Up to now we have considered acceleration to be only a change in speed. This, however, is not a comprehensive description, as accelerations defined as a rate of change in velocity. Velocity designates both a speed and a direction, so a car rounding a curve at constant speed but changing direction, should be considered accelerating. But should it?

As good, skeptical physicists, we must ask ourselves if this makes sense. It's fine to come up with a result by mathematics, but we must always return to physical reality and see if it means anything. Here's how we'll check: recall that F = ma. If there is an acceleration, there must be a force and vice-versa. Now think about this: when you round a corner in a car is the force applied to the car? Of course. If it weren't, the car would slip right off the curve.

Here's another example. If you tie a rock to a string and swing it around in a horizontal circle, you have to pull the rock always to the center (i.e. you must keep a tension in the string). So the mathematics does come up with a meaningful answer. To move an object at constant speed in a circle does require a force, and it must be directed radially inward to the center of the circle.

Now we must look back at the mathematics to find the magnitude of the acceleration. Consider a mass traveling in a circle of radius r at a speed v.

Look at the mass's position at two times separated by an interval t:

The mass will have traveled a distance s = v t, so the string will have covered an angle Ø = s/r = vÇt/r.

Now concentrate on the two velocity vectors, v₁ and v₂, so that their difference, v, can be determined:

For very small angles (which will occur in very small t's), v will approximate a piece of arc length whose radius is v. We may say, then, that v = vq is the magnitude of the velocity change.

But what is Ø? Ø = vÇt/r
If we combine this with our last equation, Çv = vØ = v(vÇt/r) = v²Çt/r
But acceleration is the change in velocity per time, a = Çv/Çt = (v²Çt/r)/Çt = v²/r

This is called the centripetal acceleration. It is inwardly directed.

By Newton's second law, then, the centripetal force is directed toward the center, and has a magnitude F_c = mv²/r.

Here it is necessary to distinguish the centripetal force from the centrifugal force. The centripetal force is REAL, and it acts toward the center. The centrifugal force is not real and it seems to act outward, away from the center, when you're riding in the coordinate system of the moving object.

Here's an example. As you sit in a car the car pushes you toward the center of its turn. If you didn't get pushed toward the center (as when the door suddenly flies open), you would move in a straight line while the car continued to turn. While the door presses you toward the center of the circle, you feel pressed outwardly against the door. It's just like what happens when the car accelerates from rest. You feel pressed into the seat while in reality the seat is pressing into you and pushing you forward.

Think about these forces the next time you change your direction, whether you're in a car, on a bicycle, a skateboard, or on foot. You will begin to perceive reality.

1. Draw free body diagrams for each mass. That is, show the forces that are acting on the mass.
a) A ball on a string swings in a circle in space.
b) Ardy Choke rides his bike in a circle.

a) note: the force is directed toward the center

b) note: the vertical forces balance, and the unbalanced force points inward

2. Penny Antee twirls a ball in a vertical circle. Diagram the forces at points a, b, and c.

a) mg + T = mv2/r

b) T - mg = mv2/r

c) T = mv2/r note: mg will slow the ball

3. A mass on a string swings in a horizontal circle. Draw the forces.

note: T must be large enough to balance mg vertically and to supply mv2/r centrally

4. Diagram the forces in each case:
a) A car rounds a corner with friction on its wheels.
b)A car slides around a banked corner without friction.

a) vertical forces balance while friction exerts the inward force, F = mv2/r

b) vertical forces balance [mg = NcosØ] horizontal component of N is centripetal force [NsinØ = mv2/r]

5. A 0.35 kg model plane travels at 7.0 m/sec on the end of a 3.1 meter wire. What is the tension in the wire?

FC = mv2/r FC = [(.35)(7)2]/(3.1) FC = 5.53 nt

6. A 0.8 kg rock is spun in a circle on a 1.3 m string. If the string breaks at 12 nt tension, how fast must the rock be moving? (Neglect the effects of gravity for the moment.)

FC = mv2/r v2 = Fr/m v2 = [(.12)(1.3)]/(.8) v = 4.42 m/sec

7. A 5200 lb car rounds a curve of 130 ft radius at 61 ft/sec. What force must the road exert on its tires to hold the car on course?

F = mv2/r F = [(5200/32)(61)2]/(130) F = 4.65 * 103 lb

8. A centrifuge spins a test tube in a circle of 18 cm radius. If it spins at 11 revolutions/sec, what force is exerted on a 2.7 gm particle in the tube?

First, the velocity: v = s/t v = [(2)(3.14)r]/t v = [(2)(3.14)(18 cm)]/(1/11 sec) v = 1244 cm/sec F = mv2/r F = [(2.7)(1244)2]/(18) F = 2.32 * 105 dyne

9. A 1.3 gm cockroach stands on the edge of a 12 inch diameter record. If the record turns at 78 rev/min, what coefficient of friction is needed by the roach to keep it from slipping off?

Force required is FC = mv2/r Maximum force friction can supply is µmg thus: µmg = mv2/r but: v = rw v = [6 * (2.54 cm/1 in)][(78 * 2(3.14) rad)/(60 sec)] v = 124.5 cm/sec µ = v2/rg µ = (124.52)/[(6 * (2.54 cm/1 in))(980)] µ = 1.04 note: this is a high µ value

10. Luke Autbeloe twirls a 750 gram mass on the end of a 0.82 meter string. Find the tension at the top and bottom of the orbit. Assume the mass's velocity at both top and bottom is 3.7 m/sec.

At both the top and bottom, there must be a net inward force of mv2/r mg + T1 = mv2/r T1 = mv2/r - mg T1 = [(.75)(3.7)2]/(.82) - [(.72)(9.8)] T1 = 5.17 nt T2 - mg = mv2/r T2 = mv2/r + mg T2 = 19.9 nt

11. Mary Derly drives her 2400 lb Porsche at 76 ft/sec around a curve of 285 ft radius. At what angle should the road be banked to give Mary the most comfortable ride?

tanØ = [mv2/r]/(mg) tanØ = v2/rg tanØ = (762)/[(285)(32)] Ø = 32.3 note: the sum of the two forces acting must be an inwardly directed force, mv2/r

12. If an engineer banks a curve of 47 m radius at 12°, what does she anticipate a typical speed to be for comfortable riding?

By the logic and diagrams of problem #11: tanØ = v2/rg 1/2 v = [(4.7)(9.8)tan(12)]1/2 v = 9.89 m/sec

We interrupt this text for an important technological announcement: The modern washing machine uses the idea of centripetal force to do a preliminary spin-dry on clothes. By getting clothes and water moving fast and simultaneously yanking the clothes toward the center of the washing machine, the water is left to go straight. (Did you ever notice those little holes around the basket?) Accelerating the clothes away from the water would work linearly. A small rocket-mounted basket in the basement could be fired off, leaving the water at the point of ignition.

13. A boy swings a rock on a string in a horizontal circle. If he swings it so that it makes a 25° angle with the horizontal, and it is held by a 1.4 m piece of string, how fast must it be moving? (Think: the radius of the rock's orbit is NOT 1.4 meters.)

V) mg = Tsin(25) (vertical forces balance) H) Tcos(25) = mv2/r (the net inward force) But r = 1.4cos(25) (actual radius of orbit) tan(25) = [Tsin(25)]/[Tcos(25)] tan(25) = mg/[(mv2)/(1.4cos(25))] tan(25) = [g(1.4cos(25))]/v2 v = [(9.8 * 1.4cos(25))/(tan(25))]1/2 v = 5.16 m/sec

14. A 0.5 kg mass on a 3.2 meter string moves at 2.5 rad/sec in a horizontal circle. What angle does the string make with the vertical?

Note: mv2/r = m(wr)2/r = mw2r tanØ = mw2r/mg tanØ = w2r/g but: r = 3.2sinØ tanØ = w2r/g tanØ = [(2.5)2(3.2sinØ)]/(9.8) divide both sides by sinØ: 1/cosØ = [(2.52)(3.2)]/(9.8) Ø = 60.7°

15. Jack D. Ripper wishes to do a loop-the-loop on his Harley-Davidson. If the loop is 12 feet in diameter and Jack plus motorcycle masses 35 slugs, what minimum speed must he attain to complete this feat?

Since no other force acts, mg is the centripetal force. mv2/r = mg v = (rg)1/2 v = [(6)(32)]1/2 v = 13.9 ft/sec

16. A pendulum of length L and mass M is released at angle q from the vertical. What is the tension in the string when it reaches the bottom of its swing?

PE = KE mgh = (1/2)mv2 v2 = 2gh v2 = 2g(L - LcosØ) T - mg = mv2/r T = mg + [(m * 2gL(1 - cosØ))/L] T = mg[1 + 2(1 - cosØ)] T = mg(3 - 2cosØ)

17. For years NASA has been salivating over the possibility of a large permanent space station. In order to simulate gravity, the station is to be shaped like a huge Quaker Oats carton and made to rotate about its central axis.
a) Assume a radius of 322 m for the station. What rotational speed w would be required to simulate gravity?
b) If a man jogged in the direction of rotation at 5 m/sec, what "g" would he feel?

a) centripetal acceleration = g g = w2r w = (g/r)1/2 w = [(9.8)/(322)]1/2 w = .174 rad/sec

b) Linear speed of rim: v = wr v = (.174)(322) v = 56.2 m/sec An additional 5 m/sec brings this to: v2 = 56.2 + 5 v2 = 61.2 m/sec so "g" becomes: v2/r = (61.2)2/(322) "g" = 11.6 m/sec2

18* An L-shaped support rotates at w, pulling the mass m toward its axis.
a) Express Ø in terms of m, L, w, g, and A.
b) Find the total energy of the mass. (KE + PE)

a) r = A + LsinØ tanØ = mw2r/mg tanØ = w2r/g tanØ = w2(A + LsinØ)/g w = [(gtanØ)/(A + LsinØ)]1/2

b) E = KE + PE E = (1/2)mv2 + mgh E = (1/2)mw2r2 + mg(L - LcosØ) E = (1/2)m[(gtanØ)/(A + LsinØ)] (A + LsinØ)2 + mgL(1 - cosØ) E = mg[(tanØ/2)(A + LsinØ) + L(i - cosØ)]

19. A 0.2 kg mass swings on a 1.4 m string at 4.1 m/sec in a vertical circle. What tension is there in the string at the top of the swing? What tension at the bottom?

At the top: mg + T = mv2/r T = mv2/r - mg T = [(.2)(4.1)2]/(1.4) - (.2)(9.8) T = .441 nt At the bottom: T - mg = mv2/r T = mg + mv2/r T = (.2)(9.8) + [(.2)(4.1)2]/(1.4) T = 4.36 nt

20. Famed stunt pilot, Cleon da Dishes, pulls out rapidly from a dive. If he is traveling at 215 mi/hr at the bottom of his trajectory, and at that moment is traveling on a curve of 752 ft radius, what acceleration does he feel?

Convert units: (215 mi/hr)(5280 ft/1 mi)(1 hr/60 min)(1 min/60 sec) = 315 ft/sec ac = Fc/m ac = (mv2/r)/m ac = v2/r ac = (3152)/(752) ac = 132 ft/sec2 afelt = ac + g afelt = 132 + 32 afelt = 164 ft/sec2 note: (164)/(32)= 5.1 g's

21. A mass m is on a spring of length L and stiffness constant k. When spun out in deep space at angular speed w, how far will the spring stretch?

Fc = Fs mw2r = kx mw2(L + x) = kx x = (mw2L)/(k - mw2)

Now we'll look at circular motion from a different angle. Before we do this it will be necessary to define a new quantity, the angular acceleration.

Recall how in the past we've used the expression s = rØ to determine the arc length of a circular path. Then we divided both sides by t to find that v = rw. Note that the first equation relates linear distance to angular distance. The second relates linear velocity to angular velocity. Our third equation will relate linear acceleration to angular acceleration. Since v = rw, we may say that for a changing angular velocity, Çv/Çt = rÇw/Çt or:

NOTE: CHANGE SYMBOL
a = ra where a = Çv/Çt as before, and a = Çw/Çt

Alpha, a, is the angular acceleration, which has units rad/sec². We find it in situations like a circular saw starting up from rest, a phono turntable slowing to a stop, or a ball rolling down a hill.

Now we have a set of angular equations that are mathematically analogous to our set of linear kinematic equations:

linear	angular
s = vt	Ø = wt
v = at	w = at
s = 1/2 at2	Ø = 1/2 at2
v2 = 2as	w2 = 2aØ

All the problems that were asked in the linear kinematics section may now be rephrased in angular terms. Instead of engaging in this painful endeavor, however, let us use this new-found revelation to uncover a new Truth about Nature.

Recall the rotational analog to force .... TORQUE! Remember further that torque is defined as t = Fr, force times lever arm.

Now we will try to generate a rotational analog to everyone's favorite relationship, F = ma. Imagine the mass on a string, tied to some pivot point.

We will apply a force to it, and observe a resulting linear acceleration a. Simply plug in the rotational analogs:

The linear relationship is F = ma, but a = ra and F = t/r
so we get (t/r) = m(ar)
which we may express as t = (mr²)a.

Note the similarity between this equation and F = ma! Torque is a rotational force, angular acceleration is a rotational acceleration, and so mr² must be a rotational mass. That is, it is the quantity that resists acceleration, in this case, angular acceleration.

Just as an astronaut can discover the mass of a weightless object by shaking it and sensing its resistance to acceleration, you may find the mr² of an object by trying to twist it back and forth rapidly.

Notice that it's not simply the mass of an object that will cause it to resist angular acceleration. It is also the distribution of mass. Thus a wheel that concentrates its mass on the rim, at large r, has a greater mr² than a uniform disk.

Our word for mr² is moment of inertia or rotational inertia. It is symbolized by I, so we get t = Ia.

We have derived our equation for a mass entirely concentrated at a radius r, and for things like hoops, dumbbells and bicycle wheels this is true to a high degree of accuracy. Try some problems:

22. A torque of 11 nt-m is applied to a 3 kg bicycle wheel whose radius is 0.35 m. What is the angular acceleration of the wheel?

I = mr2 I = (3 kg)(.35 m)2 I = .368 kg-m2 t = Ia a = t/I a = (11 nt-m)/(.368 kg-m2) a = 30.0 rad/sec2

23. Two 5 kg masses are placed on the ends of a massless stick, 3 meters long. The dumbbell is pivoted about its center, and a force of 7 nt is applied to one mass. What is the resulting angular acceleration?

t = F * r t = (7 nt)(1.5 m) t = 10.5 nt-m a = t/I a = 10.5/22.5 a = .467 rad/sec2

24. A 1.2 kg hoop, spinning in a pool of water, gradually slows to a stop. If its acceleration is -0.4 rad/sec² and its radius is 0.3 m, what torque is the water applying to the hoop?

I = mr2 I = (1.2)(.3)2 t = Ia t = [(1.2)(.3)2](-.4) t = -.0432 nt-m

25. A 4 kg bicycle wheel of radius 0.2 m is held on a fixed support, while a 1.1 kg mass on a string falls as shown. What is the acceleration of the mass?

Two equations are needed. Translational: F = ma (1.1 * 9.8) - T = (1.1)a Rotational: t = Ia t(.2) = [(4)(.2)2][a/.2] Solve each equation for T and equate. (1.1 * 9.8) - (1.1)a = [[(4)(.2)2][a/.2]]/(.2) a = 2.11 m/sec2

26. An Atwood machine is constructed using a massive hoop (mass concentrated at rim) of 22 cm radius. Find the acceleration, and keep in mind that the tensions will no longer be equal on the two sides. (How else could the wheel accelerate?)

F = ma (1.5 * 9.8) - T1 = 1.5a F = ma T2 - (1 * 9.8) = 1.0a t = Ia (T1 - T2)(.22) = (2)(.22)2(a/.22) T1 - T2 = 2a Solve for T1 and T2, then plug into rotational equation: (1.5 * 9.8) - 1.5a - (1 * 9.8) - 1a = 2a (1.5 * 9.8) - (1 * 9.8) = 2a + 1.5a + 1a a = 1.09 m/sec2

27. A bicycle wheel of radius 0.7 m and mass 3 kg has a small hub of radius 0.13 m. When the 2 kg mass drops as shown, what is its acceleration?

F = ma (2 * 9.8) - T = 2a t = Ia T(.13) = (3 * .72)(a/.13) T = 87a substitute: (2 * 9.8) - 87a = 2a (2 * 9.8) = 89a a = .220 m/sec2

28. An Atwood machine is constructed using two (count 'em) wheels as shown. What is the acceleration of the masses?

F = ma m3g - T1 = m3a F = ma T3 - m4g = m4a t = Ia (T1 - T2)r1 = (m1r12)(a/r1) T1 - T2 = m1a t = Ia (T2 - T3)r2 = (m2 - r22)(a/r2) T2 - T3 = m2a Adding the four equations, T's cancel: m3g - m4g = m3a + m4a + m1a + m2a a = [(m3 - m4)g]/[m3 + m4 + m1 + m2]

29. Find the acceleration of the mass:

F = ma (4 * 9.8) - T1 = 4a F = ma T2 - (4 * 9.8sin26) - (.3 * 5 * 9.8cos26) = 5a t = Ia (T1 - T2)(.2) = (3)(.22)(a/.2) T1 - T1 = 3a Adding the three equations, the T's cancel: (4 * 9.8) - (5 * 9.8sin26) - (.3 * 5 * 9.8cos26) = 4a + 5a + 3a a = .376 m/sec2

30* A 2-disk Atwood machine with radii of 15 cm and 38 cm, has a moment of inertia of 4 kg-m². What is the acceleration of the mass on the right?

equation 1: F = ma T2 - (3 * 9.8) = 3a2 Equation 2: F = ma (2 * 9.8) - T1 = 2a1 Equation 3: T = Ia T1(.38) - T2(.15) = 4(a1/.38) note: a also equals (a2/.15) Equation 4: Finally, a is the same for both disks: (a1/.38) = (a2/.15) Substitute equations 1 and 2 into 3: (2 * 9.8 - 2a1)(.38) - (3 * 9.8 + 3a2)(.15) = (4)(a1/.38) 3.038 = 11.29a1 + .45a2 Substitute equation 4: 3.038 = 11.29[(.38/.15)a2] + .45a2 a2 = .105 m/sec2 a1 = .265 m/sec2

So far we've used only I = mr² for the moment of inertial This will work for a mass which is concentrated at a rim of radius r, but will NOT work when the mass is distributed in some other way. In particular, consider the disk. Here the mass is spread uniformly from 0 to r. Calculus is needed to rigorously determine the moment of inertia, but intuition can see that I = 1/2 mr² is not unreasonable. After all, much of the mass is concentrated at less than radius r.

31. An Atwood machine consists of a disk of mass M, and radius r, and two masses m₁ and m₂ hanging from each side. Find the acceleration.

F = ma m1g - T1 = m1a F = ma T2 - m2g = m2a T = Ia (T1 - T2)r = [(1/2)mr2](a/r) T1 - T2 = (1/2)ma Adding the three equations: m1g - m2g = m1a + m2a + (1/2)ma a = [(m1 - m2)/[(1/2)m + m1 + m2]]g

32. A round object of 27 cm radius is mounted to spin about its axis. A string wrapped around it is pulled with a 5.4 nt force, causing the object to spin up at 14 rad/sec². What is the moment of inertia of the object?

T = Ia (5.4)(.27) = I(14) I = .104 kg-m2

33. A string is wrapped around a 3.5 kg disk of 0.3 meter radius. If the disk is released, what is its linear acceleration downward?

F = ma mg - T = ma T = Ia Tr = [(1/2)mr2](a/r) T = (1/2)ma Think: -about the center, T = Tr -about the center, a = a/r Substituting: mg - (1/2)ma = ma mg = (3/2)ma a = (2/3)g a = (2/3)(9.8) a = 6.53 m/sec2 note: this is true for all disks that ever were and ever will be!

34. A yo-yo consists of two disks of total mass M and radius R connected by a massless shaft of radius r. How fast will the yo-yo accelerate when released?

F = ma mg - T = ma T = Ia Tr = [(1/2)mr2](a/r) Substituting for T: (mg - ma)r2 = (1/2)mr2a a = [r2/[(r2/2) + r2]]g note: T = tension applied at lever arm r [(1/2)mr2] = mass with radius r (a/r) = rim of r accelerating at a

Things begin to get tricky when we look at objects rolling down inclines. You can save yourself a lot of wasted effort if you spend some time thinking about what happens when you release a basketball down a ramp. If the ramp were frictionless, the ball would just slip along and not rotate. We know that doesn't happen, so the ramp must apply a frictional force to the ball (and vice-versa). Study the diagrams:

These are the forces on the ball

IMAGE

Since the normal component of mg exactly balances N, the remaining forces are as shown here.

IMAGE

I use f rather than Ff because normally f will not need to be the maximum force friction can supply. That is, usually f < µN.

35. A disk of mass m and radius r is released on an incline of angle Ø. Write a translational and a rotational equation of motion for the disk.

Translational: F = ma mgsinØ - f = ma Rotational: T = Ia f * r = [(1/2)mr2](a/r)

36. A sphere of mass m and radius r is released down a ramp inclined at angle Ø. Solve for its acceleration down the ramp. (The moment of inertia for a sphere is 2/5 mr².)

F = ma mgsinØ - f = ma T = Ia f * r = [(2/5)mr2](a/r) f = (2/5)ma Substituting: mgsinØ - (2/5)ma = ma mgsinØ = (7/5)ma a = (5/7)gsinØ

38. A 4.7 lb object with a 3.1 inch radius rolls down a 34° slope at 14 ft/sec². What is its moment of inertia?

F = ma mgsinØ - f = ma T = Ia f * r = I(a/r) f = I(a/r2) Substituting: mgsinØ - (Ia/r2) = ma I = [m(gsinØ - a)/(a)]r2 I = [(4.7/32)[(32sin34) - 14]/(14)][3.1/12]2 I = .00273 slug-ft2

39. A 0.3 kg disk of 0.2 m radius is rolled down a ramp with friction coefficient µ = 0.2. At what angle will the disk just start to skid?

Skidding will occur when f > µN, so that point is the breaking point: f = µN f = µ(mgcosØ) F = ma mgsinØ - f = ma T = Ia f * r = [(1/2)mr2](a/r) f = (1/2)ma Substituting f: mgsinØ - f = 2f f = (1/3)mgsinØ f = µN f = µmgcosØ (sinØ/cosØ) = [(µmg)/(1/3)mg] tanØ = 3µ tanØ = 3 * .2 Ø = 31.0°

40. A solid sphere (2/5 mr²) presses up against a block and rubs against it with a coefficient of friction µ. Draw free body diagrams, write an appropriate set of equations and solve for a.

Forces on the Sphere: Since the sphere-block contact slips, f1 = µN Equation 1: mgsinØ - N - f2 = ma Equation 2: (f2 - f1)r = [(2/5)mr2](a/r) f2 - f1 = (2/5)ma f2 - µN = (2/5)ma Forces on the Block: Equation 3: F = ma mgsinØ + N - µ(mgcosØ + f1) = ma Combime equations 1, 2 , and 3: a = [m(sinØ - µcosØ)(1 + µ) + msinØ(1 + µ2)]g/ [m(1 + µ) + (7/5)m(1 + µ2)]

Now we'll try a different approach to rotational problems.

Look at where we are. We have seen rotational analogs to the kinematic equations (Ø = wt), there is a rotational analog to mass (moment of inertia), and even the rotational analog to F = ma (t = Iµ).

Surely, then, there must be a rotational kinetic energy, and it shouldn't be hard to fig-ure out. Consider again our mass on a string, tied at one end and moving with a velocity v:

Its kinetic energy is, of course, KE = 1/2mv². But v = wr, so:

KE = 1/2 m(wr)² = 1/2 (mr²)w² = 1/2 Iw²

To get the total kinetic energy, we'll have to add on the 1/2 mv² of translational motion, but that won't be hard.

41. What is the kinetic energy of a 3 kg disk of 0.2 m radius rolling on its rim at 4 m/sec?

KE = KEtranslational + KErotational KE = (1/2)mv2 + (1/2)Iw2 KE = (1/2)(3)(4)2 + (1/2)(1/2 * 3 * .22)(4/.2)2 KE = 36.0 J

42. What's the kinetic energy of a 3 kg hoop of 0.2 m radius rolling at 4 m/sec?

KE = KEtranslational + KErotational KE = (1/2)Iw2 + (1/2)mv2 KE = (1/2)(3 * .22)(4/.2)2 + (1/2)(3)(4)2 KE = 48.0 J

43. A 5.0 kg disk of 0.13 m radius rolls so that its total KE is 47 joules. How fast is it moving?

KE = (1/2)Iw2 + (1/2)mv2 KE = (1/2)[(1/2)mr2](v/r)2 + (1/2)mv2 KE = (3/4)mv2 47 - (3/4)(5)v2 v = 3.54 m/sec

44. A bicycle has a 5.8 kg frame and two wheels, each at 1.3 kg. What is its total kinetic energy when it's rolling at 6 m/sec?

KE = KEframe + KEwheels KE = (1/2)mv2 + 2[(1/2)Iw2 + (1/2)mv2] KE = (1/2)(5.8)(6)2 + 2[(1/2)(1.3r2)(6/r)2 + (1/2)(1.3) (6)2] KE = 198 J

45. A 2 kg disk starts at the top of a ramp 5 m high and rolls down to the bottom. How fast is it moving on the level plane?

For a disk: KE = (1/2)mv2 + (1/2)[(1/2)mr2](v/r)2 KE = (3/4)mv2 PE = KE mgh = (3/4)mv2 v = [(4/3)gh]1/2 v = [(4/3)(9.8)(5)]1/2 v = 8.08 m/sec

46. A cart has a body of mass M and 4 disk wheels each of mass m. A force F pushes on it for a distance S. How fast is it moving at the end of this time?

KE = (1/2)mv2 + 4[(1/2)Iw2 + (1/2)mv2] KE = (1/2)mv2 + 4[(1/2)(1/2)(mr2)(v/r)2 + (1/2)mv2] KE = (1/2)mv2 + 3mv2 KE = [(m/2) + 3m]v2 But KE = W [(m/2) + 3m]v2 = F * s v = [(F * s)/[(m/2) + 3m]]

47. The 470 gm sphere of 5.2 cm radius rolls at 881 cm/sec. How fast will it roll on the plateau?

KEo - PEg = KE (1/2)mv2 + (1/2)Iw2 - mgh = (1/2)mv2 + (1/2)Iw2 For a sphere: KE = (1/2)Iw2 + (1/2)mv2 KE = (1/2)[(2/5)mr2](v/r)2 + (1/2)mv2 KE = (7/10)mv2 (7/10)mvo2 - mgh = (7/10)mv2 (7/10)(881)2 - (980)(200) = (7/10)v2 v = 704 cm/sec

48. How fast will the disk roll after the spring is released?

PEspring = KEdisk (1/2)kx2 = (1/2)mv2 + (1/2)[(1/2)mr2](v/r)2 (1/2)kx2 = (3/4)mv2 v = [(2kx2)/(3m)]

49. A solid sphere of mass m and radius r rolls with initial velocity v_o. After p assing through the wind tunnel it goes over the ledge. How far from the base does it land?

Recall that for spheres: KE = (1/2)[(2/5)mr2](v/r)2 + (1/2)mv2 KE = (.7)mv2 (.7)mv2 - mgh1 + FL = (.7)mv2 v = [(.7mvo2 - mgh + FL)/(.7m)]1/2 s = (1/2)at2 t = [(2h2)/g] s = vt s = [[(.7mvo2 - mgh + FL)/(.7m)][(2h2)/g]]1/2

One last topic we have neglected is what to do about momentum. In linear dynamics momentum is conserved in collisions, explosions, or in any other type of interaction. The same is true for angular momentum. Without proof, here is the value of angular momentum:

L = Iw

Does it look a lot like momentum = mv? Hardly a coincidence. If, as in linear motion, momentum is conserved, then a rotating body can alter its rate of rotation by changing its moment of inertia. Thus a spinning ice skater can spin faster by bring his arms in, or a star in the sky can spin faster by contracting.

50. What is the angular momentum of a 7 kg disk, 1.3 m in radius, spinning at 3 rad/sec?

L = Iw L = [(1/2)mr2](w) L = (1/2)(7)(1.32)(3) L = 17.8 kg-m2/sec

51. A hula-hoop of 0.2 kg mass and 0.4 m radius spins in space at 5 rad/sec. How fast will it be spinning if it starts to get very cold and the radius shrinks to 0.3 m?

In the absence of outside torques, angular momentum is conserved: L1 = L2 I1w1 = I2w2 mr12w1 = mr22w2 w2 = r12/r22 w2 = [(.42)/(.32)](5) w2 = 8.89 rad/sec

52. An ice skater, spinning at 5 rad/sec has a moment of inertia of 16 kg-m². If, after pulling her arms in, her moment of I is 10 kg-m², what is h er new angular velocity?

I1w1 = I2w2 (16)(5) = (19)w2 w2 = 8.0 rad/sec

53. A 2 kg toad sits on the edge of a 3 kg lazy susan (a disk) which has a radius of 0.34 m. If the system rotates initially at 4 rad/sec and the toad hops to a point 0.10 m from the center, what is the new angular velocity?

I1 = (1/2)mr2 + mr2 I1 = (1/2)(3)(.34)2 + (2)(.34)2 I1 = .405 kg-m2 I2 = (1/2)mr2 + mr2 I2 = (1/2)(3)(.34)2 + (2)(.1)2 I2 = .193 kg-m2 I1w1 = I2w2 (.405)(4) = (.193)w2 w2 = 8.37 rad/sec