phys 205 chapter 5 solutions 5 - cloud object...

Phys 205 Chapter 5 Solutions

5.3 The Earth has a radius of 6.4× 106 m ad completes one revolution about its axis in 24 h. (a) Find the speedof a point on the equator. (b) Find the speed of New York City.

This is a kinematics of uniform circular motion problem.

Given: Radius of the Earth: rE = 6.4× 106 mPeriod of rotation: T = 24h = 86, 400 s

(a) We want to find the tangential speed of a point at the equator. Use Equation 5.1.

vEq =2πrET

vEq =2π(6.4× 106 m

)86, 400 s

= 465 m/s

(b) New York City has a latitude of 40.7◦ so the radius of its orbit will not be the full radius of the Earth (seethe figure for Problem 5.9 below). The radius of its orbit will be: r = rE cos 40.7◦ = 4.85× 106 m.

vNY =2πr

TvNY =

2π(4.85× 106 m

)86, 400 s

= 353 m/s

5.6 Consider the motion of the hand of a mechanical clock. If the minute hand of the clock has a length of 6.0 cm,what is the centripetal acceleration of a point at the end of the hand?


Given: Period T = 60 min = 3600 sRadius: r = 0.060 m

We want to find the centripetal acceleration, ac. Use Equations 5.5 and 5.1.

ac =v2

r=

1

r

(2πr

T

)2

→ ac =4π2r

T 2

This last equation is useful to use for problems like this one when you are given the period & radius and wantthe centripetal acceleration.

ac =4π2 (0.060 m)

(3600 s)2 = 1.8× 10−7m/s

2

5.9 Consider points on the Earth’s surface as sketched in the figure. Because of the Earth’srotation, these points undergo uniform circular motion. Compute the centripetal acceler-ation of (a) a point at the equator, and (b) at latitude of 30◦.


Given: Radius of the Earth: rE = 6.4× 106 mPeriod of rotation: T = 24h = 86, 400 s

(a) We want to find the centripetal acceleration, ac, of a point at the equator. Use theequation from the previous problem.

!"#$

ac =4π2r

T 2ac =

4π2(6.4× 106 m

)(86, 400 s)

2 = 0.034 m/s2

(b) At a latitude of 30◦ the radius of orbit will be: r = rE cos 30◦ = 5.54× 106 m.

ac =4π2r

T 2ac =

4π2(5.54× 106 m

)(86, 400 s)

2 = 0.029 m/s2

1

5.12 When a fighter pilot makes a quick turn, he experiences a centripetal acceleration. When this accelerationis greater than about 8 × g, the pilot will usually lose consciousness (“black out”). consider a pilot flying ata speed of 900 m/s who wants to make a sharp turn. What is the minimum radius of curvature she can takewithout blacking out?


(a) Given: Maximum centripetal acceleration: ac = 8× g = 78.4 m/s2

Tangential speed: v = 900 m/s

We want to find the minimum radius of curvature which corresponds to the maximum centripetal acceleration.Use Equation 5.5.

ac =v2

r→ r =

v2

acr =

(900 m/s)2

78.4 m/s2 = 1.0× 104 m

The minimum radius is roughly 10 km.

5.14 The Daytona 500 stock car race is held on a track that is approximately 2.5 mi long, and the turns are bankedat an angle of 31◦. It is currently possible for cars to travel through the turns at a speed of 180 mi/h. Assumingthese cars are on the verge of slipping into the outer wall of the racetrack (because they are racing!), find thecoefficient of static friction between the tires and the track.This is a dynamics of uniform circular motion problem.

Given: Track length: L = 2.5 mi = 4025 mTangential velocity: v = 180 mi/h = 80.5 m/s

Bank angle: θ = 31◦

We want to find the coefficient of static friction, µs, assuming the friction force isat a maximum at this speed. A diagram will help (see right). In order to find therequired centripetal acceleration, we need to know the radius of curvature of thebanked turns. If we assume a circular track, the radius would be L/2π = 0.398 mi.The actual track shape is much more oval so I’ll estimate r = 0.20 mi = 322 m.The centripetal acceleration is:

ac =v2

r=

(80.5 m/s)2

322 m= 20.1 m/s

2= 2.05g

!"#$ !!

!

! F g

!

! F fr

!

! F N

!

y

!

x

Applying Newton’s 2nd Law:

∑~F = m~ac →

x component: − FN sin θ − Ffr cos θ = −mac

y component: FN cos θ − Ffr sin θ −mg = 0

If the static friction force is at a maximum, we have Ffr = µsFN , and this gives us:

FN sin θ + µsFN cos θ = mac FN cos θ − µsFN sin θ = mg

FN (sin θ + µs cos θ) = mac FN (cos θ − µs sin θ) = mg

This gives us two equations with three unknowns (FN , m, and µs). We want to solve this for µs. In order toisolate µs, I’m going to divide the equation on the left by the equation on the right:

FN (sin θ + µs cos θ)

FN (cos θ − µs sin θ)=macmg

→ (sin θ + µs cos θ)

(cos θ − µs sin θ)=acg

→ sin θ+µs cos θ = (cos θ − µs sin θ)acg

Both the mass and the normal force cancel out and I’m left with one equation for µs. Isolate µs on one side:

µs (cos θ + 2.05 sin θ) = 2.05 cos θ − sin θ → µs =2.05 cos 31◦ − sin 31◦

cos 31◦ + 2.05 sin 31◦= 0.65

Your answer will vary depending on your estimate of the radius of curvature.

2

5.16 Consider the motion of a rock tied to a string of length 0.50 m. The string is spunso that te rock travels in a vertical circle as shown in figure. The mass of the rockis 1.5 kg, and it is twirling at constant speed with a period of 0.33 s.

(a) Draw free-body diagrams for the rock when it is at the top and when it is atthe bottom of the circle. Your value should include the tension in the stringbut the value of FT is not yet known.

(b) What is the total force on the rock directed towards the center of the circle?

(c) Find the tension in the string when the rock is at the top and when it is atthe bottom of the circle.

!"#$

!!

!

! F g

!

! F fr

!

! F N

!

y

!

x

!

r = 0.50m

!

m =1.5kg

This is a dynamics of uniform circular motion problem.

Given: Radius: r = 0.50 mPeriod: T = 0.33 sMass: m = 1.5 kg

Calculate:Centripetal acceleration: ac = 4π2r

T 2 = 181 m/s2

Centripetal force: mac = 272 N

(a) Free body diagrams for the rock at the top and at the bottom of the circle.

Free body diagram at the top:

h

v0

FS FS FS FS FS FS FS FS FS FS


FN

FT

mg

m2

m3 m1

F

N

Ff,max

mg

Fyou

Fleg

T

T

m

FT mg Free-body diagram at the bottom:

h

v0



FN

FT

mg

m2

m3 m1

F

N

Ff,max

mg

Fyou

Fleg

T

T

m

FT mg

(b) If the rock is in uniform circular motion, then the force directed towards the center of the circle is alwaysmac = 272 N.

(c) At the top both FT and Fg contribute to the centripetal force. At the bottom, FT has to overcome Fg toprovide the centripetal force.

Top: FT +mg = mac → FT = mac −mg FT = (1.5 kg)(

181 m/s2 − 9.80 m/s

2)

= 257 N

Bottom: FT −mg = mac → FT = mac +mg FT = (1.5 kg)(

181 m/s2

+ 9.80 m/s2)

= 286 N

5.20 A roller coaster track is designed so that the car travels upside down on a certainportion of the track as shown in the figure. What is the minimum speed the rollercoaster can have without falling from the track? Assume the track has a radius ofcurvature of 30 m.

This is a dynamics of uniform circular motion problem.!

r

!

! v

Given: Radius of curvature: r = 30 m

We want to find the minimum speed that keeps the roller coaster car on the track. At high speeds, the normalforce from the track on the car keeps the car in circular motion. At slower speeds, the normal force becomessmaller. At the minimum speed, gravity provides all of the centripetal acceleration at the top. At slowerspeeds, the car would fall. ∑

~F = m~a → mg = mac → g =v2

r

v =√gr v =

√(9.80 m/s

2)(30 m) = 17 m/s

3

5.26 Spin out! An interesting amusement park activity involves a cylindrical room thatspins about a vertical axis (see figure). Participants in the “ride” are in contactwith the wall of the room, and the circular motion of the room results in a normalforce from the wall on the riders. When the room spins sufficiently fast, the flooris retracted and the frictional force from the wall keeps the people “stuck” to thewall. Assume the room has a radius of 2.0 m and the coefficient of static frictionbetween the people and the wallis µs = 0.50.

!

r

!

! v

(a) Draw pictures showing the motion of a “rider.” Give both a side view and a top view.

(b) What are all the forces acting on the rider? Add them to your pictures in part (a). Then draw a free-bodydiagram for a rider.

(c) What are the components of all forces directed towards the center of the circle (the radial direction)?

(d) Apply Newton’s second law along both the vertical and the radial directions. Find the minimum rotationrate for which the riders do not slip down the wall.


(a) and (b) View from the side:

!

r

!

! v

!

! F g

!

! F fr

!

! F N

!

! F fr

!

! F N

!

! F g

View from the top:

!

r

!

! v

!

! F g

!

! F fr

!

! F N

!

! F fr

!

! F N

!

! F g

!

r

!

m

!

! F g

!

! F fr

!

! F N

!

! F fr

!

! F N

!

! F g

The forces on the person are the normal and frictional forces from the wall, and the force of Earth’s gravity.

(c) The only one with a component towards the center is the normal force from the wall.

(d) We want to find the minimum rotation rate to stay stuck to the wall. This will be when the friction forceis at a maximum, Ffr = µsFN .

Given: Radius: r = 2.0 mCoefficient of friction: µs = 0.50

Apply Newton’s second law to the person.∑~F = m~a → FN = mac Ffr −mg = 0

Using the fact that the static friction force is at a maximum, Ffr = µsFN = µsmac, we have:

µsmac −mg = 0 → µs

(4π2r

T 2

)= g → T =

√4π2rµsg

T =

√4π2 (2.0 m) (0.50)

9.80 m/s2 = 2.0 s

So the room must complete a minimum of a half rotation every second in order for the person to stay “stuck.”

5.28 A rock of mass m = 2.5 kg is tied to the end of a string of length L = 1.2 m. Theother end of the string is fastened to a ceiling, and the rock is set into motion sothat it travels in a horizontal circle of radius r = 0.70 m as sketched in the figure.

(a) Draw a picture showing the motion of the rock. Give both a top view and aside view.

(b) What are all the forces acting on the rock? Add them to your pictures in part(a). Then draw a free-body diagram for the rock.

(c) Find the vertical and horizontal components of all of the forces on the rock.

(d) Apply Newton’s second law in both the vertical (y) and the horizontal direc-tions. What is the acceleration along y? Find the tension in the string.

!

r

!

! v

!

! F g

!

! F fr

!

! F N

!

! F fr

!

! F N

!

! F g

!

r

!

m

4


(a) and (b) View from the side:

!

r

!

! v

!

! F g

!

! F fr

!

! F N

!

! F fr

!

! F N

!

! F g

!

r

!

m

!

! F g

!

! F

T

!

! F

T

!

! F g

!!View from the top:

!

r

!

! v

!

! F g

!

! F fr

!

! F N

!

! F fr

!

! F N

!

! F g

!

r

!

m

!

! F g

!

! F

T

!

! F

T

!

! F g

The only forces on the rock are the Earth’s gravitational force and the tension force from the string.

(c) The components of the gravitational force: Fgx = 0 and Fgy = −mg = −24.5 N

We’ll find the components of the tension force, FTx = FT cos θ and FTy = FT sin θ, when we apply Newton’ssecond law to the rock.

(d) Applying Newton’s second law to the rock.∑~F = m~a → FT cos θ = mac FT sin θ −mg = 0

The acceleration along y is zero as long as it is in uniform circular motion that is horizontal. We can find theangle knowing that the length of the string (the hypotenuse) is 1.2 m and the horizontal component is 0.70 m.

cos θ =0.70 m

1.2 m→ θ = cos−1 (0.5833) = 54.3◦

Now I’ll use the vertical component of Newton’s 2nd law to find the tension.

FT sin θ = mg FT =mg

sin θFT =

(2.5 kg)(9.80 m/s2)

sin 54.3◦= 30.2 N

5.30 Consider a Ferris wheel in which the chairs hang down from the main wheelvia a cable. The cable is 2.0 m long, and the radius of the wheel is 12 m(see figure). When a chair is in the orientation shown in the figure (the “3o’clock” position), the cable attached to the chair makes an angle of θ = 20◦

with the vertical. Find the speed of the chair.


At this position, the tension provides the centripetal force and balances grav-ity.

Given: Radius: r = 12 m

!!

!

r = 0.50m

!

m =1.5kg

∑~F = m~a → FT sin θ =

mv2

rFT cos θ = mg

Divide the horizontal equation by the vertical equation (to get rid of the tension):

FT sin θ

FT cos θ=mv2/r

mg→ tan θ =

v2

gr→ v

√gr tan θ

v =

√(9.80 m/s

2)(12 m) tan 20◦ v = 6.5 m/s

5.34 NASA has built centrifuges to enable astronauts to train in conditions in which the acceleration is very large.The device in Figure P5.34 shows one of these “human centrifuges.” If the device has a radius of 8.0 m andattains accelerations as large as 5.0× g, what is the rotation rate?


Given: Radius: r = 8.0 mCentripetal acceleration: ac = 5.0× g = 49.0 m/s

2

5

We want to find the rotation rate. I’ll find the period then take the inverse.

ac =4π2r

T 2→ T =

√4π2r

acT =

√4π2(8.0 m)

49.0 m/s2 T = 2.54 s

So the wheel rotates at 1/T = 0.39 rev/s.

5.35 Two small objects of mass 20 kg and 30 kg are a distance 1.5 m apart. What is the gravitational force of oneof these objects on the other?

This is a Newton’s Law of Gravitation problem.

Given: Separation: r = 1.5 mMasses: m1 = 20 kg, m2 = 30 kg

We simply need to use Equation 5.16 to find the magnitude of the force.

FG =Gm1m2

r2→ FG =

(6.67× 10−11 N ·m2/kg2)(20 kg)(30 kg)

(1.5 m)2 FG = 1.8× 10−8 N

5.38 Travel and lose pounds! Your apparent weight is the force you feel on the bottoms of your feet when youare standing. Due to the Earth’s rotation, your apparent weight is slightly more when you are at the SouthPole than when you are at the equator. What is the ratio of your apparent weight at these two locations?Carry three significant figures in your calculation.


At the equator, the radius of orbit is the Earth’s radius, r = 6/4 × 106 m. The normal force does not quitebalance the gravitational force at the equator.∑

~F = m~a → mg − FNequator= mac → FNequator

= mg − 4π2mr

T 2

At the pole, the radius of the orbit has gone to zero and the normal force balances your weight.∑~F = m~a → mg − FNpole

= 0 → FNpole= mg

Taking the ratio of these two normal forces:

FNpole

FNequator

=mg

mg − 4π2mr/T 2→

FNpole

FNequator

=1

1− 4π2r/gT 2

The change is going to be small so I will calculate the factor by itself:

4π2mr

gT 2=

4π2(6.4× 106 m)

(9.80 m/s2) (86, 400 s)

2 = 0.00345

Using this, the ratio of normal forces is:FNpole

FNequator

= 1.003

So your weight at the pole will be 0.3% larger than at the equator.

5.46 Suppose the density of the Earth was somehow reduced from its actual value to 1000 kg/m3 (the density ofwater). Find the value of g, the acceleration due to gravity, on this new planet. Assume the radius does notchange.


Given: Radius: r = 6.4× 106 mDensity: ρ = 1000 kg/m

3

We need to find the new mass to find the new value of g using Equation 5.21.

Mnew = ρV = ρ

(4

3πr3)

= 1.10× 1024 kg g =GMnew

r2= 1.79 m/s

2

6

5.48 In Section 5.4, we showed that the radius of a geosynchronous orbit about the Earth is 4.2× 107 m, comparedwith the radius of the Earth, which is 6.4× 106 m. By what factor is the force of gravity smaller when you arein geosynchronous orbit than when you are on the Earth’s surface?


Given: Radii: rE = 6.4× 106 m, rgs = 4.2× 107 m

Use Newton’s Law of Gravitation to find the factor relating the forces.

FGgs

FG=GmEm/r

2gs

GmEm/r2E→

FGgs

FG=r2Er2gs

FGgs

FG=

(6.4× 106 m

4.2× 107 m

)2

= 0.023

5.50 The region of the solar system between Mars and Jupiter contains many asteroids that orbit the Sun. Consideran asteroid in a circular orbit of radius 5.0× 1011 m. Find the period of the orbit.

This is a Kepler’s laws problems.

Given: Radius of orbit: r = 5.0× 1011 m

I’ll use the Earth as our reference object which is also orbiting the Sun.

(I’ll use ES to indicate the period and radius of the Earth’s orbit around the Sun. )

T 2

T 2ES

=r3

r3ES→ T = TE

(r

rE

)3/2

T = (1.00 yr)

(5.0× 1011 m

1.5× 1011 m

)3/2

T = 6.09 yr

The orbit takes a little over six years or about 2220 days.

5.52 In recent years, a number of nearby stars have been found to posses planets. Suppose the orbital radius ofsuch a planet is found to be 4.0× 1011 m, with a period of 1100 days. Find the mass of the star.

This is a Kepler’s laws problems.

Given: Radius of orbit: r = 4.0× 1011 mPeriod of orbit: T = 1100 days = 9.50× 107 s

We want to find the mass of the star that this planet is orbiting. We can use Equation 5.26 with the mass ofthe star replacing the mass of the Sun.

T 2 =

(4π2

GMstar

)r3 → Mstar =

4π2r3

GT 2

Mstar =4π2(4.0× 1011 m)3

(6.67× 10−11 N ·m2/kg2)(9.50× 107 s)2Mstar = 4.2× 1030 kg

This star is more than twice as massive as our Sun.

5.54 In our derivation of Kepler’s laws, we assumed the only force on a planet is due to the Sun. In a real solarsystem, however, the gravitational forces from the other planets can sometimes be important. Calculate thegravitational force of Jupiter on the Earth and compare it to to the magnitude of the force from Sun. Do thecalculations for the cases when Jupiter is both closest to and farthest from Earth (see figure).

!!

!

r = 0.50m

!

m =1.5kg

!"#$%&'( )*'%+(

,"-(

!"#$%&'()*'%+(

,"-(

!!

!

r = 0.50m

!

m =1.5kg

!"#$%&'( )*'%+(

,"-(

!"#$%&'()*'%+(

,"-(

7


Given: Radii of orbit: rES = 1.50× 1011 m, rJS = 7.78× 1011 mMasses: ME = 5.98× 1024 kg, MJ = 1900× 1024 kg, MS = 1.99× 1030 kg

We want to find the gravitational force of Jupiter on the Earth and compare it to the gravitational force of theSun on the Earth. I’ve used Table 5.1 on page 147 to fill in the relevant information for the Earth, Jupiter andSun. The figure above is not to scale but does show the orientation of the Earth and Jupiter in theirs orbitswhen they are closest and farthest apart.

rnear = rJS − rES = 6.28× 1011 m rfar = rJS + rES = 9.28× 1011 m

We just need to calculate three gravitational forces: (1) FG, the Sun on the Earth; (2) FGnear, Jupiter on the

Earth when closest; and (3) FGfar, Jupiter on the Earth when farthest apart.

FG =GmSmE

r2ES=

(6.67× 10−11 N ·m2/kg2)(1.99× 1030 kg)(5.98× 1024 kg)

(1.50× 1011 m)2 FG = 3.53× 1022 N

FGnear=GmJmE

r2near=

(6.67× 10−11 N ·m2/kg2)(1.90× 1027 kg)(5.98× 1024 kg)

(6.28× 1011 m)2 FGnear = 1.92× 1018 N

FGfar=GmJmE

r2far=

(6.67× 10−11 N ·m2/kg2)(1.90× 1027 kg)(5.98× 1024 kg)

(9.28× 1011 m)2 FGfar

= 8.80× 1017 N

When the Earth and Jupiter are closest, the gravitational force from Jupiter on the Earth relative to that fromthe Sun is:

FGnear

FG= 5.44× 10−5

When the Earth and Jupiter are farthest apart, the gravitational force from Jupiter on the Earth relative tothat from the Sun is:

FGfar

FG= 2.49× 10−5

While the gravitational force from Jupiter on the Earth may seem large (∼ 1018 N), it is still very smallcompared to the gravitational force from the Sun on the Earth.

8

phys 205 chapter 5 solutions 5 - cloud object...

Documents