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Presentation on theme: "UNIFORM CIRCULAR MOTION"— Presentation transcript:

An object traveling with a constant speed in a circle is accelerating because the object’s velocity is changing in direction. The object is said to have “uniform” circular motion and it is undergoing centripetal acceleration (centrum- centre, petere- to seek (Newton) p ). These two triangles are similar and for any time interval this would be so.

2 This proof is not on any test.
[towards the center] Notice on the first diagram that half way through the interval Dv is pointing at the center of the circle. This is an instantaneous acceleration so the direction of the acceleration is constantly changing. This is velocity! For the final equation remove the absolute value notation.

3 For uniform circular motion . . .
T is the period of revolution. remember f is the frequency An increase in the velocity of an object that is centripetally accelerating (with no change to r) is called tangential ac-celeration. A parallel push causes tangential acceleration and velocity’s magnitude changes. A perpendicular push causes centripetal acceleration and only the direction of velocity changes.

4 What is the centripetal acceleration of a stone being whirled in a circle, at the end of a 1.5 m string, on a smooth sheet of ice, with a frequency of 1.25 Hz? The planet Mercury moves in an approximately circular path at an average distance of 5.8 x 1010 m, accelerating centripetally at 0.04 m/s2. What is its period of rotation about the Sun?

5 CENTRIPETAL FORCE Centripetal acceleration (center seeking) can be extended to centripetal force. Remember centripetal acceleration is the observed motion while centripetal force is responsible for this acceleration. Finally, centripetal force is always a resultant force. This direction is constantly changing. textbook: p , p , p , 11-13, 26-28

6 A 1000 kg car enters a level curve at 20 m/s
A 1000 kg car enters a level curve at 20 m/s. If the curve has a radius of 80 m, what centripetal force must be supplied by friction to keep the car from skidding? FN Fg Ff Notice this would give a ms of 0.5

7 A 1000 kg ball travels around a frictionless, banked curve having a radius of 80 m. If the bank is 20o to the horizontal, at what speed must the ball travel to maintain a constant radius? 20o

8 A 3.5 kg steel ball is swung at a constant speed in a vertical circle of radius 1.2 m on the end of a light, rigid steel rod. If the ball has a frequency of 1.0 Hz, calculate the tension in the rod at the top and bottom of the circle. top bottom + is up

FG -gravitational force (in 2 directions) G -universal gravitation constant 6.67x10-11 Nm2kg-2 r -distance between the objects m1 -mass of the larger object near the earth’s surface . . . This equations could be applied to the surface of any planet or to the acceleration you would experience at any distance from on object.

10 What happens to the gravitational attraction between two particles if one mass is doubled, the other tripled and the distance between them cut in half? Planet X has a radius that is 3.5 times the radius of the earth and a mass that is 2.0 times the earth’s. Compare the acceleration due to gravity at the surface of each planet. read p p extra p p

11 SATELLITES A satellite is an object or a body that revolves around another object, which is usually larger in mass. Planets, moons, space shuttles, space stations, comets, and “satellites” are satellites. Satellites remain in a constant orbit because they are acted upon by a centripetal force and display centripetal acceleration.

12 remember m1 is the larger mass and the central object
Calculate the orbital speed for a satellite at Earth’s surface, and two Earth radii above Earth’s surface. Show the simpler calculation for the satellite at Earth’s surface.

13 What is the radius of the orbit of a geosynchronous satellite?
What is the period of rotation of the moon about the earth? read p , extra p , 6 p

14 GRAVITATIONAL FIELDS A force field exists in the space surrounding an object in which a force is exerted on objects (e.g. gravitational, electric, magnetic). The strength of gravitational force fields is deter-mined by the Law of Universal Gravitation. If two or more gravitational fields are acting on an object then the net field is the sum of all the individual fields. read p p

15 KEPLER’S LAWS In 1543 Copernicus proposes the heliocentric model of the solar system in which planets revolve around the sun in circular orbits. Slight irregularities show up over long periods of study. Tycho Brahe takes painstaking observations for 20 years with large precision instruments but dies (1600) before he can analyze them properly. A young mathematician continues Brahe’s work. From his analysis the kinematics of the planets is fully understood. Kepler’s First Law of Planetary Motion Each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus of the ellipse. Kepler’s Second Law of Planetary Motion The straight line joining a planet and the Sun sweeps out equal areas in space in equal intervals of time.

16 Planets move faster when they are closer to the Sun (centripetal force is stronger).
orbits are elliptical but are not very elongated equal areas equal times

17 Kepler’s Third Law of Planetary Motion
The cube of the average radius of a planet is directly proportional to the square of the period of the planet’s orbit. We have already proved this a few slides back. Recall. constant For our solar system m1 is the mass of the sun. Mars’ average distance from the sun is x1011 m while its period of rotation is 5.94 x 107 s. What is Jupiter’s average distance from the sun if its period of rotation is 3.75 x 108 s ? read p p , 9

Recall the Law of Universal Gravitation for constant masses, a graph of force vs. radius would be . . .

19 The graph above is a F vs. d graph which means the shaded area is the work required to move an object from r1 to r2. The shaded area is not easy to calculate but can be done with a geometric mean. In this case the work done by the lifter is equal to DEp. Another method involves calculus and integration over a range from r1 to r2. geometric mean of force

20 Know these two equations, you are not required to know the previous development.
Which preceding equation can be simplified to mgDh, the potential energy change near the earth’s surface? Potential energy is a negative function! It increases until it is zero. PE stops here because the objects come into contact and cannot get closer.

21 Recall so . . . read p p

22 Escape from a Gravitational Field
To escape a gravitational field an object must have at least a total mechanical energy of zero!! for escape Escape energy - the minimum EK needed to project a mass (m2) from the surface of another mass (m1) to escape the gravitational force of m1 Escape speed - the minimum speed needed to project a mass (m2) from the surface of another mass (m1) to escape the gravitational force of m1 Binding energy - the additional EK needed by a mass (m2) to escape the gravitational force of m1 (similar to escape energy but applies to objects that possess Ek i.e. satellites).

23 To calculate the escape energy or the escape velocity of a mass (m2):
To calculate the binding energy of a mass (m2): binding energy Calculate the escape velocity of any object on the Earth’s surface. The escape velocity is the same for all objects on the Earth’s surface while the escape energy is different for different massed object.

24 What is Ek and EM of an orbiting body (satellite)?
this is always true of satellites

25 for an orbiting satellite !!
Note that the total energy is negative since the satellite is “bound” to the central body. read p p #12 is interesting! extra p p ,26 look fun

26 a) What is the speed of Earth in orbit about the Sun?
b) What is the total energy of Earth? c) What is the binding energy of Earth? d) If Earth was launched from the surface of the Sun to its present orbit then what velocity must it be launched with (Ignore the radius of Earth.)? e) If Earth came to rest and fell to the Sun then what velocity would it have when it hit the Sun (Ignore radius of Earth.)? me= 5.98x1024 kg ms= 1.99x1030 kg re= 1.49x1011 m (of orbit) rs= 6.96x108 m (of the body) G= 6.67x10-11 Nm2kg-2

27 a) or b) c) The binding energy is x 1033 J

28 d) e)


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