CHAPTER 5 - Circular Motion; Universal Gravitation



Physics 3311

Chapter 5

NOTES

Circular Motion; Gravitation

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PERIOD

CHAPTER 5 - Circular Motion; Universal Gravitation

Circular Motion is a class of PERIODIC MOTION. Motion that repeats itself; that completes a cycle; that finishes where it starts; that oscillates; motion that repeats itself; that completes a cycle; that finishes where it starts; that oscillates; motion that repeats itself; that completes a cycle; that finishes where it starts; that oscillates…

Recall:

[pic]

Conversions:

[pic]

Uniform Circular Motion

• Uniform in this case means constant speed.

• Can an object moving with a constant speed have acceleration?

• If so, we physicists are going to want to find a value for it, but how?

NOTE: The magnitude of the velocity is not changing, but the direction is changing.

In Figure 2 triangle made by the two radii and Δl

will be similar to the triangle made by v1 ,v2 , Δv.

PROOF:

By definition:

The angle between the radius and the instantaneous velocity is always 90o. (See figure 1)

For this to remain true, the angle between r1 and r2 must be equal to the angle between v1 and v2. (See figure 2 (a) and (b))

Therefore, [pic]

And [pic]

If [pic]

Note: For small values of Δθ, Δl is ≈ s (arc length).

See Figure 2

Therefore since [pic], and [pic] is our definition of linear speed, v.

Then [pic] (Equation 1)

Hopefully, it is clear that this formula can only give positive answers, and therefore only can find the magnitude of the acceleration.

Acceleration is a VECTOR so… in what direction is the acceleration?

Centripetal vs. Centrifugal

Centri from the Latin Centrum: the middle point of a circle, the center.

Petal from the Latin Peto – to seek, strive after, endeavor to obtain.

Fugal from the Latin Fugito – to flee, to fly from, avoid or shun.

By looking at the direction of the change in velocity, you should be able to see the direction of the acceleration.

Is it Centripetal or Centrifugal?

[pic]

The acceleration of an object moving with Uniform Circular Motion is known as centripetal acceleration or radial acceleration.

How do we find the speed of an object going in a circle?

• Measure the time it takes for the object to complete a certain number of revolutions.

EXAMPLE: We know an object makes 10 revolutions in 2 seconds.

With the information given we know the frequency of each revolution.

[pic]

The time needed to complete one revolution is known as the Period (T).

[pic]

How far does an object move in one revolution? (circumference)

1 rev =2πr

(Equation 2)

More Useful Formulas…

If we substitute Equation 2 into Equation 1…

[pic]

THEN SIMPLIFY

[pic] (Equation 3)

Problems:

1. (I) A jet plane traveling 1800 km/h (500 m/s) pulls out of a dive by moving in an arc of radius 6.00 km. What is the plane’s acceleration?

2. (I) Calculate the centripetal acceleration of the Earth in its orbit around the Sun and the net force exerted on the Earth. What exerts this force on the Earth? Assume that the Earth’s orbit is a circle of radius 1.50 x 1011 m.

3. Will the acceleration of a car be the same when it travels around a sharp curve at 60 km/h, as when it travels around a gentle curve at the same speed? Explain.

DYNAMICS of Circular Motion

Revisiting Newton’s 2nd Law

Newton stated that if a body is accelerating, there must be a NET force acting on it, in the direction of the acceleration.

The NET force the cause circular motion is called the CENTRIPETAL force.

Note: This is not a new type of force, we will treat it very much the same as we did any NET force.

[pic]

A centripetal (center seeking) force causes circular motion NOT a centrifugal (center fleeing) force.

• The force that is acting on the ball causing circular motion is inward (centripetal).

• The force acting on the hand is outward, and NOT causing circular motion.

• A centrifugal force is a “fictitious” or “pseudo” force caused by viewing the situation from an accelerating reference frame.

Problems:

1. (I) A horizontal force of 280 N is exerted on a 2.0-kg discus as it is rotated uniformly in a horizontal circle (at arms length) of radius 1.00 m. Calculate the speed of the discus.

2. (II) A 0.40-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 60 N, what is the maximum speed the ball can have? How would your answer be affected if there were friction?

EXAMPLE 5-7 (from book)

A Car rounding a bend

A 1000 kg car rounds a curve on a flat road of radius 50 m at a speed of 50 km/hr (14 m/s). Will the car make the turn, or will it skid, if: (a) the pavement is dry and the coefficient of static friction is μs = 0.60; (b) the pavement is icy and μs = 0.25?

11. (II) What is the maximum speed with which a 1050-kg car can round a turn of radius 70 m on a flat road if the coefficient of friction between tires and road is 0.80? Is this result independent of the mass of the car?

More Practice (mixed problems)

1. A race car makes one lap around a track of radius 50 m in 9.0 s.

a) What is the average speed?

b) What was the car’s centripetal acceleration?

2. Normie Neutron swings a rubber ball attached to a string over his head in a horizontal, circular path. The piece of string is 1.5 m long and the ball makes 120 complete turns each minute.

a) What is the average speed of the ball?

b) What is the ball’s centripetal acceleration?

3. A car goes around a curve at 20. m/s. If the radius of the curve is 50 m, what is the centripetal acceleration of the car?

4. Professor Brown holds on to the end of the minute hand of a clock atop city hall. If the minute hand is 4.0 m long, what is the professor’s centripetal acceleration?

5. A flea gets its thrills by riding on the outer edge of a golden oldies record album of radius 15 cm as it is being played with a rotational period of 1.8 seconds.

a) What is the flea’s average speed?

b) What is the flea’s centripetal acceleration?

6. A 0.100 kg mass is attached to a string 75 cm long and swings in a horizontal circle, revolving once every 0.80 s. Calculate:

a) the centripetal acceleration of the mass.

b) the tension in the string.

7. A 0.50 kg mass is attached to a string 1.0 m long and moves in a horizontal circle completing 1 revolutions in 0.5 seconds. Calculate:

a) the centripetal acceleration of the mass.

b) the tension in the string.

8. It takes a 900. kg racing car 12.3 s to travel at a uniform speed around a circular racetrack of radius 90.0 m. What is the centripetal force acting on the car, and which force provides it?

9. A 2.0 kg object is tied to the end of a cord and whirled in a horizontal circle of radius 4.0 m completing 2 revolutions in 6 seconds. Determine:

a) the velocity of the object.

b) the acceleration of the object.

c) the pull of the object.

d) what happens if the cord breaks.

10. (II) What is the maximum speed with which a 1000-kg car can round a turn of radius 78 m on a flat road if the coefficient of friction between tires and road is 0.60?

[pic]

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[pic]

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Figure 1 a small object moving in a circle. Note the instantaneous velocity is always tangent to the circular path.

Figure 2 Determining the chanθ

Figure 1 a small object moving in a circle. Note the instantaneous velocity is always tangent to the circular path.

Figure 2 Determining the change in the velocity, Δv, for a particle moving in a circle. The length Δl is the cord from A to B. As Δθ approaches zero, Δl approaches the arc length, s, from A to B.

1

2

The direction of acceleration is also towards the center of the circle! THUS, centripetal!

Notice the direction of (v is towards the center of the circle!

a=(v/(t

(v

vf

-vi

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