Mechanical Engineering Chapter 13 Dyproblem Mms And What Are The Magnitudes The Velocity And Acceleration Point

Problem 13.89 If y=150 mm, dy

dt =300 mm/s, and

d2y

dt2=0, what are the magnitudes of the velocity and

acceleration of point P?

P

y

Solution: The equation for the location of the point Pis R2=

x2+y2, from which x=(R2−y2)

1

2=0.2598 m, and

dx

dt =−y

xdy

dt =−0.1732 m/s,

dt22

Problem 13.90 A car travels at a constant speed of

100 km/h on a straight road of increasing grade whose

vertical proﬁle can be approximated by the equation

shown. When the car’s horizontal coordinate is x=

400 m, what is the car’s acceleration? y = 0.0003x2

y

x

dx

Problem 13.91 Suppose that a projectile has the initial

conditions shown in Fig. 13.12. Show that in terms of

the x′y′coordinate system with its origin at the high-

est point of the trajectory, the equation describing the

trajectory is

y′=− g

2v2

0cos2θ0

(x′)2.

x

yy′

x′

Solution: The initial conditions are t=0, x(0)=0, y(0)=0,

Substitute xp=v2

0cos θ0sin θ0

50

Problem 13.92 The acceleration components of a

point are ax=−4 cos 2t,ay=−4 sin 2t,az=0. At

t=0, its position and velocity are r=i,v=2j. Show

that (a) the magnitude of the velocity is constant; (b) the

velocity and acceleration vectors are perpendicular; (c)

the magnitude of the acceleration is constant and points

toward the origin; (d) the trajectory of a point is a circle

with its center at the origin.

Problem 13.93 When an airplane touches down at

t=0, a stationary wheel is subjected to a constant angu-

lar acceleration α=110 rad/s2until t=1s.

(a) What is the wheel’s angular velocity at t=1s?

(b) At t=0, the angle θ=0. Determine θin radians

and in revolutions at t=1s.

θ

Solution:

Problem 13.94 Let Lbe a line from the center of the

earth to a ﬁxed point on the equator, and let L0be a

ﬁxed reference direction. The ﬁgure views the earth from

above the north pole.

(a) Is dθ/dt positive or negative? (Remember that the

sun rises in the east.)

(b) Determine the approximate value of dθ/dt in rad/s

and use it to calculate the angle through which the

earth rotates in one hour.

L

L0

u

Solution:

Problem 13.95 The angular acceleration of the line L

relative to the line L0is given as a function of time by

α=2.5−1.2trad/s2.Att=0,θ =0 and the angular

velocity of Lrelative to L0is ω=5 rad/s. Determine θ

and ωat t=3s.

L

L0

u

Solution:

52

Problem 13.96 In Active Example 13.8, suppose that

the angular acceleration of the rotor is α=−0.00002ω2,

where ωis the angular velocity of the rotor in rad/s. How

long does it take the rotor to slow from 10,000 rpm to

1000 rpm?

L

u

Solution: Let α=kω2, where k=−0.0002.

Then

a=dω

dt =kω2,

ω2

ω1

dω

ω2=t

0

kdt,

−1

Problem 13.97 The astronaut is not rotating. He has

an orientation control system that can subject him to

a constant angular acceleration of 0.1 rad/s2about the

vertical axis is either direction. If he wants to rotate

180◦about the vertical axis (that is, rotate so that he

is facing toward the left) and not be rotating in his new

orientation, what is the minimum time in which he could

achieve the new orientation?

Problem 13.98 The astronaut is not rotating. He has

an orientation control system that can subject him to

a constant angular acceleration of 0.1 rad/s2about the

vertical axis in either direction. Refer to problem 13.97.

For safety, the control system will not allow his angular

velocity to exceed 15◦per second. If he wants to rotate

180◦about the vertical axis (that is, rotate so that he

is facing toward the left) and not be rotating in his new

orientation, what is the minimum time in which he could

achieve the new orientation?

Problem 13.99 The rotor of an electric generator is

rotating at 200 rpm when the motor is turned off. Due

to frictional effects, the angular acceleration of the rotor

Solution: Let α=kω, where k=−0.01 s−1. Note that 200 rpm =

20.9 rad/s.

(a) One minute after the motor is turned off

54

Problem 13.100 The needle of a measuring instrument

is connected to a torsional spring that gives it an angu-

lar acceleration α=−4θrad/s2, where θis the needle’s

angular position in radians relative to a reference direc-

tion. The needle is given an angular velocity ω=2 rad/s

in the position θ=0.

(a) What is the magnitude of the needle’s angular

velocity when θ=30◦?

(b) What maximum angle θdoes the needle reach

before it rebounds?

θ

Problem 13.101 The angle θmeasures the direction of

the unit vector erelative to the xaxis. The angular veloc-

ity of eis ω=dθ/dt =2 rad/s, constant. Determine the

derivative de/dt when θ=90◦in two ways:

(a) Use Eq. (13.33).

(b) Express the vector ein terms of its xand ycom-

ponents and take the time derivative of e.

x

y

e

θ

Problem 13.102 The angle θmeasures the direction

of the unit vector erelative to the xaxis. The angle θis

given as a function of time by θ=2t2rad. What is the

vector de/dt at t=4s?

n=icos θ+π

2+jsin θ+π

2

is a unit vector in the direction of positive θ. The angular rate of

change is

dθ

dt t=4=[4t]t=4=16 rad/s.

Problem 13.103 The line OP is of constant length R.

The angle θ=ω0t, where ω0is a constant.

(a) Use the relations vx=dx

dt and vy=dy

dt to deter-

mine the velocity of Prelative to O.

(b) Use Eq. (13.33) to determine the velocity of Prel-

ative to O, and conﬁrm that your result agrees with

the result of (a).

Strategy: In part (b), write the position vector of P

relative to Oas r=Rewhere eis a unit vector that

points from Otoward P.

x

y

O

P

R

θ

56

Problem 13.104 In Active Example 13.9, determine the

motorcycle’s velocity and acceleration in terms of nor-

mal and tangential components at t=5s.

Solution: We are given the tangential acceleration

Problem 13.105 The armature starts from rest at t=0

and has constant angular acceleration α=2 rad/s2.At

t=4 s, what are the velocity and acceleration of point

Prelative to point Oin terms of normal and tangential

components?

80 mm

P

O

Solution: We can ﬁnd the angular velocity at t=4s.

Problem 13.106 Suppose you want to design a medi-

cal centrifuge to subject samples to normal accelerations

of 1000 g’s. (a) If the distance from the center of the

centrifuge to the sample is 300 mm, what speed of rota-

tion in rpm is necessary? (b) If you want the centrifuge

to reach its design rpm in 1 min, what constant angular

acceleration is necessary?

300 mm

Solution:

(a) The normal acceleration at a constant rotation rate is an=Rω2,

giving

Problem 13.107 The medical centrifuge shown in

Solution: α=3,ω=3t, θ =1.5t2

58

Problem 13.108 A centrifuge used to subject engineer-

ing components to high acceleration has a radius of 8 m.

It starts from rest at t=0, and during its two-minute

acceleration phase it is programmed so that its angular

acceleration is given as a function of time in seconds

by α=0.192 −0.0016trad/s2.Att=120 s, what is

the magnitude of the acceleration a component is sub-

jected to?

Solution: We will ﬁrst calculate the angular velocity

The normal and tangential components of acceleration are

Problem 13.109 A powerboat being tested for maneu-

verability is started from rest at t=0 and driven in

a circular path 12 m in radius. The tangential compo-

nent of the boat’s acceleration as a function of time is

at=0.4tm/s2.

(a) What are the boat’s velocity and acceleration in

terms of normal and tangential components at t=

4s?

(b) What distance does the boat move along its circular

path from t=0tot=4s?

Solution:

Problem 13.110 The angle θ=2t2rad.

(a) What are the velocity and acceleration of point P

in terms of normal and tangential components at

t=1s?

(b) What distance along the circular path does point P

move from t=0tot=1s? 4 m

P

O

θ

d2θ

dt2=4rad

s2=α

s=rθ =4θ=8t2

vt=16tm/s

v=rω =4(4t) =16t

at=dv

dt =16 m/s2

(a) v=16(1)etm/s =16 et(m/s)

a=Rαet+Rω2eN

a=(4)(4)et+(4)(42)eN(m/s2)

a=16et+64eN(m/s2)

P

O

4 m

θ

eN

Problem 13.111 The angle θ=2t2rad. What are the

velocity and acceleration of point Pin terms of normal

and tangential components when Phas gone one

revolution around the circular path starting at t=0?

Solution: From the solution to Problem 13.110,

We want to know vand awhen θ=2π. Substituting into the ﬁrst eqn,

60

Problem 13.112 At the instant shown, the crank AB

is rotating with a constant counterclockwise angular

velocity of 5000 rpm. Determine the velocity of point B

(a) in terms of normal and tangential components; (b) in

terms of cartesian components.

C

y

Solution:

Problem 13.113 The crank AB in Problem 13.112

is rotating with a constant counterclockwise angular

velocity of 5000 rpm. Determine the acceleration

Solution:

Problem 13.114 Suppose that a circular tunnel of

radius Rcould be dug beneath the equator. In principle,

a satellite could be placed in orbit about the center of the

earth within the tunnel. The acceleration due to gravity in

the tunnel would be gR/RE, where gis the acceleration

due to gravity at sea level and REis the earth’s radius.

Determine the velocity of the satellite and show that the

time required to complete one orbit is independent of

the radius R. (See Example 13.10.)

Equato

r

Tunnel

RER

Problem 13.115 At the instant shown, the magnitude

of the airplane’s velocity is 130 m/s, its tangential

Solution:

Problem 13.116 In the preliminary design of a sun-

the car starts from rest at Aand the tangential compo-

nent of its acceleration is at=0.6 m/s2. What are the

car’s velocity and acceleration in terms of normal and

tangential components when it reaches B?

B

A

200 m

Solution:

ds =0.6 m/s2⇒v

0

v2=2(0.6 m/s2)s

At point B

B

vB=(18.28et)m/s

aB=(0.6et+6.68en)m/s2

the students estimate that the tangential component of

in terms of normal and tangential components when it

reaches B?

at=vdv

vdv

ds

50 m =4.03 m/s2

62

Problem 13.118 Suppose that the tangential compo-

nent of acceleration of the car described in Prob-

lem 13.117 is given in terms of the car’s position by

at=0.4−0.001sm/s2, where sis the distance the car

travels along the track from point A. What are the car’s

velocity and acceleration in terms of normal and tangen-

tial components at point B?

Solution:

ds

From Fig. P13.116, SB=200 +2πρ/4 where ρ=50 m, so SB=

278.5m

Problem 13.119 A car increases its speed at a constant

rate from 40 mi/h at Ato 60 mi/h at B. What is the

magnitude of its acceleration 2 s after the car passes

point A?

80 ft

y

x

30°

120 ft

100 ft

B

A

Solution: Use the chain rule to obtain

vdv

C.At

a=v2−C

2s

The velocity is as a function of time is v(t) =v(0)+at =58.67 +

7.817tft/s. The distance from Ais

At a point 2 seconds past A, the distance is s(2)=132.97 ft, and the

R=(74.3)2

120 =46.0 ft/s2.

The magnitude of the acceleration is

Problem 13.120 The car increases its speed at a con-

stant rate from 40 mi/h at Ato 60 mi/h at B. Determine

the magnitude of its acceleration when it has

traveled along the road a distance (a) 120 ft from Aand

(b) 160 ft from A.

Solution: Use the solution in Problem 13.119.

(b) The velocity at distance 160 ft from Ais

Problem 13.121 Astronaut candidates are to be tested

in a centrifuge with 10-m radius that rotates in the hor-

izontal plane. Test engineers want to subject the can-

didates to an acceleration of 5 g’s, or ﬁve times the

acceleration due to gravity. Earth’s gravity effectively

exerts an acceleration of 1 g in the vertical direction.

Determine the angular velocity of the centrifuge in rev-

olutions per second so that the magnitude of the total

acceleration is 5 g’s.

10 m

Solution:

Problem 13.122 In Example 13.11, what is the heli-

copter’s velocity in turns of normal and tangential com-

ponents at t=4s?

Solution: In Example 13.11 we ﬁnd the xand ycomponents of

acceleration and velocity at t=4s.

64

Problem 13.123 The athlete releases the shot with

velocity v=16 m/s at 20◦above the horizontal.

(a) What are the velocity and acceleration of the shot

in terms of normal and tangential components when

it is at the highest point of its trajectory?

(b) What is the instantaneous radius of curvature of

the shot’s path when it is at the highest point of its

trajectory?

20°

Problem 13.124 At t=0, the athlete releases the shot

with velocity v=16 m/s.

(a) What are the velocity and acceleration of the shot

in terms of normal and tangential components at

t=0.3s?

(b) Use the relation an=v2/ρ to determine the instan-

taneous radius of curvature of the shot’s path at

t=0.3s.

Solution: From the solution to Problem 13.123,

|an|=v2/ρ

Problem 13.125 At t=0, the athlete releases the shot

with velocity v=16 m/s. Use Eq. (13.42) to determine

the instantaneous radius of curvature of the shot’s path

at t=0.3s.

Solution:

We now have y(x)

dy

Problem 13.126 The cartesian coordinates of a point

moving in the xy-plane are x=20 +4t2m, y=10 −

t3m. What is the instantaneous radius of curvature of

the path of the point at t=3s?

66

Problem 13.127 The helicopter starts from rest at

t=0. The cartesian components of its acceleration are

ax=0.6tm/s2and ay=1.8−0.36tm/s2. Determine the

tangential and normal components of its acceleration at

t=6s.

y

x

Solution: The solution will follow that of Example 13.11, with

The tangential acceleration component is given by aT=a·eT

Problem 13.128 Suppose that when the centrifuge in

Example 13.12 is turned on, its motor and control sys-

tem give it an angular acceleration (in rad/s2)α=12 −

0.02ω, where ωis the centrifuge’s angular velocity.

Determine the tangential and normal components of the

acceleration of the sample at t=0.2 s.

300 mm

Solution: We will ﬁrst integrate to ﬁnd the angular velocity at

At this time, the angular acceleration is

Problem 13.129* For astronaut training, the airplane

shown is to achieve “weightlessness” for a short period

of time by ﬂying along a path such that its acceleration

is ax=0 and ay=−g. If the velocity of the plane at

Oat time t=0isv=v0i, show that the autopilot must

ﬂy the airplane so that its tangential component of the

acceleration as a function of time is

at=ggt

v0

1+gt

v02.

x

y

O

Solution: The velocity of the path is v(t ) =v0i−gtj. The path

angle is

Problem 13.130* In Problem 13.129, what is the air-

plane’s normal component of acceleration as a function

of time?

Solution: From Problem 13.129, the velocity is v(t) =v0i−gtj.

The ﬂight path angle is β, from which

68