eq: why does an object in circular motion have a constant velocity yet an acceleration and...

EQ: Why does an object in circular motion have a constant velocity yet an

acceleration and unbalanced force that is directed toward the center of the circle?

Uniform Circular Motion

https://www.youtube.com/watch?v=VSDRqV1p63A



Linear/Tangential Velocity

• Objects moving in a circle still have a linear velocity = distance/time.

• This is often called tangential velocity, since the direction of the linear velocity is tangent to the circle.

v

Acceleration• As an object moves around a circle, its direction

of motion is constantly changing.• Therefore its velocity is changing.• Therefore an object moving in a circle is

constantly accelerating.

Centripetal Acceleration

• The acceleration of an object moving in a circle

points toward the center of the circle.

• This is called a centripetal (center pointing) acceleration.

a

Centripetal Force• Newton’s Second Law says that if an object is

accelerating, there must be a net force on it.• For an object moving in a circle, this is called the

centripetal force.

• The centripetal force points toward the center of the circle.


Motion along a circular path in which there is no change in speed, only a change in direction.

vFc

Question: Is there an outward force on the ball?

https://www.youtube.com/watch?v=KvCezk9DJfk



The question of an outward force can be resolved by asking what happens when the string breaks!

When central force is removed, ball continues in straight line.

v

Ball moves tangent to path, NOT outward as might be expected.

Centripetal force is needed to change direction.


https://www.youtube.com/watch?v=rf6GC746SOg



Examples of Centripetal Force

• Car going around a curve.

You are sitting on the seat next to the outside door. What is the direction of the resultant force on you as you turn? Is it away from center or toward center of the turn?

You are sitting on the seat next to the outside door. What is the direction of the resultant force on you as you turn? Is it away from center or toward center of the turn?

Force ON you is toward the center.

Fc

Car Example Continued

There is an outward force, but it does not act ON you. It is the reaction force exerted BY you ON the door. It affects only the door.

There is an outward force, but it does not act ON you. It is the reaction force exerted BY you ON the door. It affects only the door.

The centripetal force is

exerted BY the door ON you.

(Centrally)

Fc

F’Reaction

Closing task:

Why is an object moving in a circle at a constant speed accelerating?

How does velocity, acceleration, and force keep an object in circular motion?

Universal Law of

Gravitation

http://ed.ted.com/lessons/jon-bergmann-how-to-think-about-gravity

EQ: How does the mass and distance between objects affect the

gravitational force?





So because of Newton’s 3rd law every body in the

universe exerts a force of attraction on every other

body.• This is Newton’s Universal Law of Gravitation

The force between two objects, due to their masses, is called the

gravitational force (Fg)- in this case it’s not Earth specific and is not 9.81

m/s2).

FGmM

rg 2

FGmM

rg 2

Masses of the 2 objects

Distance between

the objects

Universal Gravitational Constant-

6.67 x10-11 N

m

kg

2

2

Example 1:What is the gravitational force between the Earth and the Moon?

mEarth = M = 6.0 x 1024 kg

mMoon = m = 7.4 x 1022 kg

r = 3.8 x 108 mG = 6.67 x 10-11 N

m

kg

2

2

Fg = 2.1 x 1020

Example 2: What is the gravitational force between the

Earth and Venus?

mEarth = M = 6.0 x 1024 kg

mVenus = m = 5.0 x 1024 kg

r = 3.8 x 1010 mG = 6.67 x 10-11 N

m

kg

2

2

Answer:

F = 1.386x1018 N

Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance

of 6.38 x 106 m from earth's center.

Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the

student is in an airplane at 40000 feet above earth's surface. This would place the

student a distance of 6.39 x 106 m from earth's center.

There is also a way to determine the

gravitational field around one object:

FGmM

rg 2 w

GmM

r 2 mg

GmM

r 2

mgGmM

r 2g

GM

r 2

This is now

the gravitational

Field Strength(GFS)

Example 3: What is the Gravitational Field Strength in

Earth?

• Radius of the Earth – 6.37 x 106 m

Is Gravity Diluted?

The distance d is in the denominator of this relationship, it can be said that the force of gravity

is inversely related to the distance. This mathematical relationship is sometimes referred to

as an inverse square law since one quantity depends inversely upon the square of the other

quantity

Check your understanding

Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is doubled, what is the new force of attraction between

the two objects?

Answer: F = 4 units

If the distance is increased by a factor of 2, then force will be decreased by a factor of 4 (22). The new force is then 1/4 of the

original 16 units.

F = (16 N) / 4 = 4 units

Check your understanding

Suppose that two objects attract each other with a gravitational force

of 16 units. If the mass of both objects was doubled, and if the distance between the objects

remained the same, then what would be the new force of attraction

between the two objects?Answer: F = 64 units

If each mass is increased by a factor of 2, then force will be increased by a factor of 4 (2*2). The new force is then 4 times

the original 16 units.

F = (16 units ) • 4 = 64 units

Both the ULG and the GFS follow the Inverse square law:

• ULG-If the distance between two objects is doubled the gravitational attraction is (1/4) of the original.

OR• GFS-If we travel beyond the Earth by a distance

that is double it’s radius than we will only feel a quarter of Earth’s gravitational pull (9.81 m/s2/4 = 2.45 m/s2).

2x 3x 4x

¼ re=

2.24 m/s2

1/9 re =

1.09 m/s2

1/16 re =

.61 m/s2

1x

re = 6.4 x 106 m

1 re =

9.81 m/s2

Manipulation PracticeExample 5: The gravitational attraction

between the Earth and Mars is 8.7 x 1016 N. The distance between the two planets is 5.5 x 1010m. Earth has a mass of 6.0 x 1024 kg. What’s the mass of Mars?