Mechanical Engineering Chapter 14 Problem The Circular Disk Lies The Horizontalplane And Rotates With Constant Counterclockwise

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Problem 14.71 The circular disk lies in the horizontal

plane and rotates with a constant counterclockwise angu-

lar velocity of 4 rad/s. The 0.5-kg slider Ais supported

horizontally by the smooth slot and the string attached at

B. Determine the tension in the string and the magnitude

of the horizontal force exerted on the slider by the slot.

4 rad/s

0.6 m

90°

0.6 m

Solution:

m=0.5kg

α=0

Fn:Ncos 45◦+Tsin 45◦=mRω2

Ft:−Nsin 45◦+Tcos 45◦=mRα =0

Solving, N=T=3.39 N

0.6 m

enet

Problem 14.72 The 32,000-lb airplane is ﬂying in the

vertical plane at 420 ft/s. At the instant shown the angle

θ=30◦and the cartesian components of the plane’s

acceleration are ax=−6 ft/s2,a

y=30 ft/s2.

(a) What are the tangential and normal components

of the total force acting on the airplane (including

its weight)?

(b) What is dθ/dt in degrees per second?

150

Problem 14.73 Consider a person with a mass of 72 kg

who is in the space station described in Example 14.7.

When he is in the occupied outer ring, his simulated

weight in newtons is 1

2(72 kg)(9.81 m/s2)=353 N. Sup-

pose that he climbs to a position in one of the radial

tunnels that leads to the center of the station. Let rbe

his distance in meters from the center of the station.

(a) Determine his simulated weight in his new position

in terms of r. (b) What would his simulated weight be

when he reaches the center of the station?

Solution: The distance to the outer ring is 100 m.

Problem 14.74 Small parts on a conveyor belt moving

with constant velocity vare allowed to drop into a bin.

Show that the angle θat which the parts start sliding

on the belt satisﬁes the equation cos θ−1

µs

sin θ=v2

gR ,

where µsis the coefﬁcient of static friction between the

parts and the belt.

Problem 14.75 The 1-slug mass mrotates around the

vertical pole in a horizontal circular path. The angle α=

30◦and the length of the string is L=4 ft. What is the

magnitude of the velocity of the mass?

Strategy: Notice that the vertical acceleration of the

mass is zero. Draw the free-body diagram of the mass

and write Newton’s second law in terms of tangential

and normal components.

F↑:Tcos 30◦−mg =0

Fn:Tsin 30◦=mv2

ρ=mv2

Lsin 30◦

Solving we have

T=mg

cos 30◦,v

2=g(L sin 30◦)tan 30◦

v=(32.2 ft/s2)(4ft)sin230◦

cos 30◦=6.10 ft/s

30°

Problem 14.76 In Problem 14.75, determine the mag-

nitude of the velocity of the mass and the angle θif the

tension in the string is 50 lb.

Solution:

Fn:Tsin θ=mv2

Lsin θ

Solving we ﬁnd θ=cos−1mg

T,v=(T 2−m2g2)L

Using the problem numbers we have

θ=cos−11 slug 32.2 ft/s2

50 lb =49.9◦

v=[(50 lb)2−(1 slug 32.2 ft/s2)2]4 ft

(50 lb)(1 slug)=10.8 ft/s

152

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Problem 14.77 The 10-kg mass mrotates around the

vertical pole in a horizontal circular path of radius R=

1 m. If the magnitude of the velocity is v=3 m/s, what

are the tensions in the strings Aand B?

35°

55°

−imv2

R. Separate components to obtain the two simultane-

ous equations: |TA|cos 125◦+|TB|cos 145◦=−90 N|TA|sin 55◦+

|TB|sin 35◦=98.1 N. Solve:

|TA|=84 N.|TB|=51 N

Problem 14.78 The 10-kg mass mrotates around the

vertical pole in a horizontal circular path of radius R=

1 m. For what range of values of the velocity vof the

mass will the mass remain in the circular path described?

Problem 14.79 Suppose you are designing a monorail

transportation system that will travel at 50 m/s, and you

decide that the angle θthat the cars swing out from the

vertical when they go through a turn must not be larger

than 20◦. If the turns in the track consist of circular arcs

of constant radius R, what is the minimum allowable

value of R? (See Active Example 14.6)

Problem 14.80 An airplane of weight W=200,000 lb

makes a turn at constant altitude and at constant velocity

v=600 ft/s. The bank angle is 15◦. (a) Determine the

lift force L. (b) What is the radius of curvature of the

plane’s path?

15°

0.9659 =207055 lb.

(b) The radius of curvature is obtained from Newton’s law:

154

Problem 14.81 The suspended 2-kg mass mis sta-

tionary.

(a) What are the tensions in the strings Aand B?

(b) If string Ais cut, what is the tension in string B

immediately afterward?

45°

Solution:

TB−mg cos 45◦=mv2

ρ.

But v=0 at the instant of release, so

TB=mg cos 45◦=13.9N.

Problem 14.82 The airplane ﬂies with constant veloc-

ity valong a circular path in the vertical plane. The

radius of the airplane’s circular path is 2000 m. The

mass of the pilot is 72 kg.

(a) The pilot will experience “weightlessness” at the

top of the circular path if the airplane exerts no

net force on him at that point. Draw a free-body

diagram of the pilot and use Newton’s second law

to determine the velocity vnecessary to achieve

this condition.

(b) Suppose that you don’t want the force exerted

on the pilot by the airplane to exceed four times

his weight. If he performs this maneuver at v=

200 m/s, what is the minimum acceptable radius

of the circular path?

Problem 14.83 The smooth circular bar rotates with

constant angular velocity ω0about the vertical axis AB.

The radius R=0.5 m. The mass mremains stationary

relative to the circular bar at β=40◦. Determine ω0.

Solution:

Problem 14.84 The force exerted on a charged particle

by a magnetic ﬁeld is F=qv×B, where qand vare

the charge and velocity of the particle, and Bis the

magnetic ﬁeld vector. A particle of mass mand positive

Solution: (a) The force F=q(v×B)is everywhere normal to the

velocity and the magnetic ﬁeld vector on the particle path. Therefore

the tangential component of the force is zero, hence from Newton’s

second law the tangential component of the acceleration dv

qB0. Since the term on the right is a

156

Problem 14.85 The mass mis attached to a string that

is wrapped around a ﬁxed post of radius R.Att=0, the

object is given a velocity v0as shown. Neglect external

forces on mother that the force exerted by the string.

Determine the tension in the string as a function of the

angle θ.

Strategy: The velocity vector of the mass is perpen-

dicular to the string. Express Newton’s second law in

terms of normal and tangential components.

Problem 14.86 The mass mis attached to a string

that is wrapped around the ﬁxed post of radius R.At

t=0, the mass is given a velocity v0as shown. Neglect

external forces on mother than the force exerted by the

string. Determine the angle θas a function of time.

Problem 14.87 The sum of the forces in newtons

exerted on the 360-kg sport plane (including its weight)

during an interval of time is (−1000 +280t)i+(480 −

430t)j+(720 +200t)k, where tis the time in seconds.

At t=0, the velocity of the plane’s center of gravity

is 20i+35j−20k(m/s). If you resolve the sum of the

forces on the plane into components tangent and normal

to the plane’s path at t=2 s, what are their values of

Ftand Fn?

Problem 14.88 In Problem 14.87, what is the instanta-

neous radius of curvature of the plane’s path at t=2s?

The vector components of the sum of the forces in the

directions tangenial and normal to the path lie in the

osculating plane. Determine the components of a unit

vector perpendicular to the osculating plane at t=2s.

Strategy: From the solution to problem 14.87, we

know the total force vector and acceleration vector acting

on the plane. We also know the direction of the velocity

vector. From the velocity and the magnitude of the

normal acceleration, we can determine the radius of

curvature of the path. The cross product of the velocity

vector and the total force vector will give a vector

perpendicular to the plane containing the velocity vector

and the total force vector. This vector is perpendicular

to the plane of the osculating path. We need then only

ﬁnd a unit vector in the direction of this vector.

158

Problem 14.89 The freeway off-ramp is circular with

60-m radius (Fig. a). The off-ramp has a slope β=15◦

(Fig. b). If the coefﬁcient of static friction between the

tires of a car and the road is µs=0.4, what is the

maximum speed at which it can enter the ramp without

losing traction? (See Example 14.18.)

60 m

(a)

(b)

Solution:

Problem 14.90* The freeway off-ramp is circular with

60-m radius (Fig. a). The off-ramp has a slope β(Fig. b).

If the coefﬁcient of static friction between the tires of

a car and the road is µs=0.4 what minimum slope β

is needed so that the car could (in theory) enter the off-

ramp at any speed without losing traction? (See Example

14.8.)

Solution:

Problem 14.91 A car traveling at 30 m/s is at the top

of a hill. The coefﬁcient of kinetic friction between the

tires and the road is µk=0.8. The instantaneous radius

of curvature of the car’s path is 200 m. If the driver

applies the brakes and the car’s wheels lock, what is the

resulting deceleration of the car tangent to its path?

Problem 14.92 A car traveling at 30 m/s is at the bot-

tom of a depression. The coefﬁcient of kinetic friction

between the tires and the road is µk=0.8. The instan-

taneous radius of curvature of the car’s path is 200 m.

If the driver applies the brakes and the car’s wheel lock,

what is the resulting deceleration of the car in the direc-

tion tangential to its path? Compare your answer to that

160

Problem 14.93 The combined mass of the motorcy-

cle and rider is 160 kg. The motorcycle starts from rest

at t=0 and moves along a circular track with 400-m

radius. The tangential component of acceleration as a

function of time is at=2+0.2tm/s2. The coefﬁcient of

static friction between the tires and the track is µs=0.8.

How long after it starts does the motorcycle reach the

limit of adhesion, which means its tires are on the verge

of slipping? How fast is the motorcycle moving when

that occurs?

Strategy: Draw a free-body diagram showing the tan-

gential and normal components of force acting on the

motorcycle.

Solution:

Problem 14.94 The center of mass of the 12-kg object

moves in the x–yplane. Its polar coordinates are

given as functions of time by r=12 −0.4t2m,θ =

0.02t3rad. Determine the polar components of the total

force acting on the object at t=2s.

Solution:

Problem 14.95 A 100-lb person walks on a large

disk that rotates with constant angular velocity ω0=

0.3 rad/s. He walks at a constant speed v0=5 ft/s along

a straight radial line painted on the disk. Determine the

polar components of the horizontal force exerted on him

when he is 6 ft from the center of the disk. (How are

these forces exerted on him?)

Solution:

162

Problem 14.96 The robot is programmed so that the

0.4-kg part Adescribes the path

r=1−0.5 cos(2πt) m,

θ=0.5−0.2 sin(2πt) rad.

Determine the radial and transverse components of the

force exerted on Aby the robot’s jaws at t=2s.

Problem 14.97 A 50-lb object Pmoves along the spi-

ral path r=(0.1)θ ft, where θis in radians. Its angular

position is given as a function of time by θ=2trad,

and r=0att=0. Determine the polar components of

the total force acting on the object at t=4s.

Solution:

Problem 14.98 The smooth bar rotates in the horizon-

tal plane with constant angular velocity ω0=60 rpm. If

the radial velocity of the 1-kg collar Ais vr=10 m/s

when its radial position is r=1 m, what is its radial

velocity when r=2 m? (See Active Example 14.9).

Problem 14.99 The smooth bar rotates in the horizon-

tal plane with constant angular velocity ω0=60 rpm.

The spring constant is k=20 N/m and the unstretched

length of the spring is 3 m. If the radial velocity of the

1-kg collar Ais vr=10 m/s when its radial position is

r=1 m, what is its radial velocity when r=2 m? (See

Active Example 14.9.)

3 m

Solution: Notice that the only radial force comes from the spring.

Write the term

164

Problem 14.100 The 2-kg mass mis released from rest

with the string horizontal. The length of the string is L=

0.6 m. By using Newton’s second law in terms of polar

coordinates, determine the magnitude of the velocity of

the mass and the tension in the string when θ=45◦.

Solution:

−T+mg sin θ=md2L

dt2−Lw2

Fθ=maθ

mg cos θ=m2dL

dtw+Lα

However dL

dt=d2L

dt2=0

Lα =Ldw

dθ w=gcos θ

w

wdw =g

Lπ/4

cos θdθ

2=g

Lsin θ

π/4

0=g

Lsin π

w=4.81 rad/s

|v|=Lw =2.89 m/s

−T+mg sin θ=−mLw2

T=m(g sin θ+Lw2)

T=41.6N

Problem 14.101 The 1-lb block Ais given an initial

velocity v0=14 ft/s to the right when it is in the position

θ=0, causing it to slide up the smooth circular surface.

By using Newton’s second law in terms of polar coor-

dinates, determine the magnitude of the velocity of the

block when θ=60◦.A

4 ft

Solution: For this problem, Newton’s second law in polar coordi-

nates states

In this problem, ris constant. Thus (dr/dt ) =(d2r/dt2)=0, and the

equations reduce to N=mrω2+mg cos θand rα =−gsin θ. The

ﬁrst equation gives us a way to evaluate the normal force while the sec-

ond can be integrated to give ω(θ). We rewrite the second equation as

α=dω

dt =dω

dθ

dt =ωdω

dθ =−g

rsin θ

and then integrate ω60

ω0ωdω =−g

r60◦

0sin θdθ. Carrying out the

integration, we get

ω2

2−ω2

2=−g

r(−cos θ)|60◦

0=−g

r(1−cos 60◦).

Noting that ω0=v0/R =3.5 rad/s, we can solve for ω60 =2.05 rad/s

and v60 =Rω60 =8.20 ft/s.

Problem 14.102 The 1-lb block is given an initial

velocity v0=14 ft/s to the right when it is in the position

166

Problem 14.103 The skier passes point Agoing

17 m/s. From Ato B, the radius of his circular path

is 6 m. By using Newton’s second law in terms of

polar coordinates, determine the magnitude of the skier’s

velocity as he leaves the jump at B. Neglect tangential

forces other than the tangential component of his weight.

45°

Solution: In terms of the angle θshown, the transverse component

Problem 14.104* A 2-kg mass rests on a ﬂat horizon-

tal bar. The bar begins rotating in the vertical plane about

Owith a constant angular acceleration of 1 rad/s2. The

mass is observed to slip relative to the bar when the bar

is 30◦above the horizontal. What is the static coefﬁcient

of friction between the mass and the bar? Does the mass

slip toward or away from O?

1 rad/s

1 m

2 kg

−2Rθ. For α=1, R=1, this reduces to ±µs(1+gcos θ) =−2θ+

gsin θ.Deﬁne the quantity FR=2θ−gsin θ.IfFR>0, the block

will tend to slide away from O, the friction force will oppose the

motion, and the negative sign is to be chosen. If FR<0, the block

will tend to slide toward O, the friction force will oppose the motion,

and the positive sign is to be chosen. The equilibrium condition

is derived from the equations of motion: sgn(FR)µs(1+gcos θ) =

(2θ−gsin θ), from which µs=sgn(FR)2θ−gsin θ

1+gcos θ=0.406 .

Since Fr=−3.86 <0, the block will slide toward O.

Problem 14.105* The 1/4-lb slider Ais pushed along

the circular bar by the slotted bar. The circular bar lies in

the horizontal plane. The angular position of the slotted

bar is θ=10t2rad. Determine the polar components of

the total external force exerted on the slider at t=0.2s.

2 ft

Solution: The interior angle βis between the radius from Oto the

168