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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?

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Presentation on theme: "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?"— Presentation transcript:

1 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

2 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

3 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.

4 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

5 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.

6 Uniform Circular Motion Motion along a circular path in which there is no change in speed, only a change in direction. v FcFc Question: Is there an outward force on the ball? https://www.youtube.com/watch?v=KvCezk9DJfk

7 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. Uniform Circular Motion https://www.youtube.com/watch?v=rf6GC746SOg

8 Uniform Circular Motion

9 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? Force ON you is toward the center. FcFc

10 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. The centripetal force is exerted BY the door ON you. (Centrally) FcFc F’Reaction

11 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?

12 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?

13 So because of Newton’s 3 rd 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 (F g )- in this case it’s not Earth specific and is not 9.81 m/s 2 ).

14 Masses of the 2 objects Distance between the objects Universal Gravitational Constant- 6.67 x10 -11

15 Example 1:What is the gravitational force between the Earth and the Moon? m Earth = M = 6.0 x 10 24 kg m Moon = m = 7.4 x 10 22 kg r = 3.8 x 10 8 m G = 6.67 x 10 -11 F g = 2.1 x 10 20

16 Example 2: What is the gravitational force between the Earth and Venus? m Earth = M = 6.0 x 10 24 kg m Venus = m = 5.0 x 10 24 kg r = 3.8 x 10 10 m G = 6.67 x 10 -11 Answer: F = 1.386x10 18 N

17 Determine the force of gravitational attraction between the earth (m = 5.98 x 10 24 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.38 x 10 6 m from earth's center. Determine the force of gravitational attraction between the earth (m = 5.98 x 10 24 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 10 6 m from earth's center.

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19 There is also a way to determine the gravitational field around one object: This is now the gravitational Field Strength(GFS)

20 Example 3: What is the Gravitational Field Strength in Earth? Radius of the Earth – 6.37 x 10 6 m

21 Radius of the Moon – 1.7 x 10 6 mRadius of the Moon – 1.7 x 10 6 m Example 4: What is the Gravitational field strength on the moon?

22 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

23 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 (2 2 ). The new force is then 1/4 of the original 16 units. F = (16 N) / 4 = 4 units

24 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

25 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/s 2 /4 = 2.45 m/s 2 ).

26 2x3x4x ¼ r e = 2.24 m/s 2 = 1/9 r e = 1.09 m/s 2 1/16 r e =.61 m/s 2 1x r e = 6.4 x 10 6 m 1 r e = 9.81 m/s 2

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28 Manipulation Practice Example 5: The gravitational attraction between the Earth and Mars is 8.7 x 10 16 N. The distance between the two planets is 5.5 x 10 10 m. Earth has a mass of 6.0 x 10 24 kg. What’s the mass of Mars?


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