Dynamics Sample Problems

Newton’s Law o f Motion for a Particle

Free falling Bodies, Air resistance is Neglected.

S A

C

D -S

Sample Problems

1. A ball is dropped down a well and 5 seconds later the sound of the splash is heared. If the velocity of wound is 330 m/sec., what is the depth of the well?

330

 

S

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Solution: T1 = time for the ball to travel a distance distance S T2 = time for the sound to travel a distance S 1) T1 + T2 = 5 2) S = ½ gT1 ² T1 = √2s/g

3) S = 330T2 T2 = S/330 Substitute in equation 1: √2S/9.81 + S/330 = 5 0.452√S + S/330 – 5 = 0 S = 149√S – 1650 = 0 Let y = √S

y² = S y² + 149y -1650 = 0

 0 √     √   y = 10.35 √S = 10.35









2. A stone is thrown upward from the ground with the velocity of 15 m/sec. One second later another another stone is thrown vertically vertically upward with a velocity velocity of 30 m/sec. How far above the ground will the stones be at same level?

15



S



30

 

Solution: t = time the first stone travelled t – 1 = time the second s econd stone travelled until the stone are at the same level S = Vot - ½(g)t² S = 15t – ½(9.81)t² S = 15t – 4.905t²

equation 1

S = 30(t – 1) – ½(9.81)(t – 1)²

equation 2

Equate (1) and (2): 15t – 4.905t² = 30(t – 1) – 4.905(t² 4.905(t² - 2t 2t + 1)









t = 1.4 sec. S = 15(1.4) – 4.905(1.4)² S = 11.4 m. from theground

3.A ball is thrown thrown vertically upward upward with an initial initial velocity of 3 m/s from the window of a tall building. building. The ball strikes the sidewalk at the ground level 4 seconds later. Determine the velocity with which the ball hits the ground and the th e height of the window above the ground level.





3





h



H S





Solution: Vf² = Vo² - 2gh = (3)² - 2(9.81)h









t = 0.31 t2 = 4 secs. – 0.31 secs. t2 = 3.69 secs. H = ½ g(t2)² H = ½ (9.81)(3.61) H = 66.79 m Height of the window = H – h S = 66.79 – 0.459 = 66.331 V3² = V1² + 2gh V3 = 0 + 2(9.81)(66.79) V3 = 36.20 m/s Further Application of Kinematics of Translation

Sample Problems

 300                 0 0  30           0    300

1. The position of a pa particle rticle is given by

. Where t is is in sec.















030 030 30  0  





   00     0  0      0  0     0   0            00       [ 3  3 0]    0

2. The velocity of a particle moving along x-axis is given by

0

. Evaluate the position, velocity and acceleration of particle at













, when t = 0,

.











3.the acceleration of a particle is given by

30 .

v=

3 

when t=0,s=-40m, and

Determine the position and velocity after t=2 sec.

     3     3    3    30   3 30   0         3      30       30 ∫ ∫  









Projectile Motion

SAMPLE PROBLEM:

1. The car shown is just to clear the water filled gap. Find the take off velocity? velocity?

Solution:

  .    .3 









.(0 )   3.0  3. 0  00   .  2. refer to figure. Find the ß to cause the projectile to hit the point ß exactly in 4 sec..what is the distance











30. 3. A ball id thrown so that it just clears a 10ft fence 60 ft. away. If it left the hand 3 ft. above the ground and at an angle of

0          

?

Given x=60ft. Θ=60° y=3ft.

10 ft. 60°

5 ft.

60 ft.

Solution









  .  MOTION DIAGRAM

1.) A train is to commute between station A and station B with a top speed of 250 kph but cannot accelerate accelerate nor decelerate faster than 4 m/sec². what is the minimum minimum distance between the two stations in order for the train to be able to reach its top speed?

  0000 300   . .    . .3 .     ..3   0.      0.      0. 0.











t 4

-4 a

t V

V t

t

S

S 2S

S

2) A train starts from rest at station P and stops at station s tation Q which is 10 km from P. The maximum possible possible acceleartion of the train is 15 km/hr/min and the maximum deceleration when the brakes are applied is 10 km/hr/min. if the maximum allowable speed is 60 kph., what is the least time the train can go from P to Q?









Accelearation Accelearation = 15(60) = 900km/hr² Deceleration Deceleration = 10(60) = 600 km/hr² 900t1 = 60 t1 =

= 4 min 

600t3 = 60

= 6 min.    S1= (t1) =    t3 =

S1 = 2 km S2 – S1 = 60t2 S2 – 2 = 60t2









1.8 m/s

ACCELERATION DIAGRAM t2

t1

-1.8 m/s

t3

VELOCITY DIAGRAM

V

V

t2 VELOCITY DIAGRAM

t1

1°

t3 2°

2°

S3 = 540 S2 S1









S1 = Vt1/2

eq. 5

S2 = S1 + Vt2 S2 = Vt1/2 + Vt2 S3 = S2 + Vt2/2 But t3 = t1 ; S3 = 540 m 540 = (Vt1/2 + Vt2) + Vt1/2 540 = Vt1 + Vt2 From(4) 540 = Vt1 + V(40 – 2t1)









Rectilinear Motion with Constant Acceleration.

SAMPLE PROBLEM: 1. On a certain stretch of track, trains run at 60 mph. How for back of a stopped train should a warning torpedo be placed to signal an on coming train? Assume that the brakes are applied at once & retard the train at the uniform rate of 2ft per Given:



.

              

a=-2ft/



=-2(2)S S= 1936ft

2. A stone is thrown vertically upward & returns to earth in 10sec. What its initial velocity











 01

Dynamics Sample Problems

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