Answers/Solutions: Giancoli Chapter 11

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Q1: The Earth's rotational axis is a line through the North and South poles. Moving people to the equator moves them farther away from that axis, which increases the rotational inertia. Since angular momentum must be conserved, an increase in I must be balanced by a decrease in \omega \ . So the day would get longer (rotation would slow down).




Q5: The turntable's rotational velocity would increase. Since there are no external torques acting, the angular momentum of the turntable+you system can't change. When you walk toward the center, the rotational inertia of the system decreases, thus \omega \ must increase.




Q8: Either vector could be zero, or the vectors could be parallel, or they could be anti-parallel (i.e., in opposite directions).




Q17: Independent of choice of origin: displacement, velocity, acceleration, momentum

Dependent on choice of origin: angular momentum, torque

Independent of velocity of coordinate system: displacement, acceleration, torque

Dependent on velocity of coordinate system: velocity, momentum, angular momentum (these all have a "v" in their definition, so if v changes so does the quantity!)




P2: Use L = I\omega \ . For a uniform cylinder rotating about its symmetry axis, I = (1/2)MR2. So since 1300 rpm = 136 rad/sec, the angular momentum is:

L = (1/2)(2.8 kg)(0.18)2(136 rad/s) = 6.17 kg-m2/s




P5: While the diver is in the air clearly no external torque acts about any axis through her CM. Thus angular momentum is conserved - comparing the "tuck" and the "straight" positions: LT = LS or:

IT\omega \ T = IS\omega \ S

We are given that IS = (3.5)IT and that \omega \ T = (2.0 revolutions)/(1.5 seconds) = 1.333 rev/sec. It is a simple matter to solve for \omega \ S. The answer is 0.381 rev/sec. (The problem requests the answer in these units rather than radians/sec which is the SI unit.)




P11:(a) (See Q5 above.) In the absence of rotational friction (from air resistance, and in the bearings supporting the MGR) there are no external torques and so the angular momentum of the MGR+Person cannot change. Therefore

Iinitial\omega \ initial = Ifinal\omega \ final

The rotational inertia I is the sum of IMGR and IP, but when the person is at the MGR center (i.e., the axis of rotation), he does not contribute to I, since IP = Mr2 and r = 0 at the center. So Iinitial = IMGR = 920 kg-m2.

When the person moves to the rim of the MGR, there is a contribution of IP = MR2, where R = 3.0 m is the radius of the MGR. Thus

Ifinal = IMGR + MR2 = 920 + (75)(3)2 = 1595 kg-m2.

Thus we have

(920 kg-m2)(0.95 rad/s) = (1595 kg-m2)\omega \ final

The answer is: \omega \ final = 0.548 rad/s

(b) (KE)initial = (1/2)Iinitial\omega \ initial2 = (1/2)(920)(0.95)2 = 415 Joules

KE)final = (1/2)Ifinal\omega \ final2 = (1/2)(1595)(0.548)2 = 241 Joules

What happened to the missing energy? The answer is that it was used to do work: a force was necessary in order to "speed up" the person as he moved from the center out to the rim. The force was provided by friction between the MGR surface and the person's shoes - if there were no friction she could not have gained the rotation necessary to stay at a fixed point on the MGR and speed up. (Think about it: in the complete absence of friction, she might have been able to "push away" - say from a post fixed at the MGR axis) and move outward, but he would move straight outward, not rotating with the MGR platform!) So: to do what she is described as doing, she had to gain speed (i.e., KE) and this required work to be done on her; the source of the work being the original KE of the platform, which then decreases.)




P20: Imagine (or better: draw on paper) a 3-dimensional (x,y,z) coordinate system. Choose the usual directions: +x axis to the right ("east"), +y axis up the page ("north"), +z-axis vertically upward perpendicular to the page ("skyward"). In each case below, we use the "right-hand rule:" using your right hand, straighten your fingers (not including your thumb) and orient them along the direction of the first vector in the cross-product. THEN rotate (if necessary) your wrist so that you can bend your fingers (without rotating anything else!) so as to align them with the second vector. Then you thumb, when straightened, points in the direction of the cross-product vector. Note that your thumb will always point in a direction perpendicular to each of the two original vectors.

So if A points along -x ("west") and B points along +z ("skyward):

(a) A\times \ B points "north" in the +y direction. (b) B\times \ A points "south" in the -y direction.

(Note that

B\times \ A = - A\times \ B

always, which means that reversing the order simply reverses the direction of the cross-product vector.)

(c) Since the two vectors point in opposite directions, the angle between them is 180\circ \ and since sin(180) = 0, the cross-product will be zero.




P24: Here we must use the determinant rule of Equation 11-3(b). The torque is given by

\tau \ = r\times \ F

where r is a vector from the origin to the point of application of the force. So we just "plug in" to Equation 11-3(b) where (i, j, k) is the top row, the components of r (4.0, 3.5, 6.0) are the middle row, and the components of F (9.0, 0, -4.0) are the third row, and work out the answer, which is

\tau \ = (- 68, + 16, + 36)

or, in unit vector notation:

- 68i + 16j + 36k

Nothing like this one will be on a test or final exam!




P34: The formula is

L = r\times \ p

where the vector r points from the origin to the particle.

(a) Draw a sketch like Figure 11-33 and connect point O with the particle m: this is r. Note the right-triangle, where r is the hypotenuse. The magnitude of angular momentum will be

L = (r)(p)sin(\theta \ )

where \theta \ is the angle between r and p (which is also the direction of v).

This gets wordy, but stay with me, referring frequently to your drawing: The angle \theta \ is an obtuse angle (greater than 90\circ \ and is "external" to the right-triangle you have constructed. But the sine of \theta \ is the same as the sine of the interior angle at m. This is because sin(180 - \theta \ ) = sin(\theta \ ) for any angle (so long as you use degrees, of course. - you must use \pi \ instead of 180 if \theta \ is in radians!).

Comment: It is worth using your calculator to test this for a couple of random cases. Assuming we are using degrees, verify that sin(180 - 40) = sin(140) = 0.6427876097... If you visualize the unit circle and consider that the sine function gives the height above the x-axis of the tracking point on the circle, it will become obvious why sin(180 - \theta \ ) is the same sin(\theta \ ).

BUT, looking at the diagram:

(r)sin(\theta \ ) = d

The conclusion is that

L = (p)(d) = mvd

(b) Now the vector r is anti-parallel to the vector p so the angle between them is 180\circ \ whose sine is zero. So L = 0 in this case.




P38: Refer to Figure 11-16 on p. 295, but also to Figure 10-57 on p. 279. This problem is essentially the same as #51 in Chapter 10, but with numbers. It is also essentially identical with Example 11-8 on p. 295.

The answers:

(a) a = 0.754 m/s2

(b) If the pulley is massless (I = 0) we get a = 0.774 m/s2. Thus

% error = (0.774 - 0.754)/0.754 = 0.026x100 = 2.6%




P47: The frictionless pivot tells us that there are no "external" torques on the rod+putty system; therefore angular momentum is conserved in the collision. The original angular momentum is all in the putty, since the rod is not moving. At the end, both rod and putty are rotating (the putty sticks to the rod's mid-point). We know that:

Initial L of system = Initial L of putty, which is (see #34(a) above) = mvo(d/2)

(where I use d for the length of the rod rather than the l - "ell" - used by Giancoli, because I don't have a "script l" character and the lower-case L in this typeface looks too much like the number 1 - "one" - as well as the capital I - "eye"!)

Final angular momentum = IF\omega F

We know that IF is the sum of IRod about one end and the contribution of the putty, which is a "point mass" a distance of d/2 from the axis. Thus, using case (g) of Figure 10-20 (p. 260) for IRod:

IF = (1/3)Md2 + m(d/2)2

so conservation of angular momentum requires that

[(1/3)Md2 + m(d/2)2]\omega \ F = mvo(d/2)

Solving, we obtain:

\omega \ F = mvo/[(2M/3 + m/2)d]

Now, AFTER the collision, mechanical energy is conserved. The initial (here "initial" means immediately after the collision) KE of the rotating rod+putty is converted to gravitational PE.

(1/2)IF\omega \ F2 = (M + m)gh

where h is the rise in the height of the CM (i.e., the center of the rod) when it momentarily stops. The lower end of the rod rises by twice this amount, so the answer will be 2h.

After working out the algebra (and factoring to simplify the denominator), I get:

2h = m2vo2/[g(M + m)(4M/3 + m)]




P67: This problem is very similar to #101 in Chapter 10. Examine Figure 5-23 on p. 126. Assume the SUV is turning to its left, as in the Figure, and is at the "tipping speed."

Under these conditions, the SUV's left-side tires will be just at the "lifting off" point and will have no force on them from the roadbed. The right-side tires will have a normal force FN and a frictional force (to the SUV's left side) Ffr. Gravity acts downward through the CM, which we assume is in the center (right-to-left) of the vehicle. The tire-to-tire width of the SUV is w, and h is the height of its CM above the road-bed.

\Sigma \ FHorizontal = Ffr = Mvc2R

\Sigma \ FVertical = FN - Mg = 0

And, summing torques about the CM with CCW as the positive rotational direction, and equating the net torque to zero (under borderline tipping conditions at the critical speed):

\Sigma \ \tau \ CM = (Ffr)(h) - (FN)(w/2) = 0

These are trivial to solve for vc, with the result:

vc = sqrt[gwR/(2h)]

(b) Solving the above expression for R, we get

R = 2hvc2/(gw)

And using the definition SSF = w/(2h), we can write this as

R = vc2/[(g)(SSF))

Finally, if (SSF)SUV = 1.05 and (SSF)Car = 1.40, then the minimum no-tipping turn radii at the same critical speed vc will be

RSUV/RCar = (SSF)SUV/(SSF)Car = 0.750