Answers/Solutions: Giancoli Chapter 10

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Q1: The odometer counts revolutions, and the bicycle moves a distance of 2\pi \ R each turn. The "27 inch" refers to the wheel's diameter, so R27 = 13.5 inches and R24 = 12 inches. The ratio is R24/R27 = 0.889. So, if the odometer is calibrated for 27-inch wheels and you use it with 24-inch wheels, the bike has in fact traveled only 0.889 of what the odometer reads.

Q4: Of course, provided that the smaller force is applied sufficiently far from the rotational axis.

Q7: No: two equal but opposite forces can produce a torque if they are not co-linear.

No: An object can certainly be accelerated by several forces without being rotated if the net force acts along a line through the rotational axis.

Q9: Measure their rotational inertia: apply a torque and observe the resulting angular acceleration about a diameter. The hollow shell (of the same mass) will have a greater rotational inertia, and therefore a smaller angular acceleration for a given torque.

Q14: Greater. In fact, one can easily figure out the rotational inertia about this new axis using the parallel axis theorem (Equation 10-17 on p. 264). Since all quantities in this equation are positive, we can conclude that the rotational inertia about an axis through the CM (in a given direction) is always the least possible for any axis parallel to that direction.

P3: Approximate the diameter of the spot on the Moon as an arc-length of a circle of radius D. Then D = R\theta \ (with \theta \ in radians). The answer is D = 5320 m.

P6: The wheel radius is 0.34 m, and one rotation moves the bicycle forward a distance 2\pi \ R = 2.136 m. So in traveling 7200 m, the wheel rotates (7200 m)/(2.136 m/rotation) = 3371 rotations.

P9: All three points have the same angular velocity, \omega \ = 2\pi \ radians/day (or 7.272x10-5 radians/sec). Therefore if the point moves in a circle of radius R, the total velocity will be v = R\omega \ .

(a) at the Earth's equator, R = RE = 6.38x106 m, so v = 464 m/s.

(b) at 66.5 degrees latitude (N or S) the point moves in a circle whose radius is REcos(66.5) = 2.544x106 m. Thus v = 185 m/s.

(c) at 45 degrees latitude the point moves in a circle of radius REcos(45) = 4.490x106 so v = 327 m/s

At the north pole, Santa Claus moves in a circle of radius REcos(90) = 0, so his linear velocity is 0.

P13: Over a time interval of 720 seconds, we have \omega \ o = 0, \omega \ f = (1.0 rev/min)(2\pi \ rad/rev)(1 min/60 sec) = 0.105 rad/sec.

(a) So the angular acceleration \alpha \ is given by \Delta \ \omega \ /\Delta \ t = 1.45x10-4 rad/s2.

(b) A point on the ship's skin moves in a circle of radius 4.25 m. The tangential component of the linear acceleration is

aT = R\alpha \ = (4.25)(1.45x10-4 = 6.16x10-4 m/s2

The radial component of the linear acceleration is our old friend v2/R, but v = R\omega \ , so we can write aR = R\omega \ 2. We need a value for \omega at a time 420 seconds after the start of the rotation:

\omega \ = \omega \ o + \alpha \ t = 0.0609 rad/sec

at which time

aR = R\omega \ 2 = 0.0156 m/s2

P19: We have \omega \ o = 850 rev/min = 89.0 rad/sec and \omega \ f = 0. During this time the wheel slows to a stop, it rotates through 2700\pi \ radians.

(a) We can obtain \alpha \ using Eqn 10-9(c): \omega \ f2 = \omega \ o2 + 2\alpha \ (\Delta \ \theta \ )2. The answer is \alpha = - 0.467 rad/s2. (The minus denotes that the rotation is slowing down.)

(b) There are several equations that will give us the time required for the wheel to stop. The quickest is probably Eqn. 10-9(a), which gives (since \omega \ f = 0):

\Delta \ t = - \omega \ o/\alpha \ = 191 seconds.

P25: There are three applied forces and therefore three torques to add. The axis is through the center of the wheel perpendicular to the page. To compute torque we can use any of Equations 10-10 on page 257. For each force, the "R" vector goes from the axis to the point where the force is applied, and for all three forces in this problem, R is perpendicular to F [and therefore in Eqn. 10-10c the angle sin(\theta \ ) = sin(90\circ \ ) = 1].

We arbitrarily choose CCW to be the positive direction for rotation. Then

\tau \ 28 = (28 N)(0.24 m) = 6.72 N-m

\tau \ 35 = - (35 N)(0.12 m) = - 4.20 N-m

\tau \ 18 = - (18 N)(0.24 m) = - 4.32 N-m

So the net applied torque is 1.8 N-m CW. The frictional torque, then is 0.40 N-m, and will act CCW to oppose the CW rotation. The final net torque is, then, \tau \ = 1.4 N-m CW.

P30(a): The axis is at C perpendicular to the page. For each force, the vector R is from C to the point where the force is applied. We choose CCW as positive rotational direction. Then

\tau \ 56 = - (56 N)(1.0 m)sin(30\circ \ ) = - 28 N-m

\tau \ 65 = (65 N)(0 m) = 0

\tau \ 52 = (52 N)(1.0 m)sin(30\circ \ ) = +45 N-m

So \tau \ Net = 17.0 N-m CCW.

(b) Do this for yourself! The answer is 10.0 N-m CW.

P40: The moment of inertia (or "rotational inertia") is defined as \Sigma \ MiRi2. Here we have four point masses, each a known distance from the rotational axis.

(a) About the vertical axis shown:

Iy-axis = (m)(0.50)2 + (M)(0.50)2 + (m)(1.00)2 + (M)(1.00)2 = 1.25m + 1.25M = (1.25)(2.2) + (1.25)(3.1) = 6.625 kg-m2

(b) About the horizontal axis shown, the answer is Ix-axis = 0.66 kg-m2.

(c) Since \tau \ Net = I\alpha \ , the torque required for a specified angular acceleration is larger if I is larger. Here, Iy-axis is almost ten times Ix-axis. Thus, to have the same angular acceleration, we would need almost ten times as much torque about the y-axis as about the x-axis. Clearly it is "harder" to accelerate about the y-axis in case (a).

P46: Draw free-body diagrams for both blocks and for the pulley (as in Figure 10-21) There is no sliding friction, and therefore I will not write the equations for the forces perpendicular to the inclines, since we are not interested in the normal forces. The relevant forces are:

Gravity on MA: MAg downward

Tension (in the rope) on MA = FTA

Gravity on MB: Mbg downward

Tension (in the rope) on MB = FTB

Tension (in the rope) CCW on pulley = FTA, resulting torque \tau \ CCW = FTAR

Tension (in the rope) CW on pulley = FTB, resulting torque \tau \ CW = FTBR

The equations of motion that result are:

FTA - MAg sin(32\circ \ ) = MAa

- FTB + MBg sin(61\circ \ ) = MBa

FTBR - FTAR = I\alpha \ = I(a/R)

where I have been consistent with signs: "down-slope" is positive for MB, "up-slope" is positive for MA and CW is positive for rotation of the pulley. Thus, as the blocks slide "toward the right" the pulley rotates CW, and a is positive.

We are given R for the pulley, along with (on the Figure) the masses MA and MB, and the acceleration a. The unknowns in the three equations are FTA, FTB, and I, the rotational inertia of the pulley. The answers are:

FTA = 49.6 N

FTB = 75.7 N

I = 0.59 kg-m2

P51: This is almost the same problem as P46, but simpler since no angles are involved. We will get three equations: vertical motion of MA, vertical motion of MB, and rotational motion of pulley. You should draw free-body diagrams for all three objects. The equations of motion are:

FTA - MAg = MAa

MBg - FTB = MBa

FTBR - FTAR = I\alpha \ = I(a/R)

The unknowns are FTA, FTB, and a. The answers will contain the given quantities: the masses MA and MB. the rotational inertia I and the radius R of the pulley. The two forces are not requested and thus solving the equations involves eliminating them and solving for a. The answer(s) are on p. A-24 of Giancoli.

P58:(a) The rotational inertia of a point-mass M a distance Ro from the rotational axis AB is IAB = MRo2.

(b) The rotational inertia of a solid sphere of mass M and radius r1 rotating about a diameter is (from Figure 10-20 on p. 260):

ICM = (2/5)Mr12.

Then, by the parallel axis theorem, the rotational inertia about the axis AB is:

IAB = ICM + MRo2 = (2/5)Mr12 + MRo2

(c) The fractional error incurred in using the expression in part (a) would be:

fractional error = (Ib - Ia)/Ib = (2/5)MRo2/[(2/5)Mr12 + MRo2]

The mass divides out and the expression simplifies to

fractional error = 1/[1 + (5/2)(Ro/r1)2]

Using the values Ro = 1.0 m and r1 = 0.09 m, we get a fractional error of 0.0032 or a percent error of 0.32%.

P65: The amount of work on the MGR is, of course, the increase in its KE. Since it starts at rest, this is just the amount of the MGR's KE at the end. So

W = KE = (1/2)I\omega \ 2

For a "solid cylinder" rotating about its axis of symmetry (see Figure 10-20, p.260), I = (1/2)MR2. The rotational velocity given is 1.00 rev/8 seconds = \pi \ /4 radians/sec. Using the values given for M and R, the answer is

W = 14,226 Joules

P67: This is the same physical system as P51, but different information is given and a different quantity required. It is also much easier because we are going to use conservation of energy.

Let the original energy be zero: nothing is moving, and we can set the zeros of gravitational PE at the starting heights of each mass (this means there is a different zero for each mass; this is OK).

After mass B has fallen a distance of 2.5 m, mass A has raised by this same height and they both have acquired KE. The pulley has also acquired KE. Conservation of energy requires that the total energy is still zero:

MAg(+2.5) + MBg(- 2.5) + (1/2)MAv2 + (1/2)MBv2 + (1/2)Ip\omega \ 2 = 0

Also, we know that

\omega \ = v/R

and, for a "uniform cylinder" rotating about its axis of symmetry

Ip = (1/2)MpR2

We can plug these into the energy equation and solve for v. The answer is

v = 1.40 m/s

P73: We are given values for ro, m, and the length and angle of the incline. We will use conservation of energy and the problem will be simple. The total vertical distance the sphere drops is h = (10.0 m)sin(30\circ \ ) = 5.0 m. Set the zero of gravitational PE at the bottom of the ramp. Then initial energy is mgh and the final energy is all KE (translational + rotational):

mgh = (1.2)mv2 + (1/2)I\omega \ 2

and we know (from Figure 10-20 on p. 260) that

I = (2/5)mr2

and we use the fact that

\omega \ = v/r

(a) Substitute these into the energy equation and solve for v. Note that both the mass m and the radius r of the sphere drop out. The answer is v = sqrt(10gh/7), and since h = 5.0 m

v = 8.37 m/s (independent of both m and r)

Then \omega \ = v/r = 34.2 rad/s (here r does matter)

(b) The ratio (KE)trans/(KE)rot = 5/2 (again, independent of m and r)

(c) As noted in the above answers, the answers for v and for the ratio of KEs do not depend on either m or r. The value for \omega \ does depend on r, but not on m.

P86: If the acceleration is 1.00 m/s2 and the starting speed is zero, then after 2.5 seconds, the speed of the axle must be

v = 0 + (1.00 m/s2)(2.5 s) = 2.5 m/s

This is the speed of the axle over the ground. But from the viewpoint of the axle, the ground is moving to the left at this speed, and a point at the top of the wheel is moving to the right at this same speed. To the speed of the top-most point is moving over the ground at a speed of 5.0 m/s.

P89: Let the distance between teeth on both front and rear sprockets be d. The circumferences are, then (using subscript F for "Front" and B for "Back"):

CF = 2\pi \ RF = NFd

CB = 2\pi \ RB = NBd

from which we obtain:

RF = NFd/(2\pi \ )

RB = NBd/(2\pi \ )

Now, for the two sprockets: the linear speed of the chain (call it v) around the rim of the sprockets must be the same. Thus vF = vB. But we know that, for circular motion, v = R\omega \ . Thus we arrive at:

RF\omega F = RB\omega B

which, using the equations above for RF and RB, leads immediately to:

\omega \ B/\omega \ F = NF/NB

which answers part (a). The answers for (b) and (c) then are 4 and 1.5 respectively.

P93: A Challenge! As the wheel hits the step, the "corner" of the rise (where the wheel contacts the step), becomes an axis of rotation. As the wheel lifts off the bottom surface, only two forces can exert torque about this axis: F and the gravitational force, both of which act through the CM, but in different directions.

The torque exerted by F about the axis of rotation is F(R - h) in a CW direction.

The torque exerted by gravity is CCW about the axis of rotation, and equals Mg(d) where d is the distance from the lift-off point on the bottom surface to the axis of rotation. A little geometry (involving use of the Pythagorean Theorem) should convince you that d = sqrt[R2 - (R - h)2].

The minimum force needed is the one that makes the net torque zero. So, calling CW rotation positive

F(R - h) - (Mg)sqrt[R2 - (R - h)2] = 0

which can trivially be solved for F in terms of the other quantities (M, R, h, and g). The answer is in the textbook on page A-25.

P101: Comment: Another Challenge! Also, this problem illustrates why most of the problems we do are so "unrealistic:" realistic problems can be quite complicated!

(a) Take a horizontal line through the CM (perpendicular to page) as the axis of rotation. As the front wheel lifts off, only three forces act: gravity (Mg downward through CM), the normal force (FN upward at rear wheel point of contact with ground) and the frictional force (Ffr to the right at point of rear wheel contact with ground). The frictional and normal forces (but not gravity) provide torques about the rotational axis through the CM.

We apply Newton's laws of motion: we write an equation for the x-acceleration (a; the sought-for unknown), another equation for the y-acceleration (assumed to be zero), and one for rotation about the axis. Take CCW as positive rotational direction. Then the condition for lift-off of the front wheel (normal force on it is zero) is that the total torque produced by these two forces about the CM just equals zero. (Before lift-off the torque due to just these two forces was negative, but the normal force on the front wheel caused a balancing positive torque. A net torque - either CCW or CW - would cause a rotation. For the condition of "just barely" lift off of the front wheel, we need a net torque of zero, provided by just the two forces that remain.)

\Sigma \ Fx = Ffr = Ma

\Sigma \ Fy = FN - Mg = 0

\Sigma \ \tau \ CM = Ffrx - FNy = 0

Note that x and y are considered "given" here. The unknowns are Ffr, FN, and a. Solving is easy:

a = (x/y)g

(b) To get small value of a we want a small x and a large y. The rider should position himself as high and as far to the rear as possible for the smallest possible a that will provide lift-off of the front wheel.

(c)a = (x/y)g = (35/95)(9.8) = 3.6 m/s2