Home Work Problemsand Solutions: 1-6

1-1At the instant of Fig. 11-42, a 2.0kg particle P has a position vector ofmagnitude 3.0 m and angle θ1 = 45°and a velocity vector of magnitude4.0 m/s and angleθ2 30°. Force ofmagnitude 2.0 N and angle θ3=30°,acts on P. All three vectors lie in thexy plane. About the origin, whatare the (a) magnitude and (b) directionof the angular momentum of Pand the (c) magnitude and (d) directionof the torque acting on P? (HR 11-28)

Sol: We note that the component of perpendicular to has magnitude v sin  where = 30°. A similar observation applies to .

(a) Eq. 11-20 leads to

(b) Using the right-hand rule for vector products, we find points out of the page, or along the +z axis, perpendicular to the plane of the figure.

(d) Using the right-hand rule for vector products, we find is also out of the page, or along the +z axis, perpendicular to the plane of the figure.

1-2A track is mounted on a large wheel that is free to turnwith negligible friction about a vertical axis (Fig. 11-49). Atoy train of mass m is placed on the track and, with thesystem initially at rest, the train’s electrical power is turnedon.The train reaches speed 0.15 m/s with respect to the track.What is the angular speed of the wheel if its mass is 1.1m andits radius is 0.43 m? (Treat the wheel as a hoop, and neglectthe mass of the spokes andhub.) (HR 11-49)

Sol: No external torques act on the system consisting of the train and wheel, so the total angular momentum of the system (which is initially zero) remains zero. Let I = MR2 be the rotational inertia of the wheel. Its final angular momentum is

where is up in Fig. 11-47 and that last step (with the minus sign) is done in recognition that the wheel’s clockwise rotation implies a negative value for . The linear speed of a point on the track is R and the speed of the train (going counterclockwise in Fig. 11-47 with speed relative to an outside observer) is therefore where v is its speed relative to the tracks. Consequently, the angular momentum of the train is Conservation of angular momentum yields

When this equation is solved for the angular speed, the result is

2-3 In Fig. 12-42, a 55 kg rockclimber is in a lie-back climb along a fissure,with hands pulling on one side ofthe fissure and feet pressed against theopposite side.The fissure has width w0.20 m, and the center of mass ofthe climber is a horizontal distanced =0.40 m from the fissure.The coefficient of static frictionbetween hands and rock is μ1=0.40, and between boots and rockit isμ2 =1.2. (a) What is the leasthorizontal pull by the hands andpush by the feet that will keep theclimber stable? (b) For the horizontalpull of (a), what must bethe vertical distance h betweenhands and feet? If the climber encounterswet rock, so thatμ1 andμ2 are reduced, what happens to(c) the answer to (a) and (d) theanswer to (b)? (HR 12-26)

Sol: (a) The problem asks for the person’s pull (his force exerted on the rock) but since we are examining forces and torques on the person, we solve for the reaction force (exerted leftward on the hands by the rock). At that point, there is also an upward force of static friction on his hands f1 which we will take to be at its maximum value . We note that equilibrium of horizontal forces requires (the force exerted leftward on his feet); on this feet there is also an upward static friction force of magnitude 2FN2. Equilibrium of vertical forces gives

(b) Computing torques about the point where his feet come in contact with the rock, we find

(c) Both intuitively and mathematically (since both coefficients are in the denominator) we see from part (a) that would increase in such a case.

(d) As for part (b), it helps to plug part (a) into part (b) and simplify:

from which it becomes apparent that h should decrease if the coefficients decrease.

2-4In Fig. 12-50, a uniform plank, with a length L of 6.10 mand a weight of 445 N, rests on the ground and against a frictionlessroller at the top of a wall of height

h = 3.05 m. Theplank remains in equilibrium forany value of θ≧70° but slips if θ 70°. Find the coefficient of staticfriction between the plank andthe ground. (HR 12-37)

Sol: The free-body diagram on the right shows the forces acting on the plank. Since the roller is frictionless the force it exerts is normal to the plank and makes the angle  with the vertical. Its magnitude is designated F. W is the force of gravity; this force acts at the center of the plank, a distance L/2 from the point where the plank touches the floor. is the normal force of the floor and f is the force of friction. The distance from the foot of the plank to the wall is denoted by d. This quantity is not given directly but it can be computed using d = h/tan.

The equations of equilibrium are:

The point of contact between the plank and the roller was used as the origin for writing the torque equation.

When  = 70º the plank just begins to slip and f = sFN, where s is the coefficient of static friction. We want to use the equations of equilibrium to compute FN and f for  = 70º, then use s = f/FN to compute the coefficient of friction.

The second equation gives F = (W – FN)/cos  and this is substituted into the first to obtain

f = (W – FN) sin /cos  = (W – FN) tan .

This is substituted into the third equation and the result is solved for FN:

where we have use d = h/tan and multiplied both numerator and denominator by tan . We use the trigonometric identity 1+ tan2 = 1/cos2 and multiply both numerator and denominator by cos2 to obtain

Now we use this expression for FN in f = (W – FN) tan  to find the friction:

We substitute these expressions for f and FN into s = f/FN and obtain

Evaluating this expression for  = 70º, we obtain

3-1 In Fig. 15-31, two identicalsprings of spring constant7580 N/m are attached to ablock of mass 0.245 kg. What isthe frequency of oscillation onthe frictionless floor? (HR 15-13)

Sol:When displaced from equilibrium, the net force exerted by the springs is –2kx acting in a direction so as to return the block to its equilibrium position (x = 0). Since the acceleration , Newton’s second law yields

Substituting x = xm cos(t + ) and simplifying, we find

where  is in radians per unit time. Since there are 2 radians in a cycle, and frequency f measures cycles per second, we obtain

3-2In Fig. 15-36, twosprings are joined and connectedto a block of mass 0.245kg that is set oscillating over africtionless floor. The springseach have spring constant k = 6430 N/m.What is the frequency of the oscillations? (HR 15-26)

Sol:We wish to find the effective spring constant for the combination of springs shown in the figure. We do this by finding the magnitude F of the force exerted on the mass when the total elongation of the springs is x. Then keff = F/x. Suppose the left-hand spring is elongated by and the right-hand spring is elongated by xr. The left-hand spring exerts a force of magnitude on the right-hand spring and the right-hand spring exerts a force of magnitude kxr on the left-hand spring. By Newton’s third law these must be equal, so . The two elongations must be the same and the total elongation is twice the elongation of either spring: . The left-hand spring exerts a force on the block and its magnitude is . Thus . The block behaves as if it were subject to the force of a single spring, with spring constant k/2. To find the frequency of its motion replace keff in with k/2 to obtain

With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz.

3-3 For Eq. 15-45, suppose the amplitude xm is given by

where Fm is the (constant) amplitude of the external oscillatingforce exerted on the spring by a rigid support inFig. 15-15. At resonance, what are the (a) amplitude and(b) velocity amplitude of the oscillating object? (HR 15-61)

Sol:(a) We set  = d and find that the given expression reduces to xm = Fm/b at resonance.

(b) since the velocity amplitude vm = xm, at resonance, we have vm = Fm/b = Fm/b.

3-4 In Fig. 15-60, a solidcylinder attached to a horizontalspring (k = 3.00 N/m) rollswithout slipping along a horizontalsurface. If the system isreleased from rest when thespring is stretched by 0.250 m, find (a) the translational kineticenergy and (b) the rotational kinetic energy of the cylinder asit passes through the equilibrium position. (c) Show that underthese conditions the cylinder’s center of mass executessimple harmonic motion with periodT = 2π(3M/2k)1/2

where M is the cylinder mass. (Hint: Find the time derivativeof the total mechanical energy.) (HR 15-106)

Sol:(a) The potential energy at the turning point is equal (in the absence of friction) to the total kinetic energy (translational plus rotational) as it passes through the equilibrium position:

which leads to = 0.125 J. The translational kinetic energy is therefore .

(b) And the rotational kinetic energy is .

(c) In this part, we use vcm to denote the speed at any instant (and not just the maximum speed as we had done in the previous parts). Since the energy is constant, then

which leads to

Comparing with Eq. 15-8, we see that for this system. Since  = 2/T, we obtain the desired result: .

4-1 A uniform rope ofmass m and length L hangsfrom a ceiling. (a) Show thatthe speed of a transverse waveon the rope is a function of y, the distance from the lower end,and is given by v = (gy)1/2. (b) Show that the time a transversewave takes to travel the length of the rope is given by t = 2(L /g)1/2. (HR 16-25)

Sol: (a) The wave speed at any point on the rope is given by v = , where  is the tension at that point and  is the linear mass density. Because the rope is hanging the tension varies from point to point. Consider a point on the rope a distance y from the bottom end. The forces acting on it are the weight of the rope below it, pulling down, and the tension, pulling up. Since the rope is in equilibrium, these forces balance. The weight of the rope below is given by gy, so the tension is  = gy. The wave speed is

(b) The time dt for the wave to move past a length dy, a distance y from the bottom end, is and the total time for the wave to move the entire length of the rope is

4-2 The type of rubber band used inside some baseballs andgolf balls obeys Hooke’s law over a wide range of elongationof the band. A segment of this material has an unstretchedlength L and a mass m. When a force F is applied, the bandstretches an additional length ΔL . (a) What is the speed (interms of m, ΔL , and the spring constant k) of transverse waveson this stretched rubber band? (b) Using your answer to (a),show that the time required for a transverse pulse to travel thelength of the rubber band is proportional to 1/(ΔL)1/2ifΔL < L and is constant ifΔL>L.(HR 16-89)

Sol:(a) The wave speed is

(b) The time required is

Thus if , then and if , then

4-3 Underwater illusion. One clue used by your brain to determinethe direction of a source of sound is the time delayΔtbetween the arrival of the sound at the ear closer to the sourceand the arrival at the farther ear. Assume that the source isdistant so that a wavefront from it is approximately planarwhen it reaches you, and let D represent the separation betweenyour ears. (a) If the source is located at angleθin front ofyou (Fig. 17-31), what isΔt in terms of D and the speed ofsound v in air? (b) If you are submerged in water and thesound source is directly to your right, what is Δt in terms of Dand the speed of sound vw in water? (c) Based on the time-delayclue, your brain interprets the submerged sound to arriveat an angleθfrom the forwarddirection. Evaluateθfor freshwater at 20°C. (HR 17-12)

Sol:The key idea here is that the time delay is due to the distance d that each wavefront must travel to reach your left ear (L) after it reaches your right ear (R).

(a) From the figure, we find .

(b) Since the speed of sound in water is now , with , we have

Solving for with (see Table 17-1), we obtain

4-4 In Fig. 17-42, a French submarine and a U.S. submarinemove toward each other during maneuvers in motionlesswater in the North Atlantic. The French sub moves at speedvF= 50.00 km/h, and the U.S. sub at vUS= 70.00 km/h. TheFrench sub sends out a sonar signal (sound wave in water) at1.000 ×103 Hz. Sonar waves travel at 5470 km/h. (a) What isthe signal’s frequency as detected by the U.S. sub? (b) Whatfrequency is detected by the French sub in the signal reflectedback to it by the U.S. sub? (HR 17-61)

Sol:We denote the speed of the French submarine by u1 and that of the U.S. sub by u2.

(a) The frequency as detected by the U.S. sub is

(b) If the French sub were stationary, the frequency of the reflected wave would be fr = f1(v+u2)/(v – u2). { fdetected = fsource(v+u2)/v, freflected = fdetectedv/(v – u2) }

Since the French sub is moving towards the reflected signal with speed u1, then

5-1 The orbit of Earth around the Sun is almost circular:Theclosest and farthest distances are 1.47×108 km and 1.52 ×108km respectively. Determine the corresponding variations in(a) total energy, (b) gravitational potential energy, (c) kineticenergy, and (d) orbital speed. (Hint: Use conservation of energyand conservation of angular momentum.) (HR 13-87)

Sol: (a) The total energy is conserved, so there is no difference between its values at aphelion and perihelion.

(b) Since the change is small, we use differentials:

which yieldsU 1.8  1032 J. A more direct subtraction of the values of the potential energies leads to the same result.

(d) With KdK = mv dv, wherev 2R/T, we have

which yields a difference of v0.99 km/s in Earth’s speed (relative to the Sun) between aphelion and perihelion.

5-2 The fastest possible rate of rotation of a planet is that forwhich the gravitational force on material at the equator just barelyprovides the centripetal force needed for the rotation.(Why?) (a)Show that the corresponding shortest period of rotation isT =(3π/Gρ)1/2 where is the uniform density (mass per unit volume) of thespherical planet. (b) Calculate the rotation period assuming adensity of 3.0 g/cm3, typical of many planets, satellites, andasteroids. No astronomical object has ever been found to bespinning with a period shorter than that determined by thisanalysis.(HR 13-90)

Sol:If the angular velocity were any greater, loose objects on the surface would not go around with the planet but would travel out into space.

(a) The magnitude of the gravitational force exerted by the planet on an object of mass m at its surface is given byF = GmM / R2, where M is the mass of the planet and R is its radius. According to Newton’s second law this must equalmv2 / R, where v is the speed of the object. Thus,

Replacing M with (4/3) R3(whereis the density of the planet) andv with 2R/T(where T is the period of revolution), we find

We solve for T and obtain

(b) The density is 3.0  103 kg/m3. We evaluate the equation for T:

5-3 Several planets (Jupiter, Saturn, Uranus) are encircled byrings, perhaps composed of material that failed to form a satellite.In addition, many galaxies contain ring-like structures.Consider a homogeneous thin ring of mass Mand outer radiusR (Fig. 13-55). (a) What gravitationalattraction does it exert on aparticle of mass m located on thering’s central axis a distance x fromthe ring center? (b) Suppose theparticle falls from rest as a result ofthe attraction of the ring of matter.What is the speed with which itpasses through the center of thering? (HR 13-99)

Sol:(a) All points on the ring are the same distance (r = ) from the particle, so the gravitational potential energy is simply U = –GMm/. The corresponding force (by symmetry) is expected to be along the x axis, so we take a (negative) derivative of U (with respect to x) to obtain it. The result for the magnitude of the force is GMmx(x2 + R2)3/2.

(b) Using our expression for U, then the magnitude of the loss in potential energy as the particle falls to the center is GMm(1/R1/). This must “turn into” kinetic energy ( mv2 ), so we solve for the speed and obtain

v = [2GM(R1 – (R2 + x2)1/2)]1/2 .

5-4 A certain triple-star systemconsists of two stars, each of massm, revolving in the same circular orbitof radius r around a central starof mass M(Fig. 13-54).The two orbitingstars are always at opposite endsof a diameter of the orbit. Derivean expression for the period of revolutionof the stars.(HR 13-93)

Sol:The magnitude of the net gravitational force on one of the smaller stars (of mass m) is

This supplies the centripetal force needed for the motion of the star:

Plugging in for speed v, we arrive at an equation for period T:

6-1 In Fig. 14-37, waterstands at depth D = 35.0 m behindthe vertical upstream faceof a dam of width W = 314 m.Find (a) the net horizontalforce on the dam from thegauge pressure of the waterand (b) the net torque due tothat force about a line through O parallel to the width of thedam. (c) Find the moment arm of this torque. (HRW14-24)

Sol:(a) At depth y the gauge pressure of the water is p = gy, where  is the density of the water. We consider a horizontal strip of width W at depth y, with (vertical) thickness dy, across the dam. Its area is dA = W dy and the force it exerts on the dam is dF = p dA = gyW dy. The total force of the water on the dam is

(b) Again we consider the strip of water at depth y. Its moment arm for the torque it exerts about O is D – y so the torque it exerts is

d = dF(D – y) = gyW (D – y)dy

and the total torque of the water is

14-2 Figure 14-53 shows astream of water flowingthrough a hole at depth h = 10cm in a tank holding water toheight H = 40 cm. (a) At whatdistance x does the streamstrike the floor? (b) At whatdepth should a second hole bemade to give the same value ofx? (c) At what depth should ahole be made to maximize x?(HRW14-65)

Sol:(a) Since Sample Problem 14-8 deals with a similar situation, we use the final equation from it:

The stream of water emerges horizontally (0 = 0° in the notation of Chapter 4), and setting y – y0 = –(H – h) in Eq. 4-22, we obtain the “time-of-flight”

Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find