푚/푠

VC/B

VC α VB

100.3° 19.7°

푉 /

sin(100.3°) =푉

sin(훼) =푉

sin(79.7− 훼)

=>1.5

sin(훼) =2.73

sin(79.7−훼)

=> 1.47 cos(훼) − 3 sin(훼) = 0

=> tan(훼) =1.47

3 => 0.49, (훼) = 26.1°

푉 / =1.5 × 푠푖푛100.3°

푠푖푛26.1° = 3.35푚/푠

VC/B

6.5°

C) SB/C =VB/C× 푡 = 3.35 × 10 = 33.5푚… . .퐴푛푠

6.5°

SB/C

11.108 Air plane A is flying due east at 700km/hr while air plane B is flying at 500km/hr at the same altitude and in a direction to the west of south. knowing that the speed of B with respect to A is 1125km/hr. determine the direction of the flight path of B.

Soln: (N)

(W ) B (E)

α VA=700km/hr

(S)

VB=500km/hr .

Given : VB/A=1125km/hr

VA=700km/hr

α β

VB=500km/hr

VB/A=1125km/hr

Using cosine formula:

1125 = 700 + 500 − 2(700)(500)푐표푠훽

푐표푠훽 =700 + 500 − 1125

2(700)(500)

훽 = 138.7°

훼 = 180° − 138.7° = 41.3°푠표푢푡ℎ표푓푤푒푠푡.

Alternate method:

푉 = 푉 + 푉 / -----------(1)

Algebra:푉 = 700푖

푉 = 푉 푖 + 푉 푗

푉 = 푉 푖 + 푉 푗 --------put in (1)

푉 푖 + 푉 푗 = 700푖 + 푉 푖 + 푉 푗

푉 = 700 + 푉 − −− −− (2)

푉 = 푉 − − −− −−−−(3)

2 + 3

=>푉 + 푉 = 700 + 1400푉 + 푉 + 푉

500 = 700 + 1400푉 + 1125

=> 푉 = −1075

푉 = 푉 − 푉 = 1125 + 1075 = 332

From (2)

푉 = 700 + (−1075) = −375

푓푟표푚(3)

푉 = 332푉

푡푎푛훼 = ,훼 = 41.5°푠표푢푡ℎ표푓푤푒푠푡푉 α

푉

11.110 Knowing that the instant shown assembly A has a velocity 225mm/s and an acceleration of 375mm/s2 ,both directed downward. Determine a)the velocity of block B b)the acceleration of block B.

Given 푎 = 375

푉 = 225

푆 + 2푆 = 푐표푛푠푡

Differentiating the above eqn we get

푉 + 2푉 = 0

푉 = −푉2 =

−2252 = −112.5

푚푚푠

=> 푉 = 112.5

#using cosine rule :

푉 = 225 + 112.5 − 2(225)(112.5)푐표푠50°푉 = 112.5

= 175.32 50° β

Using sine rule: VA =225 α

112.5푠푖푛훼 =

175.2푠푖푛50° => 푠푖푛훼 =

112.5푠푖푛50°175.2 푉 ′

= 0.49

60.55°훼 = 29.44°

VB’=175.32mm/s 훽 = 100.55°

B) 푎 + 2푎 = 0푎 = 187.5

푎 = −3752 = −187.5푚푚 푠 50° β

푎 = 375 + 187.5 − 2(375)(187.5)푐표푠50°푎 = 375 α 푎 ’

푎 = 292.3푚푚 푠

187.5푠푖푛훼 =

292.3푠푖푛50° => 푠푖푛훼 = 0.49,훼 = 29.43,훽 = 100.57°60.57°

푎 ′

11.112 ,11.113 A t t=0 , a=wedge A starts moving to the left with a constant acceleration of 80mm/s2 and block B starts moving along the wedge towards the right with a constant acceleration of 120mm/s2 relative to the wedge . Determine a)the acceleration of block B b) the velocity of block B when t=3s.

Soln: B B

20° A 20°

11.112 11.113

11.112 푎푡푡 = 0,푎 = 80푚푚 푠 푎 =120 y

푎 = 120푚푚 푠 20° 20° α β

푎 = 120 + 80 − 2(120)(80)푐표푠20°푎 = 80푚푚 푠

= 52.52푚푚 푠

푎 = 52.52 80푠푖푛푌

=52.52푠푖푛20°

=> 푠푖푛푌 =푠푖푛20° × 80

52.52= 0.520951.4°

푌 = 31.397°,훼 = 128.6°, 훽 = 51.4°

푉 = 푉 + 푎푡

= 0 + 52.52 × 3 = 157.6푚푚 푠⁄ 51.4°

11.113

푎 = 80 + 120 − 2(80)(110)푐표푠20°

= 52.52푚푚 푠 푎 = 80푚푚 푠 α

= 푉 = 157.6푚푚 푠 20° α

80푠푖푛훼 =

52.52푠푖푛20° => 푠푖푛훼 =

80 × 푠푖푛20°52.52 푎 = 120

푚푚푠

훼 = 31.39°

훼 = 31.39°

푎 = 52.52푚푚푠

CURVILINEAR MOTION(translation).RADIAL,TRANSVERSE,POLAR

11.120 What is the smallest radius which should be used for a highway curve if t he normal component of the acceleration of a car travelling at 72km/hr is not to exceed 0.72m/s?

soln:

푉 = 72푘푚/ℎ푟 = 20푚/푠

푎 = 0.72푚/푠

푎 =푣휌 => 휌 =

푣푎 =

200.72

퐴푛푠 ≔ 555.56푚

11.122 Determine the peripheral speed of the centrifuge test cab A for which the normal component of the acceleration is 10g.

7.5m

Given, 푎 = 98.1푚/푠

휌 = 7.5푚

푎 =푣휌 => 푉 = 푎 휌

= √9.81 × 7.5

Ans: = 27.12푚.

11.124: A motorist is travelling on a curved portion of highway of radius 350m at a speed of 75km/hr.The brakes are suddenly applied, causing the speed to decrease at a constant rate of

1.25 . Determine the magnitude of the total acceleration of the automobile (a) immediately after the brakes have been applied (b)4 sec later.

Soln

(a) ρ= 350푚, 푣 = 20푚 푠, 푎 = −1.25푚 푠

푎 =푣푟 =

20350 = 1.143푚 푠 ,푎 = −1.25푚 푠

푎 = 푎 + 푎 = √1.694푚 푠 ----ans

(푏)

푉 = 20− 1.25(4) = 15푚 푠⁄

푎 = −1.25푚 푠 ,푎 = −1.25 + 0.643 = 1.406푚 푠

11.126 The speed a racing car is increased at a constant rate from 100km/hr to 120km/hr over a distance of 180m along a curve of 240m radius. Determine the magnitude of the total acceleration of the car after it has travelled 120m along the curve.

Soln: 푉 = 27.78푚 푠⁄ over a distance 180m increased

푉 = 33.33푚 푠⁄ rate of speed is constant.

휌 = 240푚,푓푖푛푑푎? 푎푓푡푒푟120푚

푎 =푉 − 푉푡 − − −−(1)

푠 = 푢푡 +12푎푡

180 = 27.78푡 +12 (푉 − 푉

푡 )푡

180 = 27.78푡 +12

(33.33− 27.78)푡

180 = 27.78푡 + 2.775푡

푡 = 5.89푠푒푐 − −− −(2)

Putting the value of t in (1)

푎 =33.33− 27.78

5.89 = 0.942푚푠

V at 120m:

푉 − 푈 = 2푎푠

푉 − 27.78 = 2 × 0.942 × 120

푉 = 31.59푚 푠⁄ 푎푡푑푖푠푡푎푛푐푒푠 = 120푚

푎 =푣휌 =

31.59240 = 4.15

푚푠

푎 = 0.942푚푠

Total acceleration

푎 = 푎 + 푎 = 0.942 + 4.15

= 4.263 ------ans.

11.130)

A Nozzle discharges a stream of water in the direction shown with an initial velocity of 7.5 m/s. Determine the radius of curvature of the stream (a) as it leaves the nozzle (b) as the maximum height of the stream.

(a) an = 9.81 Cos 35° = 80.36 m/sec2

an= V2/ρ r = V2/an

= 7.52/8.036 = 7 m

(b) an= 9.81 m/sec2 ρ = Vo Cos 35°/9.81 = (7.5 Cos35°)2/9.81 = 3.85 m

11.132) Determine the radius of curvature of the trajection described by the projectile of as shown in figure (a) as the projectile leaves the gun (b) at the Mamimum elevation of the projectile.

55° 35°

(a) an = V2/ρ

= 1802 /ρ

ρ = 1802/9.81 Cos30°

= 2810 m ( b ) ρ = 180Cos30°/9.81 = 2480m

11.142)

The two dimentional motion of a particle is defined by the relations π = 2bCosωt and θ=ωt, when

b and ω are constant. Determine (a) the velocity and Speed of the particle at any instant (b) the

radius of curvature of its path what conclude can you draw regarding the path of the Particle?

11.128)

Automobile A is travelling along a straight highway, while B is moving along a circular exit

ramp of 150 m radius. The speed of A is being increased at the rate of 1.5 m/sec2 and the speed

of B is being decreased at the rate of 0.9 m/sec2 for the position of them, determine (a) the

velocity of A relative to B (b) The acceleration of A relative to B.

at= dv/dt=-0.9 m/Sec2

an= v2/ρ =11.112/150=0.823 m/Sec2

aB=√푎푏푡 + 푎푏푛 VA/B= VA – VB

= 1.22 m/sec2 VA/B + VB = VA

VA=20.73m/Sec2

α VA/B

180m/s at an 30° A=9.81

VB/A= VB – VA VB/A+ VA = VA

A

B

VA/B=√11.11 + 20.73 − 2(11.11)(20.73)퐶표푠32°

= 12.42 m/sec

11.11/Sinα=12.42/Sin30° VA/B=12.42 m/S Sinα=11.11Sin30°/12.42 26.57° α=26.57°

an atα=tan-1an/at =tan-10.823/0.9 = 42.3° 12.3°

a A/B 1.22/Sinα=2.7/Sin16.36° 167.7° aB=1.33m/s2 Sinα = 0.097 α = 5.57° aA=1.53 m/S2

aA/B =√1.5 +1.222-2(15)(1.33)Cos16.76° =2.7m/s2

11.140) The rotation of rod about O is defined by the relation ϴ=2t2, where ϴ is expressed in radian and t in seconds. Collar B Shades along the rod in such a way that its distance from O is π=60t2-20t3, where ‘r’ is in mm and t in sec. When t=1sec

(a) Determine the Velocity of the collar (b) The total acceleration of the collar (c) The acceleration of the collar relative to the rod.

ϴ=2t2

R=60t2-20t3

Find (a) VB (b) aB (c) aB/OA when t=1S

(a) V=√Vr2+Vϴ

2

Vr=dr/dt=120t-60t2

A r B ϴ . O

Vϴ=rdϴ/dt

=(60t2-20t3)4t

= 240t3-80t4

V=√(120t-60t2)2+(240t3-80t4)2

At t=1 sec

V=√(120-60)2+(240-80)2

=√602+1602

= 170.68 mm/s

Or V=60ϵr +180ϵϴ

(b) ar= d2r/dt2-r(dϴ/dt)2 at t=1 sec

=120-120t-(60t2-20t3)(4t)2

= -(60-20)16

= -640

aϴ= rd2ϴ/dt2+2dr/dt*dϴ/dt

=(60t2-20t3)4+2(120t-60t2)(4t) at t=1Sec

aϴ=2(120-60)4+40(4)

= 640

a=-640ϵr +640ϵϴ

Soln: (a) r=2bcosωt

θ= ωt

V =√ Vr2+Vθ

2 Vr= dr/dt Vθ=rdθ/dt

Vr= dr/dt=d(2bcosωt)/dt

=-2bsinωt(ω)

=-2bwsinωt

Vθ=rdθ/dt = 2bcosωt(w)=2bwcosωt

V = √(-2bwsinωt)2+(2bwcosωt)2

= 2bw(√δm2ωt+cos2ωt) =2bw

a = √ar2+aθ

2

ar=d2r/dt-r(dθ/dt)2

=-2bwcosωt(ω)-2bcosωt(ω)2

= -2bw2cosωt-2bw2cosωt

= -4bw2cosωt

aθ=rd2θ/dt+2dr/dt*dθ/dt

= -bcosωt(0)+2(-2bwsinωt)ω

= -4bw2sinωt

a =√(-4bω2cosωt)2+(-4bω2sinωt)2

a = 4bω2

11.156] The pin at B is free to slide along the circular slot DE and along the rotating rod OC

.Assuming the rod OC rotates at a constant rate θ°,

(a) show that the acceleration of pin B is of constant magnitude.

(b) Determine the direction of acceleration of point B

r D B C

s

O A b

b

GIVEN DE is circle with centre A ,radius b

OC is a straight rod rotating at a constant rate θ=dθ/dt= w0

Pin B remains on circle DE which is fixed and rod OC which is rotating

FIND ACCELERATION OF PIN B (1)INTERMS OF PATH PARAMETER S IS NORMAL AND LONGITUDNAL FORM LET PIN

B be at O| at t=0

And let be at position B at time t

Let S be the curvilinear coordinate Velocity

of pin at B = v= ds/dt ; S= b(2θ)

ds/dt =2bw0.

V= 2bw0

ACCELERATION OF PIN B AT ANY TIME T At = tangential component= dv/dt=0 as V = constant

An =normal component of acceleration = v2/b=(2bw0)2/b = 4bw02 = a=constant

2] IN CARTESIAN COORDINATES At any time t, coordinates of pins B are

X= OG= OA+AG= b+bcos2θ

Y=BG = bsin2θ

VELOCITY OF PIN B AT ANY TIME T Vx =dx/dt= -2bw0sin 2θ

VY =dy/dt= 2bw0cos2θ V=(Vx2+V

2)1/2 =2bw0 tan a=vx/vy=tan2θ

=>a=2θ

ACCELERATION OF PIN B AT ANY TIME T Ax= dvx/dt= d2x/dt2=-4w0

2cos2θ

Ay=dv y/dt= d2y/dt2=-4w 02sin2θ

A= (ax2+ay2)1/2

=4w02b

Tan b= tan 2θ

=>b=2θ

3] In polar coordinate / in terms of radical tranverse form:

At any time t ,pin is at B (r,θ) r=2bcosθ

dθ/dt =w0

Velocity of pin B(t) Vr =rdθ/dt= -2bw0sinθ Vθ=rdθ/dt=

2bw0cosθ

V=(Vr2+ Vθ 2)1/2 = 2bw0 Tan

a= tanθ

a=θ

ACCELERATION OF PIN B IS Ar = d2r/dt2 - r(dθ/dt)2

= -2bw02cosθ - 2bw0

2cosθ β=Ф

= -4bw02 cosθ Ф 2Ф a

A0=rd2θ/dt2 + 2(dr/dt)(dθ/dt)

=0+2[-2bw0 sinθ]w0

= -4Abw02 sinθ

a=(ar2+ao2)1/2 = 4bw02

Tan b = tan θ

⇨b= θ

V.V.I Shames problem

GIVEN; points A and B are constrained to remain in fixed slot (elliptical)and rod (movable) BA

At the instant shown velocity of rod ABC is= Vc =2 m/ s

X=1.5m ,ac =3 m/s2

Find vB and aB at the instant

At any time t pin is at B(x,y)

Then, x2/a2 + y2/b2 = 1 or x2/32 + y2/22 = 1

⇨ x2/9 + y2/4= 1 ----------1

As the pin B always remains in slot of rod ABC

⇨ X component velocity of pin = Vx=2m/s

⇨ Also , X component of acceleration of pin =Ax = 3m/s2

Differentiate 1 w.r.t t

1/9 (2x)dx/dt+ ½(y)dy/dt = 0

2/9(x)dx/dt +1/2 (y) dy/dt = 0,----------2

But, dx/dt = Vx = 2. ; dy/dt = Vy X= 1.5 m ; from (1.5)2/9 + (y)2/4= 1

⇨ y= +-1.732m

Put in (2) , 2/9(1.5) 2 + 1/2(1.732)dy/dt =0

⇨ dy/dt = Vy = -0.770

⇨ Velocity of pin B at that instant is

V = Vxi+ Vy j = 2i -0.770j = VB

= -2i+0.77.j = VA

Differentiate equation 2 w.r.t t

2/9[xd2x/dt2+ (dx/dt)2]+1/2 [yd2y/dt2+ (dy/dt)2] = 0

At the instant shown ,

2/9[1.5(3) + (2)2]+1/2 [1.732d2y/dt2+ (dy/dt)2] = 0

1.889+0.866 d2y/dt2+ 0.296 = 0 d2y/dt2 = ay

= y acceleration of pin B = -0.560

⇨Acceleration of pin B qt instant shown is

a= axi+ ay j = 3i-0.560j = a B

3i+ 0.560j =a A

Relative velocity of B w.r.t A is VB/A=VB - VA

2i - 0.770j - 2i - 0.770j = -1.540j

i.e B approaches A at a speed of 1.540 m/s a1 v1

SHAMES GIVEN at t= o

r= 1.5i+ 2j + k (m)

V= 0 At any time t

a= 2t i+1.5t2j +3k m/ s2

TO FIND At t1= 10 sec. Find V1 , a1, r1.

Let V, a, r be the velocity acceleration and position vector at ‘t’

GIVEN , a= dv/ dt= 2t i + 1.5 t2 j + 3k

Separate variables and integrals v∫dv= i

0t∫2tdt+ j 0t∫1.5t2dt+ k 0t∫ 3dt

[V]v = i[t2]t0+ j[ 0.5t3]t

0+ k [ 3t]t0

V = t2i + 0.5t3 j + 3t k

We know , V = t2i + 0.5t3 j + 3t k = dr/ dt

Separate variables and integrals

r∫dr =i0t∫t2dt+j 0t∫0.5t3dt + k 0t∫3tdt

[r]r = i[t3/3]t0+ j[0.125t4]t

0+k[1.5 t2]t0

r-r0 = t3/3 i + 0.125t4 j + 1.5 t2 k

r = ( t3/3+ 1.5) i +(0.125t4+2) j +(1.5 t2

+1)kAt t1= 10 sec

r1= rt=334.83+1252 j + 151 k

v1= vt=10= 100 i+500j+ 30 k

a1= at=10 = 20 i+ 150 j + 3 k

MIXED PROBLEM

11.170] Two airplanes A and B areflying at the same altitude plane A is flying due east at a constant speed of 800 kilometres per hour while plane B is flying southwest at constant speed of of 500 kilometres per hour determine the change in position of plane B relative relative to plane A, which takes place during a a 3 minute interval.

N

A

W E Sa

Sb 135* α

B Sb/A

S

Sol; VA = 13.33km/hr

V B= 8.33 km/hr

S B/A=[(40)2+(25)2 -2(40)(25)cos135]1/2

⇨60.3KM 25/sin a =60.3/sin 135

⇨ Sin a= 0.293 ⇨a= 17.05°

11.172] A ball is projected from point A with a velocity V0 which is perpendicular to the incline shown. Knowing that the ball strikes the incline at B , determine the range R in yerms of V0 and β .

90 V0

X O

A )β X

R B

Sol] x= V0 sin βt = Rcos β -------1

-y= V0 cos β t – 4.9t2

⇨ -Rsin β = V0cos β t- 4.9t2------- 2

Put the value of t in 1

⇨ -Rsin β = V0 cos β (R cos β / V0 sin β) -4.9 (R cos β / V0 sin β)2

⇨ - sin2 β = cos2 β - 4.9 (R cos2 β / V02 sin2 β)

⇨ 4.9 (R cos2 β V02 sin2 β)= 1

⇨ R= V02 sin2 β /4.9 cos2 β

11.176] knowing that block A movesdown to the left with a constant velocity of 80 millimetres per second determine

(a) the velocity of block B

(b) the change in position of block A relative to block B which takes place in 4 seconds. Sa Sb A 90* B 20*

Sa=320mm Sb=160mm 20* Sa/B

90*

20* α

Sol; SA+ 2SB= constant

VA+ 2VB= 0

80+ 2VB= 0

2VB=- 80

VB=-40

S A/B = [(320)2+(160)2 -2(320)(160)cos 90]1/2

⇨ 358 mm

320/sin α=358/sin 90

Sin α= 320/358

⇨ α= 63.43°

DYNAMICS _ A _ 1 Rectilinear Motion of particle

Ex 1: The motion of a particle is defined by the relation 풙 = 풕ퟑ − ퟔ풕ퟐ + ퟗ풕 + ퟓ(풎) . Determine time (a) when the velocity is zero (b) position, acceleration and total distance travelled when t=5sec?

Ans : (a) t=1 or 3 sec

(b)25m, 18풎/풔ퟐ , 28m

Ex 2: The acceleration of a particle is defined by the relation a = -4풎/풔ퟐ. If v = +24m/s and x = 0 when t= 0 , determine the velocity, position and total distance travelled when t = 8sec.

Ans : v = -8m/s,

x = 64m,

s = 82m.

Ex 3: The acceleration of a particle is defined by the relation 풂 = ퟗ − ퟑ ퟐ. The particle starts t=0 with v=0 and x= 5. Determine (a) time when the velocity is again zero (b) the position and velocity when t=4sec. (c) total distance travelled by the particle from t= 0to t=4 sec.

Ans : (a) t = 3sec

(b) x = 13 m, v = -28m/s

(c) s = 32.5 m.

Ex 4 : The acceleration of a oscillating particle is defined by the relation a = -k x. Determine (a) the value of k such that v = 15m/s where x = 0 and x = 3m when v=0. (b) The speed of particle when x = 2m.

Ans : (a) k = 25 풔 ퟐ

(b) v = 11.18m/s.

Ex 5 : The acceleration of a particle is defined by the relation a = -4v where a(풎풎/풔ퟐ) . Knowing that t = 0 the velocity is 30mm/s, determine (a) the distance the particle will travel before coming to rest (b) the time required for the particle to come to rest (c) the time required for the velocity of the particle to be reduced to 1% of its initial value .

Ans : (a) x = 75mm

(b) t = infinity

(c) t = 2.475s.

Ex 6 : The acceleration of the particle is defined by the relation 풂 = −풌풗ퟑ. The particle starts x = 0 with velocity of 25m/s and when x = 40m the velocity is found to be 20m/s. Determine the distance the particle will travel (a) before its velocity drops to 10 m/s (b) before it comes to rest.

Ans : (a) x = 240m

(b) x = infinity

Ex 7 : A particle moves in a straight path with the acceleration shown in the fig. Knowing that it starts from the origin with v = -18m/s. (a) plot the v-t and x-t curves for 0 <t<20 (b) Determine its velocity position and the total distance travelled after 12s.

Ans :74m, 20m/s, 176m.

Ex 8: The acceleration record was obtained for a small plane travelling around a straight course. Knowing that x = 0and v = 60m/s when t = 0 , determine (a) the velocity and position of the plane at t = 20sec (b) its average velocity during the interval 6s<t<14s.

Ans : 60m/s , 1192m, 59.6m/s.

Ex 9 : The maximum possible deceleration of a particle passing train under emergence conditions travelling at 90m/s determine by approximate means (a) the time required for the train to come to rest (b) the distance travelled in time.

Ans : t = 9.83sec,

S1 = 56.25m,

S2 = 116.25m

S3 = 132.92m.

Ex 10 : Given at this instant 풗푬 = 풎풔 ↓.

Find 풗푪,풗푾,풗푬,풗푪/푬,풗푾/푬.

Ans : 10m/s(↑), 5m/s(↑), 5m/s(↓), 15m/s(↑), 10m/s(↑).

Ex 11 : The slider block B moves to the right with a constant velocity of 450mm/s.

Find (a)풗푨, (b)풗푫, (c)풗푨/푩 , (d)풗푪/푫.

Ans : 675mm/s(→),900mm/s(→), 225mm/s(→), 450mm/s(←).

Ex 12 : Collar A starts from rest and moves to the left with a constant acceleration. Knowing that after 6sec velocity of collar B with respect to collar A is 450mm/s. Find 풂푨 and 풂푩; 풔푩 and 풗푩after 8 sec.

Ans : 50풎풎/풔ퟐ ←, 25풎풎/풔ퟐ →, 800mm→, 200mm/s→.

Ex 13 : The three blocks are equally spaced horizontally and move vertically with constant speed. Knowing that initially they are at same level and that the relative velocity of C w.r.t B is 200mm/s downwards, determine the velocity of each block so that the three blocks will remain aligned during their motion.

Ans : v(A) = 150mm/s↑,

v(B) = 50mm/s↓,

v(C) = 250mm/s↓.

Ex 14 : Under normal operating conditions tape is transferred between reels shown. At a speed of 720mm/s. At t = 0, portion A of the tape is moving to the right at a speed of 600mm/s and has a constant acceleration. Knowing that portion B of the tape has a constant speed of 720mm/s and that the speed of portion A reaches 720mm/s at t = 6s determine (a) the acceleration and velocity of component C, at t = 4sec (b) the distance through which C will have moved at t = 6sec.

Ans : a(A)= 20풎풎/풔ퟐ → , a(C) = 5풎풎/ ퟐ ↓, v(C) =10mm/s↑, s(C) = 90mm↑.

Ex 15 : Collar A starts from rest at t= 0 and moves upward with a constant acceleration of 90mm/s^2. Knowing that collar B moves downward with a constant velocity of 400mm/s, Determine (a) the time at which the velocity of block is zero (b) the corresponding position of block C .

Ans : t = 2.22sec.

Ex 1 : Automobile A travelling around a straight highway, while B is moving along a circular exit ramp of 150m radius. The speed of A, being increased at the rate of 1.5풎/풔ퟐ

and the speed of B is being decreased at the rate of 0.9풎/풔ퟐ. For the position shown, determine (a) the velocity of A relative to B (b) the acceleration of A relative to B.

Ans :

Ex 2 : Knowing that at the instant shown assembly A has a velocity of 225mm/s and an acceleration of 375풎풎/풔ퟐ both directed downwards. Determine (a) velocity of block B (b) acceleration of block B

Ans :

Ex 3 : The rotation of rod OA about O is defined by the relation 휽 = ퟐ풕ퟐ,where 휽 is expressed in radian and t is seconds. Collar B slides along the rod in such a way that its distance from O is : 풓 = ퟔퟎ풕ퟐ − ퟐퟎ풕ퟑ , where r is in mm and t is secs. When t = 1sec (a) find velocity of collar and (b) total acceleration of collar.

Ans : 풗⃗ = ퟔퟎ푬풓⃗ + ퟏퟖퟎ푬휽⃗,

풂⃗ = −ퟔퟒퟎ푬풓⃗ + ퟔퟒퟎ푬휽⃗.

Ex 4 : A particle moves along the spiral shown, determine the magnitude of the velocity of the particle in the terms of b, 휽̇and휽 . Hyperbola spiral 풓휽 = 풃

Ans :풗 = 풃

휽ퟐ. 풅휽풅풕

(ퟏ + 휽ퟐ),

풗 = 풆풃휽휽̇√(ퟏ+ 풃ퟐ).

Ex 5 : The pin at B is free to slide along the circular slot DE along the rotating rod OC. Assuming that the rod OC rotates at the centre A constant rate 휽(a) show that the acceleration of pin B is of constant magnitude (b) determine the direction of the acceleration of pin B. Solve by three methods .

Ans : v=2bw(constant),

a=4b풘ퟐ(constant).

Ex 6 : A rocket is fired vertically from a launching pad at B. Its flight tracked by radar from point A. Determine the velocity of rockets in terms of b,휽풂풏풅휽̇.

Ans : v = 풃(퐬퐞퐜휽)ퟐ휽̇

a =풃 (퐬퐞퐜휽)^ퟐ(ퟐ퐭퐚퐧휽. 휽̇ퟐ

+ 휽̇) .

Ex 7 : Pins A and B are constructed to remain in fixed slot (elliptical). At the instant shown velocity of the rod ABC is 풗풄 = ퟐ풎

풔,풂풄 = ퟑ풎

풔ퟐ. At x = 1.5m, find 풗푩⃗ and 풂푩⃗ at the instant.

Ans : 풗푩⃗ = ퟐ⃗− ퟎ.ퟕퟕ ⃗,

풗푨⃗ = ퟐ⃗+ ퟎ.ퟕퟕ ,⃗

풂푩⃗ = ퟑ⃗− ퟎ.ퟓퟔ ,⃗

풂푨⃗ = ퟑ⃗+ ퟎ.ퟓퟔ .⃗

Ex 8 : What is the radius which should be used for a highway curve if the normal component of the acceleration of a car travelling at 72kmph is not exceed 0.72풎/풔ퟐ.

Ans : 555.56m.

.