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UNIT 6 Circular Motion and GravitationPowerPoint Presentation

UNIT 6 Circular Motion and Gravitation

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UNIT 6Circular Motion and Gravitation

Bonnie

Klyde

ConcepTest 9.1a Bonnie and Klyde I1) same as Bonnie’s

2) twice Bonnie’s

3) half of Bonnie’s

4) 1/4 of Bonnie’s

5) four times Bonnie’s

Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds.

Klyde’s angular velocity is:

Bonnie

Klyde

ConcepTest 9.1a Bonnie and Klyde I1) same as Bonnie’s

2) twice Bonnie’s

3) half of Bonnie’s

4) 1/4 of Bonnie’s

5) four times Bonnie’s

Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds.

Klyde’s angular velocity is:

The angular velocityw of any point on a solid object rotating about a fixed axis is the same. Both Bonnie & Klyde go around one revolution (2p radians) every two seconds.

Kinematics of Uniform Circular Motion

Dynamics of Uniform Circular Motion

Newton’s Law of Universal Gravitation

Kepler’s Laws and Newton’s Synthesis

Torque and Newton’s Laws of Motion

Kinematics of Uniform Circular Motion

Uniform circular motion: motion in a circle of constant radius at constant speed

Instantaneous velocity is always tangent to circle.

Kinematics of Uniform Circular Motion

Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that

(7-1)

Kinematics of Uniform Circular Motion

This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.

Dynamics of Uniform Circular Motion

For an object to be in uniform circular motion, there must be a net force acting on it.

We already know the acceleration, so can immediately write the force:

(7-1)

Dynamics of Uniform Circular Motion

We can see that the force must be inward by thinking about a ball being whirled in a vertical circular path on a string:

Dynamics of Uniform Circular Motion

- Consider a ball of mass mthat is being whirled in a horizontal circular path of radius rwith constant speed.

- The force exerted by the string has horizontal and vertical components. The vertical component is equal and opposite to the gravitational force. Thus, the horizontal component is the net force.

- This net force, which is directed toward the center of the plane of the circle, is the centripetal force.

When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

- Centripetal Force is simply the name given to the net force on an object in uniform circular motion.
- Any type of force or combination of forces can provide this net force.
- For example, friction between the race car’s tires and a circular track is a centripetal force that keeps the car in a circular path.
- As another example, gravitational force is a centripetal force that keeps the moon in its orbit around the earth.

Dynamics of Uniform Circular Motion

There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome.

If the centripetal force vanishes, the object flies off tangent to the circle.

Centripetal Acceleration Problem

The Moon’s nearly circular orbit about the Earth has a radius r of 3.84 x 105 km and a period T of 27.3 days.

Determine the acceleration of the Moon toward the Earth.

or

- A 1000.0 kg car rounds a curve on a flat road of radius 50.0 m at a speed of 14.0 m/s. Will the car follow the curve or will it skid?
- For a dry pavement the coefficient of static friction is 0.60
- For an icy pavement the coefficient of static friction is 0.25

The forces on the car are gravity mg downward, the normal force exerted upward by the road, and a horizontal frictional force due to the road. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

In the vertical direction there is no acceleration. Newton’s 2nd law states that the normal force is equal to the weight of the car since the road is flat and level.

In the horizontal direction there is only the force of friction, and we must compare to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is …

Now we compute the maximum total static friction force to see if

it can be large enough to provide a safe centripetal acceleration.

For (a) and the maximum friction force attainable is ...

Since a force of only 3900 N is needed, and that is how much will

be exerted by the road as a static friction force, the car can follow the curve. But in (b) the maximum static friction force possible is …

The car will skid because the ground cannot exert sufficient

force (3900 N needed) to keep it moving in curve of radius 50.0 m at a speed of 14.0 m/s.

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