Mechanical Engineering Chapter 14 Homework Solving For The Positive Root 02687

456

Ans:

*14–80.

When

s=0

, the spring on the firing mechanism is

unstretched. If the arm is pulled back such that

s=100 mm

and released, determine the speed of the 0.3-kg ball and the

normal reaction of the circular track on the ball when

u=60°

. Assume all surfaces of contact to be smooth.

Neglect the mass of the spring and the size of the ball.

SOLUTION

Potential Energy. With reference to the datum set through the center of the circular

track, the gravitational potential energies of the ball when

u=0°

and

u=60°

are

When

u=0°

, the spring compress

x1=0.1 m

and is unstretched when

u=60°

.

u=0°

Conservation of Energy. Since the ball starts from rest,

T1=0

.

T1+V1=T2+V2

s

1.5 m

u

457

SOLUTION

Equation of Motion. It is required that the ball leaves the track, and this will occur

provided

u790°

. When this happens,

N=0

. Referring to the FBD of the ball, Fig. a

Potential Energy. With reference to the datum set through the center of the circular

track Fig. b, the gravitational potential Energies of the ball when

u=0°

and

u

are

When

u=0°

, the spring compresses

x1=0.1 m

and is unstretched when the ball

is at

u

for max height. Thus, the elastic potential energies in the spring when

u=0°

and

u

are

Conservation of Energy. Since the ball starts from rest,

T1=0

.

Equating Eqs. (1) and (2),

u=117.77°=118°

s

1.5 m

u

14–81.

When

s=0

, the spring on the firing mechanism is

unstretched. If the arm is pulled back such that

s=100 mm

and released, determine the maximum angle

u

the ball will

travel without leaving the circular track. Assume all surfaces

of contact to be smooth. Neglect the mass of the spring and

the size of the ball.

Ans:

458

Ans:

14–82.

SOLUTION

The work is computed by moving Ffrom position r1to a farther position r2.

Vg=- U=- LFdr

If the mass of the earth is

Me

, show that the gravitational

potential energy of a body of mass m located a distance r

from the center of the earth is Vg =−GMem>r. Recall that

the gravitational force acting between the earth and the

body is F =G(Mem>r 2), Eq. 13–1. For the calculation, locate

force.

r2

r1

r

the datum at r : q. Also, prove that F is a conservative

459

14–83.

A rocket of mass mis fired vertically from the surface of the

earth, i.e., at Assuming no mass is lost as it travels

upward, determine the work it must do against gravity to

reach a distance The force of gravity is

(Eq. 13–1), where is the mass of the earth and rthe

distance between the rocket and the center of the earth.

Me

F=GMem>r2

r2.

r=r1.

SOLUTION

F=GMem

r2

r1

r

Ans:

460

Ans:

*14–84.

The 4-kg smooth collar has a speed of

3 m>s

when it is at

s=0.

Determine the maximum distance s it travels before

it stops momentarily. The spring has an unstretched length

of

1 m

.

SOLUTION

Potential Energy. With reference to the datum set through A the gravitational

potential energies of the collar at A and B are

2

1.5 m

3 m/

s

k  100 N/m

s

A

B

14–85.

SOLUTION

yA=40 Mm>h=11 111.1 m>s

A

B

v

A

vBrB80 Mm

rA20 Mm

A 60-kg satellite travels in free flight along an elliptical orbit

such that at A, where rA = 20 Mm, it has a speed vA = 40 Mm>h.

What is the speed of the satellite when it reaches point B, where

rB = 80 Mm? Hint: See Prob. 14–82, where Me = 5.976(1024) kg

and G = 66.73(10–12) m3>(kg # s2).

Ans:

462

14–86.

The skier starts from rest at A and travels down the ramp. If

friction and air resistance can be neglected, determine his

speed

vB

when he reaches B. Also, compute the distance s to

where he strikes the ground at C, if he makes the jump

traveling horizontally at B. Neglect the skier’s size. He has a

mass of 70 kg.

SOLUTION

TA+ VA= TB+ VB

4 m

vB

s

C

B

A

50 m

30

Ans:

463

14–87.

SOLUTION

Datum at initial position:

The block has a mass of 20 kg and is released from rest

when s0.5m. If the mass of the bumpers Aand Bcan

be neglected, determine the maximum deformation of each

spring due to the collision.

k

B

= 800 N/m

s= 0.5 m

A

B

k

A

= 500 N/m

Ans:

464

Ans:

*14–88.

SOLUTION

Datum at B:

The 2-lb collar has a speed of at A.The attached

spring has an unstretched length of 2 ft and a stiffness of

If the collar moves over the smooth rod,

determine its speed when it reaches point B, the normal

force of the rod on the collar, and the rate of decrease in its

speed.

k=10 lb>ft.

5ft>s

y

A

4.5 ft

k10 lb/ft

x

2

1

2

y4.5 

465

14–89.

SOLUTION

At point B:

Datum at bottom of curve:

Substitute Eq. (1) into Eq. (2), and solving for ,

Solving for the positive root:

t=0.2687 s

vB

When the 6-kg box reaches point Ait has a speed of

Determine the angle at which it leaves the

smooth circular ramp and the distance sto where it falls

into the cart. Neglect friction.

uvA=2m>s. vA=2m/s

1.2 m

B

A

s

θ

20°

Ans:

466

14–90.

When the 5-kg box reaches point A it has a speed

vA

=10 m>s.

Determine the normal force the box exerts

on the surface when it reaches point B. Neglect friction and

the size of the box.

SOLUTION

Conservation of Energy. At point B,

y=x

Applying the energy equation,

9 m

B

y

x

A

y  x

x1/2  y1/2  3

Ans:

467

SOLUTION

Conservation of Energy. With reference to the datum set coincide with

x

axis,

the gravitational potential energy of the box at A and C (at maximum height) are

Then,

Thus

14–91.

When the 5-kg box reaches point A it has a speed

vA

=10 m>s.

Determine how high the box reaches up the

surface before it comes to a stop. Also, what is the resultant

normal force on the surface at this point and the

acceleration? Neglect friction and the size of the box.

9 m

B

y

x

A

y  x

x1/2  y1/2  3

Ans:

468

Ans:

*14–92.

SOLUTION

Datum at A:

TA+VA=TB+VB

Theroller-coaster car hasaspeed of when it is at

the crestofavertical parabolic track. Determine the car’s

velocity and the normal force it exerts on the track when it

reaches point B.Neglect friction and the mass of the

wheels.The total weight of the car and the passengers is

350lb.

15 ft

>

s

v

A

15 ft/s

200 ft

A

y

(40 000 x

2

)

1

200

y

469

14–93.

A

k 500 N/m

D

E

B

0.4 m

u

The 10-kg sphere Cis released from rest when and

the tension in the spring is .Determine the speed of

the sphere at the instant .Neglect the mass of rod

and the size of the sphere.AB

u=90°

100 N

u=0°

SOLUTION

Potential Energy: With reference to the datum set in Fig. a,the gravitational potential

energy of the sphere at positions (1) and (2) are

Conservation of Energy:

T1+V1=T2+V2

2 ks22=1

2(500)(0.42)=40 J

A

Vg

B

1=mgh1=10(9.81)(0.45) =

470

14–94.

SOLUTION

Conservation of Energy:

T1+V1=T2+V2

Aquarter-circular tube AB of mean radius rcontains a smooth

chain that has a mass per unit length of . If the chain is

released from rest from the position shown, determine its

speed when it emerges completely from the tube.

m0

A

B

O

r

Ans:

SOLUTION

14–95.

The cylinder has a mass of 20 kg and is released from rest

when

h=0.

Determine its speed when

h=3 m.

Each

spring has a stiffness

k=40 N>m

and an unstretched

length of 2 m.

kk

h

2 m 2 m

472

Ans:

*14–96.

If the 20-kg cylinder is released from rest at

h=0,

determine the required stiffness k of each spring so that its

motion is arrested or stops when

h=0.5 m.

Each spring

has an unstretched length of 1 m.

SOLUTION

kk

h

2 m 2 m

14–97.

Apan of negligible mass is attached to two identical springs of

stiffness .If a 10-kg box is dropped from a height

of 0.5 m above the pan, determine the maximum vertical

displacement d.Initially each spring has a tension of 50 N.

k=250 N>m

SOLUTION

Potential Energy: With reference to the datum set in Fig. a,the gravitational potential

1m 1m

0.5 m

k250 N/m k250 N/m

d

Conservation of Energy:

T1+V1+T2+V2

Ans: