UCM notes



Uniform Circular Motion

Uniform Circular Motion (UCM) is any circular motion for which the speed remains constant. Note that the velocity does not remain constant because the direction is changing. Therefore, there is an acceleration, and Newton's 2nd Law is in effect. And that means that there is an unbalanced force present which produces the circular motion.

Circular motion should not be unfamiliar to you. Amusement park rides, athletic events (like the discus, figure skating, and swinging a baseball bat), rounding a turn (in a car or on a bike) and orbital motion are all circular motions. What distinguishes these examples is the unbalanced force that produces the circular motion. In the case of orbital motion, the force is gravity; for the discus, the force is the tension in the arm; for a car rounding a turn, friction is present; and for an amusement park ride like the Rotor, an applied force (by the wall) enables the rider to move in a circle. The 4 forces mentioned in these examples constitute our force menu!!!

The vectors present in UCM are illustrated below. There are only 3 : velocity,

force, and acceleration. We will demonstrate the directions of these vectors and provide evidence for them with the demos & photo on the following pages.

THE UCM VECTORS

Demonstration 1 : Direction of velocity

For each of the following cases, predict the direction that the object in UCM will travel in after it leaves the circle.

(a) Circle with Gap (web video)

(b) Hammer Throw (CC 1B - 23)

(c) Air Table (CC 1B - 26)

Result : The direction of the velocity vector is .

Question : What evidence is there that the speed of a UCM is constant?

Demonstration 2 : Direction of the unbalanced force (Σ F)

For each of the following cases, identify the unbalanced force that produces the UCM.

For each case, imagine that you had to provide the force that causes the circle.

In what general direction would you have to pull/push to produce the particular circle?

(a) Dry Ice Puck (CC 1B - 22)

(b) Hammer Throw (CC 1B - 23)

(c) Rotor (CC 1B - 28)

(d) Bucket of water whirled in a circle (Gulp! Mr. E demo)

Result : The direction of the unbalanced force is

Question : Assuming that you believe Newton’s 2nd Law (ΣF = ma), what must be the direction of the acceleration of a UCM?

Question : Isn’t the unbalanced force centrifugal? What gives?

THE UCM FORMULAS

v = 2πR ac = v2

T R

ac = 4π2R Fc = mv2 = m (4π2R)

T2 R T2

T = 1/f

In these formulas, T represents the period, that is, the time that is needed to complete one complete revolution. Tension forces will be represented with a vector arrow above the T or with a bold face T.

SPECIFIC SITUATIONS :

(a) car rounds a level turn (b) Rotor at Great Adventure

ΣFy = 0 ΣFx = mv2 ΣFy = 0 ΣFx = mv2

R R

μmg = mv2 = m(4π2R) mg = Ff = μN N = Fc = mv2

R T2 R

N = Fapp by the wall!

(head-on view)

(c) mass moves in a horizontal circle (d) mass on a string whirled in vertical circle

Highest Point : ΣFy = T + mg = mv2

R

Lowest Point : ΣFy = T - mg = mv2

R

note : for Loop the Loop’s, Fseat replaces T

Newton & v2/R

Not surprisingly, Newton developed the formula for centripetal acceleration.

We'll follow his basic reasoning to show that ac = v2/R.

The circle below show a displacement vector R1 and a velocity vector v1 at a certain instant. A short time later, the displacement vector is R2 and the velocity vector is v2. Therefore, ΔR = R2 – R1 = R2 + (-R1). And, Δv = v2 –v1 = v2 + (-v1).

(Why are things changing? Because of the force that produces the UCM (like gravity or tension, for example)!!!)

(1) Show that Δv/ΔR = v/R. (Hint : Show that the triangles above are similar)

(2) Use the result above and the fact that, in general, a = Δv/Δt to show that

a = v2/R

Reference - Rounding Turns at High Speeds

If you're like most people, you know intuitively that it is difficult to round a turn at a high speed. Newton's 2nd Law for UCM explains why. In general, for any UCM,

Fc = mv2/R

In particular, for rounding a turn (which represents a "temporary" UCM),

Fc = Ffriction = μmg

Therefore, mv2/R = μmg

Notice that the expression for friction represents a constant value (for any given car on a particular road surface). But the centripetal force expression is not constant... as v increases, the centripetal force increases.

So...friction is the force that is available, and it is constant. The centripetal force, which is the amount of force needed to make a turn of radius R at a speed v, depends on v. When the speed gets to the point where the needed force (mv2/R) is greater than the available force (friction), guess what happens? Yup, the car (or whatever) goes off the road because there's not enough available force to supply the required centripetal force!!! (By the way, this is like life in general...if what you need is more than what you have, you are in trouble; think about it!)

UCM Problems

1. The Rotor at Great Adventure has a radius of 2.5 meters and a period of

2 seconds. Find the centripetal force, acceleration, speed, and frequency of a

60 kg rider.

(7.85 m/s,

24.6 m/s2

1479 N

0.5 Hz)

2. A discus weighs 1 kg. If a discus thrower turns one complete revolution in 0.35 seconds with his arm extended horizontally 1 meter from the axis of rotation, find

(a) the speed with which the discus is released

(b) the force he feels in his arm

(17.9 m/s,

322 N)

3. The earth weighs 6 x 1024 kg. It is 1.5 x 1011 meters from the sun. The period of the earth around the sun is 3.15 x 107 sec (i.e., 1 year). Find the

(a) speed of earth in its orbit

(b) its acceleration

(c) the centripetal force that the sun exerts on the earth.

(30,000 m/s

0.006 m/s2

3.6 x 1022 N)

4. A ball on a string is whirled in a horizontal circle with a horizontal force of 20 N. Its acceleration is 12 m/s2 and its speed is 4 m/s. Find the (a) mass (b) radius, (c) period, and (d) frequency of the ball.

(1.67 kg,

1.33 m,

2.09 sec

0.48 Hz)

5. A car rounds a turn of radius 10 m at a speed of 15 m/s. Find the coefficient of friction necessary to keep the car on the road.

(2.25)

6. A car rounds a turn of radius 25 m and coefficient of friction 1.6. Find the maximum speed for which the car will stay on the road.

(20 m/s)

7. A coin that is positioned 10 cm from the center of a turntable is rotating with a period of 0.5 sec. Find the coefficient of friction between the coin and the turntable.

(1.6)

8. A bucket of water weighing 3 kg is whirled in a vertical circle of radius 1 meter by a physics teacher entertaining guests at a wild party. If the speed of the bucket is

4 m/s at the top and 8 m/s at the bottom of the circle, how much tension does the teacher feel in his arm at the (a) highest point, and (b) lowest point?

(18 N, 222 N)

9. In fast pitch softball, a pitcher can wind up and release the ball in 0.2 seconds. The ball weighs 300 grams and the pitcher’s arm is 0.66 m long. Assuming that the speed of the ball remains constant throughout the windup, find (a) the speed of the ball, and (b) the tension exerted on the pitcher’s shoulder at the release point (i.e., the bottom of the circle).

(20.7 m/s,198 N)

10. A loop-the-loop ride at an amusement park has a radius of 5 meters. At the highest point a 70 kg rider moves at 8 m/s. At the lowest point, the rider moves at 13 m/s. Find what the rider feels like he weighs (aka Fseat) at those points.

(196 N, 3066 N)

-----------------------

v

v

ΣF & a

ΣF & a

note : Fc represents ΣF…it’s the unbalanced force directed toward the center!!!

R

Fc = mv2

R

In these cases䱮䱯䲍䲭䳐䳑䳒䶨䶩䶪䶫䶬䶭䶮䶯䶰䶱䶲䷉䷞䷴上下亏亐云互ñæææñ, the centripetal force can be caused by, for example, tension, gravity, an applied force (like the hammer thrower), or some combination of forces (like the conical pendulum)

ΔR

v1

-R1

R2

v2

R1

R2

v2

Δv

-v1

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