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)