Mechanical Engineering Chapter 12 Homework If it then starts to increase its speed at where tis in

121

SOLUTION

v

0

t=1=

1.83879i

+

4j

a=4 sin ri +4j

12–113.

The position of a particle is defined by r = {4(t – sin t)i

+ (2t2 – 3)j} m, where t is in seconds and the argument for

the sine is in radians. Determine the speed of the particle and

its normal and tangential components of acceleration when

t = 1 s.

Ans:

122

12–114.

The automobile has a speed of 80 ft

>

s at point A and an

acceleration a having a magnitude of 10 ft>s2, acting in

the direction shown. Determine the radius of curvature

of the path at point A and the tangential component of

acceleration.

SOLUTION

Acceleration: The tangential acceleration is

Ans:

12–115.

The automobile is originally at rest at If its speed is

increased by where tis in seconds,

determine the magnitudes of its velocity and acceleration

when t=18 s.

v

#=10.05t22ft>s2,

s=0.

SOLUTION

Therefore the car is on a curved path.

at=0.05t2

s

240 ft

300 ft

Ans:

124

*12–116.

SOLUTION

The car is on the curved path.

So that

v=0.0167(19.06)3=115.4

at=0.05 t2

The automobile is originally at rest If it then starts to

increase its speed at where tis in seconds,

determine the magnitudes of its velocity and acceleration at

s=550 ft.

v

#=10.05t22ft>s2,

s

=0.

s

240 ft

300 ft

12–117.

The two cars A and B travel along the circular path at

constant speeds vA = 80 ft

>

s and vB = 100 ft

>

s, respectively.

If they are at the positions shown when t = 0, determine the

time when the cars are side by side, and the time when they

are 90° apart.

A

B

v

A

vB

rB  390 ft

r

A  400 ft

SOLUTION

a) Referring to Fig. a, when cars A and B are side by side, the relation between

their angular displacements is

Here,

sA=vA t=80 t

and

sB=vB t=100 t

. Apply the formula

s=ru

or u=

r

s

. Then

Substitute these results into Eq. (1)

(b) Referring to Fig. a, when cars A and B are

90°

apart, the relation between their

angular displacements is

Here,

sA=vA t=80 t

and

sB=vB t=100 t

. Applying the formula

s=ru

or u=

s

r

.

Then

12–118.

Cars A and B are traveling around the circular race track. At

the instant shown, A has a speed of 60 ft

>

s and is increasing

its speed at the rate of 15 ft

>

s2 until it travels for a distance of

100p ft, after which it maintains a constant speed. Car B has

a speed of 120 ft

>

s and is decreasing its speed at 15 ft

>

s2 until

it travels a distance of 65p ft, after which it maintains a

constant speed. Determine the time when they come side by

side.

A

B

v

A

vB

rB  390 ft

r

A  400 ft

Ans:

SOLUTION

Referring to Fig. a, when cars A and B are side by side, the relation between their

angular displacements is

u

A=

u

B+

p (1)

The constant speed achieved by cars A and B can be determined from

(

v

A)c

2=(

v

A)0

2+2(

a

A)t [

s

A–(

s

0)A]

(

2=(

2+2(

(

The time taken to achieve these constant speeds can be determined from

(vA)c=(vA)0+(aA)t (tA)1

(tA)1=3.6084 s

(vB)c=(vB)0+(aB)t (tB)1

(tB)1=1.9359 s

Let t be the time taken for cars A and B to be side by side. Then, the times at

which cars A and B travel with constant speed are

(tA)2=t–(tA)1=t–3.6084

(tB)2=t–(tB)1=t–1.9359

(sA)1=100

(sA)2=(vA)c (tA)2 =

(sB)1=65

(sB)2=(vB)c (tB)2=90.96 (t–1.9359)

Substitute these results into Eq. (1),

127

12–119.

SOLUTION

n=20 Mm>h=20(106)

3600 =5.56(103)m>s

The satellite Stravels around the earth in a circular path

with a constant speed of If the acceleration is

determine the altitude h. Assume the earth’s

diameter to be 12 713 km.

2.5 m>s2,

20 Mm>h.

h

S

Ans:

128

*12–120.

The car travels along the circular path such that its speed is

increased by , where tis in seconds.

Determine the magnitudes of its velocity and acceleration

after the car has traveled starting from rest.

N

eglect the size of the car.

s=18 m

at=(0.5et)m>s2

SOLUTION

Solving,

dv =Lt

0.5etdt

s18 m

Ans:

129

12–121.

SOLUTION

Velocity:The speed of the car at Bis

Radius of Curvature:

Acceleration:

an=vB2

r

=17.282

324.58 =0.9194 m>s2

y=16 –1

625 x2

The car passes point Awith a speed of after which its

speed is defined by . Determine the

magnitude of the car’s acceleration when it reaches point B,

where and x = 50 m.s=51.5 m

v=(25 –0.15s)m>s

25 m

>

s

y16 x2

1

625

y

s

x

16 m

B

A

130

12–122.

s=101.68 m and x = 0.

SOLUTION

Velocity: The speed of the car at Cis

Radius of Curvature:

Acceleration:

at=v

#=0.5 m>s

y=16 –1

625 x2

s

x

16 m

B

A

131

12–123.

The motorcycle is traveling at when it is at A. If the

speed is then increased at determine its speed

and acceleration at the instant t=5s.

v

#=0.1 m>s2,

1m>s

x

s

A

y

y0.5x

2

SOLUTION

Solving,

at=n=0.1

Ans:

132

*12–124.

SOLUTION

y=0.4 x2

The box of negligible size is sliding down along a curved

path defined by the parabola .When it is at A

(, ), the speed is and the

increase in speed is . Determine the

magnitude of the acceleration of the box at this instant.

dv >dt =4m>s2

v=8m>syA=1.6 mxA=2m

y=0.4x2

y

x

2m

y0.4x

2

A

133

SOLUTION

Lv

s

dv=

Lt

0

0.06t dt

12–125.

The car travels around the circular track having a radius of r= 300 m

such that when it is at point A it has a velocity of 5 m

>

s, which is

increasing at the rate of

v

#

= (0.06t) m

>

s2, where t is in seconds.

Determine the magnitudes of its velocity and acceleration when it

has traveled one-third the way around the track.

y

x

r

A

Ans:

134

12–126.

The car travels around the portion of a circular track having

a radius of r = 500 ft such that when it is at point A it has a

velocity of 2 ft

>

s, which is increasing at the rate of

v

#

=

(0.002t) ft

>

s2, where t is in seconds. Determine the

magnitudes of its velocity and acceleration when it has

traveled three-fourths the way around the track.

Ans:

y

x

r

A

SOLUTION

at=0.002 s

at ds =v dv

Ls

0

0.002s ds =

Lv

2

v dv

135

12–127.

At a given instant the train engine at Ehas a speed of

and an acceleration of acting in the

direction shown. Determine the rate of increase in the

train’s speed and the radius of curvature of the path.r

14 m>s2

20 m>s

v20 m/s

a14 m/s

2

E

75

r

SOLUTION

136

SOLUTION

The distance traveled by the car along the circular track can be determined by

integrating

v dv=at ds

. Using the initial condition

v=20 m>s

at

s=0

,

The time can be determined by integrating dt =

ds

v

with the initial condition

s=0

at

t=0

.

*12–128.

The car has an initial speed v0 = 20 m

>

s. If it increases its

speed along the circular track at s = 0, a

t=

(0.8s) m

>

s

2

,

where s is in meters, determine the time needed for the car

to travel s = 25 m.

s

r  40 m

Ans:

137

s

r  40 m

12–129.

The car starts from rest at s = 0 and increases its speed at

at = 4 m

>

s2. Determine the time when the magnitude of

acceleration becomes 20 m

>

s2. At what position s does this

occur?

SOLUTION

Acceleration. The normal component of the acceleration can be determined from

Ans:

138

12–130.

SOLUTION

Lv

15

dv =L5

0

0.8tdt

A boat is traveling along a circular curve having a radius of

100 ft. If its speed at is 15 ft/s and is increasing at

determine the magnitude of its acceleration

at the instant t=5s.

v

#=10.8t2ft>s2,

t=0

Ans:

139

12–131.

SOLUTION

at=2m>s2

A boat is traveling along a circular path having a radius of

20 m. Determine the magnitude of the boat’s acceleration

when the speed is and the rate of increase in the

speed is v

#

=2m>s2.

v=5m>s

Ans:

140

*12–132.

Starting from rest, a bicyclist travels around a horizontal circular

path, at a speed of

where tis in seconds.Determine the magnitudes of his velocity

and acceleration when he has traveled s=3m.

v=10.09t2+0.1t2m>s,r=10 m,

SOLUTION

Solving,

t=4.147 s

Ls

0

ds =Lt

010.09t2+0.1t2dt