UNIT 4 | unit 4 kinematics of particle

MECH

UNIT 4

Kinematics of Particle

Question Bank

Question 1)the motion of particle is defined by x = t3 - 6t2 - 36 t - 40 in meter Determine (1) when the velocity is zero. (2) velocity , acceleration & total distance travelled when x = 0 .

Answer 1)

→given,

X=t3 - 6t2 -36t- 40

Differentiating w.r.t. t we get

∴ v =dx /dt =3t2 – 12t - 36

Again differentiating w.r.t. ‘t’

∴a= dv/dt =d2x/ d2t= 6t – 12

(1) When the velocity is zero.

For, v= 0

V= 3t2-12t - 36

0 = 3t2 - 12t - 36

Solving above equation, we get

[t = 6 sec.]

( 2) velocity , acceleration & total distance travelled at x=0 .

For x = 0,

x =t3 - 6t2 – 36t – 40

O =t3- 6t2 – 36t -40

Solving above equation, we get

T = 10sec.

∴velocity (for t =0)

V10 = 3t2 – 12 t – 36

= (3x102) - (12x10)-36

V10 = 144m/sec

∴acceleration (for t = 10)

A= 6t – 12

= (6x10)-12

A = 48m/sec2

Distance travelled =|x10 – x6| + |x6 –x0|

∴x10 = 103 – ( 6 x 102)- ( 36 x 10 ) – 40

= 1000 – 600 – 360- 40

X10 = 0m

∴x6 = 63 – (6x62) –(36x6) – 40

X6 = -256m

∴x0 = 03 – 6 x 02 -36x 0 -40

X0 =-40 m

∴distance travelled = |0 – (-256)| +|-256 –(-40)|

=256 + 216

= 472m

Distance travelled = |ss- s0|

= |-60-40|

=100m

A = 6t-12

= (6 x 5) -12

A = 18 m/s2

(d) Distance travelled from 4 to 6 sec

As at t= 5sec, v = 0

Thus,

Distance travelled = distance travelled from 4 to 5 sec + distance travelled from 5 to 6 sec.

= |s5 – s4|+ |s6 – s5|

∴at t = 6,

s6 = 63 - (6 x 62) – (15 x 6) + 40

=-50m

At t =4,

s4 = 43 - (6 x 42) – (15 x 4 ) + 40

= -52m

Distance travelled = | -60 – (-52)| + |-50 – (-60)|

= 8+10

= 18m

Question 2) The acceleration versus time for a particle moving along x axis is given in the figure given below. The time interval is 0 to 40 sec for some time interval plot

(1) V-t diagram (2) x – t diagram (3) also find max speed attained & max distance covered

Answer 2)Diagram

We know that,

Change in velocity= area under a-t diagram from the given a – t diagram

∴at T =20sec

V20 - v0 = 1/2 x 20x12

V20- v0 =120 m/sec

But at T=0 v0= 0

∴ 20= 120m/sec

Now,

At t =40 sec

∴ v40- v20= ½x 20 x12 = 120

∴ v40 = 120+v20

∴ v40 = 240m/sec

Thus v -t diagram will be as follows

Now from above v -t diagram

Change in displacement = area under v-t diagram

At t=20sec

X20 -X 0 =1/3 x 20x 120 =800m

As (X0 = 0 at t = 0)

∴X20 = 800m

At t =40sec

X40 - X20 = (20x 120) + (2/3 x20 x120)

X40 = 2400+1600+ X20

X40 = 24001600+800

X 40= 4800m

∴ x -t diagram will be as follows

Max speed attained = 240m/sec

Max distance travelled = 4800m

Question 3)A projectile is fired with a velocity of 60 m/s on Horizontal plane. Find its time of flight in the following 3 cases.

a) is Range is 4 times the max . Height

b) Its max height is 4 times Horizontal range.

c) Its max. Height & Horizontal range are equal.

Answer 3)

u = 60m/s

a) When R = 4 H

u2 sin 2 /g=4 [ 42.sin2/2g]

:. u2/g 2sincos = 242/g sin2

:. Cos = sin

:. Cos - sin =o:. = 45

Time of flight t= 2usin /g = 2* 60* sin 45/9081 = 8.65 sec.

b) When H = 4R

:. u2sin2/2g = 4[ 42 sin 2/g]

:. Sin2 = 8 sin 2

:. Sin =(2*8) cos

:. Sin = 16 cos

:. Tan= 16 & = 86.42

t= 2usin/g = 2*60*sin86.42/9.81 = 12.21 sec

c) When H = R

u2sin2/2g = u2 sin2g:. Sin2 /2 = sin2

:.Sin2 = 2*2 sin cos

:. Sin = 4 cos

:.tan =4 :.= 75.96

t= 2usin /g

= 2*60*sin75.96/9.81

t = 11.87 sec.

Question 4)A projectile is aimed at an object Qn a H.p through the point of projection and tall 8 M 8 short when the angle of projection is 15, while it overshoots the the object by 18 m when the angle of projection is 45 Determine the angle of projection to Hit the object exactly.

Answer 4)

let R = actual Range Required to hit the object.

ax I - =15

Range = R -8

:. u2* sin (2*15)/g = (R-8)

:. Multiply both sides by 2

:. 2/g = 2R-16--------(1)

Cos (2) =45

Range = R +18

u2sin2/g = R+18

u2/g. Sin 90 = R +18

:. 42/g = R =18------------ (2)

From (1) & (2) R +18 = 2 R- 16

:. 2R – R = 18+16 = 34.

:. 2R = 34m ---- Actual Range to hit the object

Actual Range

R (42/g) sin 2

34 = (R +18) sin2

34 = (34+18) sin2

34 = 52 sin 2

:. Sin 2= 0.653

= 20.38 - Angle of projection to hit the object

Question 5)A shot is fired from the gun .After 2 sec. The velocity of shot is inclined at 30 up the horizontal After 1 more second. It attains max height. Determine the initial velocity and angle of projection.

Answer 5) Let, u = initial velocity = angle of projection. Let, after. 2 second, the shot fired from gun reaches at point D Here Vo makes 30 angle with Horizontal.

& Let after one more second , shot attains max , Height at point C as shown in figure.

At point D, V0 makes 30 with Horizontal.

:. X component of velocity at ‘D’ = VD cos 30

But we know that velocity in X dirn is constant (U.m.)

:. VD cos 30 = u cos

:. 0.87 VD cos 30 = ucos ----(1)

Consider y-Motion from A to D. By substituting the value of VD ineqn (1) & (2)

This is Motion under gravity 0.87 VD = u cos

:. V = u +at : U cos = 0.87*19.62

:. VD sin 30 = usin- gtAD : ucos = 0.87* 19.62

:. 0.5 VD = usin- 9.81*2 : ucos=17.069 m/s – (3)

:.0.5 VD = usin- 19.62 – (2)

Now

Consider y motion from D-c ,(M.V.G) Also, usin= 0.5VD + 19.62

V = u +at= 0.5*19.62+ 19.62

0 = VD sin30 – g*tDC=u sin= 29.43 m/s & -------- (4)

0= 0.5 VD – 9.81 *1 from eqn (3) & (4)

:. 0.5 VD = 9.81usin/ucos = 29.43/17.069

:. VD = 19.62 m/s : tan = 1.724

:. = 59.886& u = 29.43/sin54.886 =34.02 m/s

Question 6) A projectile is fired from the edge of 150 m cliff an initial velocity of 180 m/s at 30 angle with Horizontal. Find 1) The Horizontal distance from the gun to the point where the projectile strikes the ground 2) The greatest elevation above the ground reached by projectile 3) striking velocity. Refer the given figure.

Answer 6)

Let

X= Horizontal distance between A& B

A= point of projection

B= point of striking.

We can see from the fig that A& B are not on same level. TAB = time Req = tAB

Consider the Horizontal motion from A to B (U.M)

:. Distance = velocity * time

X = 180 cos 30 * tAB

X= 155.88 tAB ------ (1)

Consider vertical motion from A to C, H+ 150+ ½*9.81* t2CB

: V = u + at 412.84+ 150 = 4.905 +tCB

Vcy = 180sin30- g*tAC : t2CB = 562.84/4.905

:0 = 90 – 9.81 tAC:. t2CB= 114.748

:tAC= 90/9.81:. tCB= 10.71 sec

:tAC= 9.17 sec.tAB = 9.17 + 10.71 = 19.88 sec
Consider vertical motion from A to E from eqn (1)

H = u2sinsin2/2g = 1802*sin2 30/2*9.81 X = 155.88 tAB = 155.88*19.88

X = 3098.9 m

H = 412.84 m.

:. Now using. Eqn of motion

S = ut + ½ gt2

Horizontal distance from the to the point of striking is

X = 3098.9 m

Time req. From A to B = tAB = 19.88 sec.

Greatest Height Reached by projectile above the ground is

Hmax H + 150 = 412.84+150

Hmax = 562.84 m

Now,

Consider that VB = striking velocity. &

Ø = angle made by striking velocity with Horizontal. As shown.

Let VBX = X component of VB.

VBy= y component of VB.

But we know that,in X dirn, motion is uniform.

Thus VBX = u cos = 180 cos 30 = 155.88 m/sec.

To find VByconsider the motion from C to B.

:. V = u + gt

VBy= Vcy + g * tCB

VBy= 0 + 9.81 * 10.71

VBy = 105.06 m/sec

:. VB = VBX2 +V2BY = 155.882 + 105.062

VB = 187.9 m/sec

Tanø = VBy/ VBX = 105.06/155.88

:. Ø = 33.97

Question 7) A particle moves along a curved path given by the relation y = starting with initial velocity. If vx = constant, determine of vy& ay at x = 3m. Also determine the magnitude of velocity and acceleration.

Answer 7)

→so, vx0 = 5m/sec and vy0 = 3m/sec

y =

Differentiating w.r.t. ‘t’

vy = (8x + 8)

= (8x + 8)vx

vy = 8x.vx + 8vx

Again differentiating w.r.t. ‘t’

= [0 + 8vx + 0]vx

Now at x = 3m,

Vy = 8.x.vx + 8vx

= (8 × 3 × 5) + (8 × 5)

= 120 + 40

= 160 m/sec

(as vx = constant)

(vx0 = vx= 5 m/sec)

= 8.vx2

= 8 × 52

= 200 m/s2

= 0

Magnitude of velocity

v =

v = 160.07 m/s

Magnitude of acceleration

= 200 m/s2

Question 8. A particle along the path (8t2)i + (t3 + 5)j. Where ‘t’ is in seconds. Determine the magnitude of particle velocity and acceleration when t = 4 sec. Find the equation of path, y = f(x).

Answer 8)

→ given: (8t2)i + (t3 + 5)j position vector

x = 8t2& y = t3 + 5

Differentiating w.r.t. t

vx = & vy = = 3t2

Again differentiating w.r.t. t

= 16t &= 6t

Now, at t = 4 sec

Vx = 16t = 16 × 4 = 64 m/sec

Vy = 3t2 = 3 × 42 = 48 m/sec

Magnitude of velocity

v =

v = 80 m/sec

= 16 m/s2

= 6t = 6 × 4 = 24 m/s2

Magnitude of acceleration

28.84 m/s2

Now as,

X = 8t2

t2 = x/8

t =

But y = t3 + 5

Put the value of t in this equation

y = []3+ 5

y = ---------path equation

Question 9)the y co-ordinate of the particle is given by y=4t3 – 3t. If ax = 12t m/s2& vx = 8 m/s at t = 0. Calculate the magnitude of velocity & acceleration of particle at time t = 2 seconds.

Answer 9)

→given: at t = 0, ax = 12t & vx = 8 m/s

At t = 2, v = ? a =?

y=4t3 – 3t ------given

Differentiating w.r.t. Time‘t’

12t2 – 3

Vy = 12t2 – 3 -------- (1)

Differentiating again

24t ------- (2)

12t -------- (3) given

= 12t

Dvx= 12t dt

Taking integration

Vx = 12 + c1

Vx = 6t2 + c1

Now using the given condition

t = 0, vx = 8 m/s

8 = 6 × 0 + c1

c1 = 8

vx = 6t2 + 8 ---------(4)

At time t = 2 sec

vx = 6t2 + 8

= 6 × 22 + 8

Vx = 32 m/s

Vy = 12t2 – 3

= 12 × 22 – 3

Vy = 45 m/s

ax = 12t

= 12 × 2

ax = 24 m/s2

ay = 24t = 24 × 2 = 48 m/s2

Magnitude of velocity at t = 2,

v =

V = 55.21 m/s2

Magnitude of acceleration at t=2

53.66 m/s2

Question 10)A particle is moving along a curve y = x - . If vx = 4 m/s and is constant. Determine the magnitude of velocity & acceleration when x = 30 m.