Physics 18 Spring 2011

Homework 4 - Solutions

Wednesday February 9, 2011

Make sure your name is on your homework, and please box your final answer. Becausewe will be giving partial credit, be sure to attempt all the problems, even if you don’tfinish them. The homework is due at the beginning of class on Wednesday, February16th. Because the solutions will be posted immediately after class, no late homeworks canbe accepted! You are welcome to ask questions during the discussion session or during officehours.

1. A particle of mass m moving along the x axis experiences the net force Fx = ct, wherec is a constant. The particle has a velocity v0x at time t = 0. Find an algebraicexpression for the particles velocity vx at a later time t.

————————————————————————————————————

Solution

Newton tells us that the force is mass times acceleration, Fx = max. So, the accelera-tion is ax = Fx/m = ct

m. To get the velocity we just need to integrate the acceleration,

since ax = vx. Doing so gives

vx =ct2

2m,

where we have set v0x = 0 as given in the initial conditions.

1

2. Estimate the widest stance you can take when standing on a dry, icy surface. That is,how wide can you safely place your feet on not slip into an undesired “split?” Let thecoefficient of static friction of rubber on ice be roughly 0.25.

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Solution

Consider the forces acting on your foot, when your feet arespread apart at an angle θ, as shown in the figure to the right.The frictional force, ~Ff , preventing you from sliding points to

the right, the normal force ~FN points vertically, while the force ofyour weight, ~FW , points at an angle, since your leg isn’t straightup and down. Note that FW 6= mg, in general, since yourweight is spread over two feet. In particular, it would likelybe FW = mg/2, but we’ll just leave it as FW for now. Let’swork out the components,∑

Fx = −FW sin θ + Ff = 0 ⇒ Ff = FW sin θ∑Fy = −FW cos θ + FN = 0 ⇒ FN = FW cos θ.

Dividing Ff/FW gives

FfFN

= tan θ.

Notice that the weight cancels out - the angle doesn’t dependon how much you weigh!

θ

FW

Ff

FN

Now, the frictional force is given by Ff = µsFN , in the static case, where µs is thecoefficient of static friction. So, Ff/FN = µs. Thus,

µs = tan θ ⇒ θ = tan−1(µs).

For the ice and shoe combination, µs = 0.25, and so

θ = tan−1(µs) = tan−1(0.25) = 14◦,

which corresponds to an angle of 28◦ between your legs.

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3. A block of mass m is pulled at a constantvelocity across a horizontal surface bya string as shown in the figure. Themagnitude of the frictional force is (a)µkmg, (b) T cos θ, (c) µk (T −mg), (d)µkT sin θ, or (e) µk (mg − T sin θ).

————————————————————————————————————

Solution

The block is being pulled at a constant velocity, so the acceleration is zero. Thismeans that all the forces balance. Since the block isn’t being pulled off the ground, thegravitational force is balanced by the vertical component of the tension in the string.The friction is fighting the motion to the left, providing a force acting to the right.The only force acting to the left is the horizontal component of the tension, −T cos θ.The frictional force has to cancel this, so Ff = T cos θ. So, answer (b) is correct.

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4. A 12 kg turtle rests on the bed of a zookeeper’s truck, which is traveling down a countryroad at 55 mi/h. The zookeeper spots a deer in the road, and slows to a stop in 12s. Assuming constant acceleration, what is the minimum coefficient of static frictionbetween the turtle and the truck bed surface that is needed to prevent the turtle fromsliding?

————————————————————————————————————

Solution

If the deceleration of the truck is too big,the turtle will start sliding. Let’s look atthe different forces on the turtle.∑

Fx = −Ff = −max∑Fy = −Fg + FN = 0,

where the minus sign on the accelerationcomes from the fact that the truck is decel-erating. So, these say that Fg = FN = mg,and that the frictional force is Ff = max.Now, the frictional force is Ff = µsFN =µsmg, where µs is the coefficient of slidingfriction. Thus,

µs =axg.

So, we just need to determine the acceler-ation of the truck. The truck starts at avelocity of 55 mph, which is the same asvi = 55 miles/hr × 1609 meters/mile × 1hr/3600 s = 25 m/s. The final velocity iszero, since the truck stops. If it takes atime t (which is 12 seconds) to stop, thenthe acceleration is ax = v/t. Thus,

Ff

Fg

FN

x

y

µs =axg

=v

gt=

25

9.8× 12= 0.21

So, we need a coefficient of at least 0.21 to keep the turtle from sliding.

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5. You and your friends push a 75.0 kg greased pig up an aluminum slide at the countyfair, starting from the low end of the slide. The coefficient of kinetic friction betweenthe pig and slide is 0.070.

(a) All of you pushing together (parallel to the incline) manage to accelerate the pigfrom rest at the constant rate of 5.0 m/s2 over a distance of 1.5 m, at which pointyou release the pig. The pig continues up the slide, reaching a maximum verticalheight above its release point of 45 cm. What is the angle of inclination of theslide?

(b) At the maximum height the pig turns around and begins to slip down the slide,how fast is it moving when it arrives at the low end of the slide?

————————————————————————————————————

Solution

(a) Let’s look at the forces acting on the pig.∑Fx = −Fk − Fg sin θ = max∑Fy = −Fg cos θ + FN = 0.

Solving the second gives FN = Fg cos θ =mg cos θ. Now, the frictional force is Fk =µkFN = µkmg cos θ. Plugging this in tothe first gives, after canceling off the mass,

−g (µk cos θ + sin θ) = ax.

This is the acceleration the pig feels as itmoves up the ramp. How far along theramp does he move? From the constant ac-celeration equation, v2

f = v2i + 2a∆x, then

if he stops moving at the end of the ramp,

vf = 0, and so d = − v2i2a

.

Chapter 5

432

(a) Substitute numerical values and evaluate F:

( ) ( )( )[ ] N24m/s1.57m/s9.810.40kg10 22 =−=F 61 ••• You and your friends push a 75.0-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070. (a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 5.0 m/s2 over a distance of 1.5 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 45 cm. What is the angle of inclination of the slide? (b) At the maximum height the pig turns around and begins to slip down once slide, how fast is it moving when it arrives at the low end of the slide? Picture the Problem The free-body diagram shows the forces acting on the pig sometime after you and your friends have stopped pushing on it but before it has momentarily stopped moving up the slide. We can use a constant-acceleration equations and Newton’s 2nd law to find the angle of inclination of the slide and the pig’s speed when it returns to bottom of the slide. The pictorial representation assigns a coordinate system and variable names to the variables that we need to consider in solving this problem.

θ

kfr

nFr

gFr

y

x

00 =xm 5.11 =x

2x02 =v

1v

00 =v

3v

02 =v

Pig going up the incline

Pig going down the incline

(a) Apply amF =∑ to the pig: xx maFfF =−−=∑ θsingk (1)

and 0cosgn =−=∑ θFFFy (2)

Substitute nkk Ff μ= and mgF =g in equation (1) to obtain:

xmamgF =−− θμ sinnk (3)

Solving equation (2) for Fn yields: θcosgn FF =

Substitute for Fn in equation (3) to obtain:

xmamgmg =−− θθμ sincosk

Solving for ax yields: ( )θθμ sincosk +−= gax (4)

Plugging in for a gives

d =v2i

2g (µk cos θ + sin θ).

Finally, if the distance along the ramp is d, then the height is h = d sin θ. Thus,the height the pig slides up is

h =v2i sin θ

2g (µk cos θ + sin θ)=

v2i

2g (1 + µk cot θ),

after dividing through by sin θ. Solving for the angle gives

tan θ =µk(

v2i2gh− 1) .

5

Now, we just need the initial velocity. Since the pig is pushed with an accelerationof 5 m/s2, then after a distance of 1.5 m, he’s traveling at a speed of vf =

√2ad =√

2× 5× 1.5 = 3.9 m/s. So, we have everything that we need,

θ = tan−1

µk(v2i2gh− 1) = tan−1

[0.070(15

2×9.8×.45− 1)] = tan−1(0.01) = 5.7◦.

(b) Now the pig is sliding back down the ramp. In this case the frictional force opposesthe slide down and changes his acceleration to

a = −g (sin θ − µk cos θ) ,

and is constant. So, since the pig starts from rest at the top, and is moving at aspeed vf at the bottom, if he moves a distance −∆x, then,

vf =√

2a× (−∆x) =√

2g (sin θ − µk cos θ) ∆x.

What’s ∆x? This is the total distance than the pig slid up, d = h/ sin θ, plus theextra 1.5 meters that he started off sliding down (call that distance d0). The totaldistance is ∆x = h

sin θ+ d0. Putting everything together gives

vf =√

2g (sin θ − µk cos θ) ∆x =

√2g (sin θ − µk cos θ)

(d0 +

h

sin θ

).

Plugging in all the numbers gives

vf =√

2g (sin θ − µk cos θ)(d0 + h

sin θ

)=

√2× 9.8 (sin 5.7◦ − 0.07× cos 5.7◦)

(1.5 + .45

sin 5.7◦

)= 1.87 m/s.

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6. A pilot with a mass of 50 kg comes out of a vertical dive in a circular arc such that atthe bottom of the arc her upward acceleration is 3.5g.

(a) How does the magnitude of the force exerted by the airplane seat on the pilot atthe bottom of the arc compare to her weight?

(b) Use Newton’s laws of motion to explain why the pilot might be subject to ablackout. This means that an above normal volume of blood “pools” in her lowerlimbs. How would an inertial reference frame observer describe the cause of theblood pooling?

————————————————————————————————————

Solution

(a) We can draw a free-body diagram for thepilot, which is seen to the right. ApplyingNewton’s laws to the system gives∑

Fy = FN − Fg = macent,

where acent is the centripetal acceleration.The weight that the pilot feels comes fromthe strength of the normal force from theseat pushing back up on her. Solving forthe normal force gives

FN = m (g + acent) .

The ratio of the normal force to her normalweight mg is

FNFg

=g + acent

g= 1 +

acent

g.

Chapter 5

454

78 •• A pilot with a mass of 50 kg comes out of a vertical dive in a circular arc such that at the bottom of the arc her upward acceleration is 3.5g. (a) How does the magnitude of the force exerted by the airplane seat on the pilot at the bottom of the arc compare to her weight? (b) Use Newton’s laws of motion to explain why the pilot might be subject to a blackout. This means that an above normal volume of blood ″pools″ in her lower limbs. How would an inertial reference frame observer describe the cause of the blood pooling? Picture the Problem The diagram shows the forces acting on the pilot when her plane is at the lowest point of its dive. nF is the force the airplane seat exerts on her. We’ll apply Newton’s 2nd law for circular motion to determine Fn and the radius of the circular path followed by the airplane.

gmFrr

=g

r

nFr

(a) Apply radialradial maF =∑ to the

pilot:

cn mamgF =− ⇒ ( )cn agmF +=

Because ga 5.3c = : ( ) mgggmF 5.45.3n =+=

The ratio of Fn to her weight is:

5.45.4n ==mg

mgmgF

(b) An observer in an inertial reference frame would see the pilot’s blood continue to flow in a straight line tangent to the circle at the lowest point of the arc. The pilot accelerates upward away from this lowest point and therefore it appears, from the reference frame of the plane, as though the blood accelerates downward. 79 •• A 80.0-kg airplane pilot pulls out of a dive by following, at constant speed, the arc of a circle whose radius is 300 m. At the bottom of the circle, where his speed is 180 km/h, (a) what are the direction and magnitude of his acceleration? (b) What is the net force acting on him at the bottom of the circle? (c) What is the force exerted on the pilot by the airplane seat?

Since the centripetal acceleration at the bottom of her arc is acent = 3.5g, thenthe ratio is 1 + 3.5g/g = 4.5. So, she feels 4.5 times heavier than usual.

(b) The pilot might black out because the blood can’t flow up to her brain. An inertialobserver watching from outside the plane doesn’t see any additional “centrifugal”force pushing the blood down to her feet. Instead, they see the blood trying tocontinue along it’s path tangent to the circular arc. Because the pilot pulls up inthe arc, her feet travel up, while the blood continues along horizontally; so, theblood seems to be “pulled” down towards her feet.

7

7. A small bead with a mass of 100 g slideswithout friction along a semicircular wirewith a radius of 10 cm that rotates abouta vertical axis at a rate of 2.0 revolutionsper second. Find the value of θ for whichthe bead will remain stationary relative tothe rotating wire.

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Solution

Let’s start by looking at the forces on the bead,as we’ve done before. The normal force pointsperpendicularly to the wire, as seen in the figure.Because the bead is spinning around in the hori-zontal plane, there is a centripetal force in the xdirection. So,∑

Fx = −FN sin θ = mv2

r∑Fy = FN cos θ −mg = 0.

Solving the bottom equation gives FN = mgcos θ

.Plugging this into the first gives for the angle

tan θ =v2

rg.

Chapter 5

464

Picture the Problem The semicircular wire of radius 10 cm limits the motion of the bead in the same manner as would a 10-cm string attached to the bead and fixed at the center of the semicircle. The horizontal component of the normal force the wire exerts on the bead is the centripetal force. The application of Newton’s 2nd law, the definition of the speed of the bead in its orbit, and the relationship of the frequency of a circular motion to its period will yield the angle at which the bead will remain stationary relative to the rotating wire.

Apply∑ = aF m to the bead: ∑ ==

rvmFFx

2

n sinθ

and ∑ =−= 0cosn mgFFy θ

Eliminate Fn from the force equations to obtain: rg

v2

tan =θ

The frequency of the motion is the reciprocal of its period T. Express the speed of the bead as a function of the radius of its path and its period:

Trv π2

=

Using the diagram, relate r to L and θ :

θsinLr =

Substitute for r and v in the expression for tanθ and solve for θ : ⎥

⎦

⎤⎢⎣

⎡= −

LgT

2

21

4cos

πθ

Substitute numerical values and evaluate θ :

( )( )( ) °=⎥

⎦

⎤⎢⎣

⎡= − 52

m0.104πs0.50m/s9.81cos 2

221θ

Now, we don’t know the speed. However, we do know the period, T = 2πr/v. So,v = 2πr/T . We are told the frequency of rotations, f = T−1. And so, plugging thisback into the expression for the angle gives

tan θ =4π2r

gT 2=

4π2f 2r

g.

Finally, the radius of the ring is L, but the bead rotates at a distance r. From thegeometry, we see that r = L sin θ. Making this substitution, gives 1/ cos θ = 4π2f 2L/g,after canceling off the mutual sine term. So, finally solving for the angle gives

θ = cos−1

(g

4π2f 2L

)= cos−1

(9.8

4π2 × 22 × 0.1

)= 52◦.

8

8. In 1976, Gerard O’Neill proposed that large space stations be built for human habi-tation in orbit around Earth and the moon. Because prolonged free-fall has adversemedical effects, he proposed making the stations in the form of long cylinders andspinning them around the cylinder axis to provide the inhabitants with the sensationof gravity. One such O’Neill colony is to be built 5.0 miles long, with a diameter of0.60 mi. A worker on the inside of the colony would experience a sense of “gravity”because he would be in an accelerated reference frame due to the rotation.

(a) Show that the “acceleration of gravity” experience by the worker in the O’Neillcolony is equal to his centripetal acceleration.

(b) If we assume that the space station is composed of several decks that are atvarying distances (radii) from the axis of rotation, show that the “acceleration ofgravity” becomes weaker the closer the worker gets to the axis.

(c) How many revolutions per minute would this space station have to make to givean “acceleration of gravity” of 9.8 m/s2 at the outermost edge of the station?

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Solution

(a) An astronaut standing on the “ground” in the spinning station will experiencetwo forces - one is the centripetal force which comes from spinning around in acircle, and the other is the normal force from the ground pushing back up on hisfeet. As long as he’s not accelerating (which is the case if he’s standing on theground), then these two forces will cancel. Since the normal force is what we feelas our weight, then the centripetal force is what he’ll feel as gravity.

(b) The acceleration of “gravity” is given by the centripetal acceleration, as we’vediscussed. So, we can write g = v2

r. The different decks are traveling at different

speeds (decks at bigger radii have to go faster to go around in the same amount oftime), but they all have the same orbital period, T . The period is just T = 2πr/v,and so T = 2πr/v. Thus, the acceleration is

g =v2

r=

(4π2

T 2

)r.

So, the acceleration gets bigger as we go further out.

(c) The number of revolutions per second, f = 1/T , and so g = (2πf)2 r. Solving forthe frequency gives f = 1

2π

√gr. The diameter is given in miles. Recalling that 1

mile = 1609 meters, we can solve for the frequency.

f =1

2π

√g

r=

1

2π

√9.8

.3× 1609= 0.023 rev/sec.

Multiplying this by 60 gives us f = 1.36 revolutions per minute, which seemsfairly slow (but it is a big station).

9

9. The position of a particle of mass m = 0.80 kg as a function of time is given by~r = xi+ yj = (R sinωt) i+ (R cosωt) j, where R = 4.0 m and ω = 2πs−1.

(a) Show that the path of this particle is a circle of radius R, with its center at theorigin of the xy plane.

(b) Compute the velocity vector. Show that vx/vy = −y/x.

(c) Compute the acceleration vector and show that it is directed toward the originand has the magnitude v2/R.

(d) Find the magnitude and direction of the net force acting on the particle.

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Solution

(a) The magnitude of the displacement is r =√x2 + y2. Plugging in the components

gives

r =√x2 + y2 =

√R2 cos2 ωt+R2 sin2 ωt = R.

So, the magnitude of the displacement doesn’t change with time - the particle isalways at a distance r. In other words, it’s moving in a circle of radius R. Becausex and y both have a range from ±R, the circle is centered at the origin - if it wascentered at some other point, say at x = 4, then the range of x would be from4 +R to 4−R. Since this isn’t the case, the circle is centered at the origin.

(b) The velocity vector is ~v = ~r, so we have to compute the time-derivative of theposition vector. Doing so gives

~r = (Rω cosωt) i− (Rω sinωt) j.

Computing vy/vx = −(Rω sinωt)/(Rω cosωt) = −R sinωt/R cosωt = −y/x.

(c) The acceleration vector is ~a = ~r = ~v, and so we just need to take one morederivative of the velocity. This gives

~a = − (Rω2 cosωt) i− (Rω2 sinωt) j

= −ω2[(R cosωt) i+ (R sinωt) j

]= −ω2~r.

So, the acceleration vector points back along the displacement vector, from thepoint, back to the origin. The magnitude of ~a = a = ω2R. But, v = ωR, and so

a = ω2R =

(v2

R2

)R =

v2

R,

(d) The net force acting on the particle is just ~F = m~a, and so

~F = m~a = −mω2[(R cosωt) i+ (R sinωt) j

],

which points along the acceleration (i.e., back towards the origin) with a magni-tude F = ma = mRω2 = .8× 4× (2π)2 = 126 N.

10

10. After a parachutist jumps from an airplane (but before he pulls the rip cord to open hisparachute), a downward speed of up to 180 km/h can be reached. When the parachuteis finally opened, the drag force is increased by about a factor of 10, and this can createa large jolt on the jumper. Suppose this jumper falls at 180 km/h before opening hischute.

(a) Determine the parachutist’s acceleration when the chute is just opened, assuminghis mass is 60 kg.

(b) If rapid accelerations greater than 5.0g can harm the structure of the human body,is this a safe practice?

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Solution

(a) From Newton’s laws, the sum of the forcesis his net acceleration. So, after his chuteopened,∑

Fy = Fd − Fg = 10bv2 −mg = maopen,

So

aopen =10bv2

m− g.

Before he opened the chute the drag forcewas only Fd = bv2. So,∑

Fy = Fd − Fg = bv2 −mg = maclosed.

If the parachutist was traveling at termi-nal velocity, vT , before he pulled the cord,then his acceleration aclosed = 0, and sob = mg/v2

T .

Applications of Newton’s Laws

521

Picture the Problem The free-body diagram shows the drag force dF exerted by the air and the gravitational force gF exerted by the Earth acting on on the parachutist just after his chute has opened. We can apply Newton’s 2nd law to the parachutist to obtain an expression for his acceleration as a function of his speed and then evaluate this expression for tvv = .

y

jbvF ˆ10 2d =r

jmgF ˆg −=r

(a) Apply yy maF =∑ to the parachutist immediately after the chute opens:

open chute210 mamgbv =−

Solving for open chutea yields:

gvmba −= 2

open chute 10 (1)

Before the chute opened:

ymamgbv =−2

Under terminal speed conditions, ay = 0 and:

02t =−mgbv ⇒ 2

tvg

mb=

Substitute for b/m in equation (1) to obtain:

gvv

gvv

ga

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

1110

110

22t

22t

open chute

Evaluating open chutea for v = vt yields:

ggvv

a 91110 2t2

topen chute =

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

(b) Because this acceleration exceeds the safe acceleration by 4g, this is not a safe practice. 135 • Find the location of the center of mass of the Earth–moon system relative to the center of Earth. Is it inside or outside the surface of Earth? Picture the Problem We can use the definition of the location of the center of mass of a system of particles to find the location of the center of mass of the Earth-moon system. The following pictorial representation is, of course, not shown to scale.

Plugging this back into the open acceleration equation gives

aopen =10bv2

m− g = g

(10

(v

vT

)2

− 1

).

So, his acceleration depends on how fast he was traveling before he pulled thechute, which makes sense. If he was traveling at terminal velocity, vT , then

aopen = g

(10

(vTvT

)2

− 1

)= g(10− 1) = 9g.

So, he’s subjected to an acceleration of 9g.

(b) This is 4g greater than what is needed to harm the body. So, this is not a verysafe practice (for a variety of reasons!).

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