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Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730

Question 1 A racing car travels with a constant speed of 240 km/hr around the elliptical race track. Determine the acceleration experienced by the driver at A.

Solution: For the driver at A moving with constant speed, then at  v  0

an 

v2



Where ρ is the radius of curvature. From calculus, the radius of curvature ρ is given as 3

  dy 2  2 1       dx    d 2y dx2

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

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From the equation of motion of the car

x2 y2  1 16 4 Differentiating the equation of motion to determine

d 2y dy and we have; dx dx2

x y  dy  0 8 2 y dy x  2 dx 8

dy 2x x   dx 8y 4y 4y

dy  x dx

Differentiating the expression one more time we have; dy dy d 2y 4  4y  1 dx dx dx2 2

d 2y  dy  4    4y  1  dx  dx2 2  d 2y   dy    1  4   4y   dx2   dx    2

d 2y dx2 But

dy x  therefore; dx 4y

 dy  1  4    dx   4y

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730 2

d 2y dx2

 x  1  4     4y   4y

 4 y2  x2      16y 3  dx2

d 2y

3

  x 2  2 3 1      2 2 2 16y  x   4y      2 2 4 4 y2  x2 4y  x



16y 3







At A x=4 km and y=0 km, hence; ρA= 1 km = 1000 m The speed of the car is given as;

v  240km 

240 x1000  66.67m / s 3600

There acceleration at A is given as

v2

66.672 aA    4.44m / s2  A 1000 aA  4.44m / s2

Question 2

Particles A and B are traveling counterclockwise around a circular track at a constant speed of 8 m/s. If at the instant shown the speed of A begins to increase by (at) A = (0.4SA) m/s2, where is SA in meters, determine the distance measured counterclockwise along the track from B to A when t = 1s.What is the magnitude of the acceleration of each particle at this instant?

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730

Solution

Given: v 0  8m / s

b  0.4 s 2

t1  1s r  5m   1200

To get the distance moved by the particle, we determine an expression in terms of velocity and distance as shown below; Recall atA 

dv A dt

(1.1)

Also VA 

dSA Hence dt

dt 

dS A vA

Substituting for Equation (1.2) into equation (1.1) we have;

v Adv A  bSAdSA

Integrating both sides with limits of integration we have;

(1.2)

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730

vA

SA1

v0

0

 vAdvA   bSAdSA

Where S A1 is the distance covered in time t1  1s

v2A v02 bS2A   2 2 2 v A  v02  bS2A Recall VA 

dSA hence dt dt 

dS A Substituting for v A , we have; vA

dSA

dt 

v02  bS2A

Integrating both sides of the equation we have; t1



SA1

1dt 

0



1





2 v02  S A b

0

dS A

The second half of the integral can be evaluated using trigonometric substitution as shown below: Let u  bSA and a  v0 therefore; SA1

t1 

 0

1 a2  u2

Considering the triangle in Figure 1 below, we have;

dSA

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730 tan 

bSA u  a v0

v SA  0 tan Therefore b v dSA  o sec2  d b

Also from the triangle;

sec 

1 u2  a2 therefore  cos a

u2  a2  a sec Substituting for dS A and

u2  a2 in the integral we have; SA1

t1 

1

 v0 sec x

v0 b

0

1 t1  b t1 

sec2  d

SA1

 sec d 0

1 S ln sec  tan 0A1 b

Substituting for sec and tan we have;

t1 

1 ln b



bSA



2

 v02

v0

SA1



bSA v0 0

t1 

1 ln b



bSA



2

 v02  bS A

SA1

v0 0

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1  1  ln t1  b 

@02696730



bSA1



2

 v02  bSA1

v0



t1 b  ln

bS A1



2

v  ln 0 v0

    

 v02  bS A1

v0

Taking exponential of both sides, we have;



et1 b 



bS A1



2

 v02  bS A1

v0



2

 v02  bSA1  v0et1 b



2

 82  0.4 SA1  8e1x 0.4

bSA1

Substituting for b , v 0 and t1 we have;



0.4 SA1



0.4 SA1



2

 82  15.06  0.4 SA1

Squaring both sides of the equation we have;



0.4 SA1





2

0.4 SA1





 82  15.06  0.4 SA1 x 15.06  0.4 SA1



2

 64  226.803  19.05SA1 

Therefore;

19.05SA1  226.803  64 SA1  8.546m



0.4 SA1



2



Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730

The distance covered by particle B in t1 is given as SB1  v0t1  8m Total distance S AB is given as;

SAB  SA1  r  SB1

r 

 360

120 x2x x 5  10.47m 360

x2 r 

Therefore;

SAB  8.546  10.47  8 SAB  11.02m Particle A will have both tangential and normal acceleration and this is given as; 2

aA 

 bSA1 

2

 v2  bS2  A1   0   r  

2

aA 

 0.4 x8.546 

2

 64  0.4  8.546 2      5  

aA  11.685  347.55 aA  18.95m / s2 Particle B will have only the normal acceleration and this is given as;

v2 82 aB  0   12.8m / s2 r 5

Question 3

The automobile travels from a parking deck down along a cylindrical spiral ramp at a constant speed of 1.5 m/s. If the ramp descends a distance of 12 m for every full revolution, θ = 2π rad, determine the magnitude of the car’s acceleration as it moves along the ramp r = 10m.

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730

Solution Given: constant speed of car is v  1.5m / s , r  10m From the Figure 1 below it can be seen that tangent to the ramp at any point is at an angle of

 12    2 x x10 

  arctan 

  10.8120

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

@02696730

The components of the car’s velocity are ;

vr  0 v  v cos  1.5x cos10.812 v  1.473m / s v z  1.5x sin10.812 v z  0.282m / s

r  10m , r  0 , r  0 Also

v0  r

1.473  10x

  0.147rad / s The components of acceleration of the car are:



ar  r  r 2



ar  0  10  0.147 

2

ar  0.216m / s2 a  r  2r

a  10 x 0  2 x 0 x 0.147

a  0 az  z az  0 The magnitude of the acceleration of the car is

Oluwamayowa Okeyoyin Advanced Dynamics Homework 1

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a  ar2  a2  a2z 

 0.216 2   0 2   0 2

a  0.216m / s2