*FAP 0015 PHYSICS I*Uniform circular motionAngular displacement, angular velocity, angular acceleration, period, frequencyCentripetal accelerationDynamic equation, Centripetal force Linear vs circular motionNewtons Law of GravitationWeight, Gravity, and satellite in circular orbit

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Lesson OutcomesAt the end of the lesson, students should be able to:define angular displacement, angular velocity, angular acceleration, period and frequency.state the relation between the linear and circular parts of the motions.apply Newtons universal laws of gravitation to determine the weight of a body.use free-body diagrams to solve problems involving centripetal forces and accelerations.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Tie a string to a stone and then swing it above your head horizontally. The motion of the stone is an example of circular motion.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Uniform Circular MotionIts a motion of a particle around a circle or circular arc at constant (uniform) speed.The velocity is always directed tangent to the circle in the direction of the motion.Period T, is the time required to travel once around the circle, that is to complete one revolution.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Uniform circular motionIn Fig. (a), at time t0 the velocity is tangent to the circle at point O and at a later time t the velocity is a tangent at point P.As the object moves from O to P, the radius traces out the angle , and the velocity vector change the direction.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I* In Fig. (b), the velocity vector at time t is redrawn with its tail at O parallel to itself. The angle between the two vectors indicates the change in the direction. Since the radii CO and CP are perpendicular to the tangent at O and P so it follows that

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*The acceleration a, is the change in v in velocity divided be elapsed time t, a = v/t. The resultant velocity vector, v, has a new direction after an elapsed time t = t - t0Fig. (c) shows two velocity vectors oriented at the angle , together with the vector v that represents the change in the velocity vectors (vt0 + v)= vt

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I* Fig. (d) shows the sector of the circle COP. When t is very small the arc length OP is straight line and equals to the distance vt that traveled by the object. In this limit, COP is an isosceles triangle with apex angle .

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I* Compare COP in Fig (d) with triangle in Fig (c). They are similar because both are isosceles triangles with apex angles labeled are same. ThusThis equation can be solved for , to show the magnitude of centripetal acceleration ac,Fig. (c)

FAP 0015 PHYSICS I

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*FAP 0015 PHYSICS I*Linear vs Circular

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*FAP 0015 PHYSICS I*When an object is moving in a uniform circular motion, there is an acceleration towards the center of the circular path. (centripetal acceleration)Dynamics of uniform circular motionThe magnitude of the acceleration is

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I* The force is called the centripetal force. The magnitude of this force can be calculated by using Newtons 2nd. law of motion.To provide this acceleration, there must be a force acts towards the center of the circular path.

FAP 0015 PHYSICS I

*FAP 0014 PHYSICS I* Fc = macEquations describing uniform circular motion

FAP 0014 PHYSICS I

*FAP 0015 PHYSICS I*Radius and Centripetal Acceleration The bobsled track contained turns with radii of 33 m and 24 m. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was achieved in the two-man event. Express the answers as multiples of g = 9.8 m/s2.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*UCM and Equilibrium,Conceptual Problem:A car moves at a constant speed, and there are three parts to the motion. It moves along a straight line toward a circular turn, goes around the turn, and then moves away along a straight line. In each of the parts, is the car in equilibrium?

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Speed and Centripetal ForceThe model airplane has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. The path of the airplane and its guideline lie in the same horizontal plane, because the weight of the plane is balanced by the lift generated by its wings. Find the tension in the guideline (length 17 m) for speed of 19 and 38 m/s.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*ExampleA 1200.0 kg car rounded a corner of a radius r = 45 m. If the coefficient of static friction s = 0.82, what is the greatest speed the car can have in the corner without skidding?

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*FAP 0015 PHYSICS I*ExampleIf a lateral acceleration of 8.9 m/s2 represents the maximum ac that can be attained without skidding out of the circular path, and if the car is traveling at a constant 45 m/s, what is a minimum radius of curve it can negotiate? If the driver rounding a flat with unbanked curve with radius R. If the coefficient of friction between the tires and road is s, what is the maximum speed v at which he can take the curve without skidding?

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Banked CurveA vehicle can negotiate a circular turn without relying on static friction to provide the centripetal force if the turn is banked at an angle relative to the horizontal.

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*FAP 0015 PHYSICS I*To provide centripetal force (without friction) :For a given speed, v, the centripetal force needed for a turn of radius r can be obtained by banking the turn at an angle , independent of the mass of vehicle.What would happen if a vehicle moves at a speed much larger or much smaller than v?

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*GRAVITYGravity is a fundamental force in sense that cannot be explained in terms of any other force.Fundamental forces are: gravitational, electromagnetic and nuclear forces.These forces seem to be responsible for everything that happens in the universe.Gravitational forces act between all bodies in the universe and hold together planets, stars and galaxies of stars.

FAP 0015 PHYSICS I

Gravity**FAP 0015 PHYSICS INewton's apple tree, Trinity College, Cambridge, England

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*FAP 0015 PHYSICS I*"Gravity cannot be held responsible for people falling in love.A quote by Albert EinsteinYou can't blame gravity for falling in love.

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*FAP 0015 PHYSICS I*Newtons Law of Universal Gravitation Newton proposed a force law saying that every particle attracts any other particle with a gravitational force. Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Gravitational Attraction of Spherical BodiesA uniform sphere with a radius R and mass M, and object of mass m is brought near the sphere at the distance r from the center.Newton showed that, the net force exerted by the sphere on the mass, m is the same as if all the masses of the sphere were concentrated at its center this force is,

FAP 0015 PHYSICS I

The force of gravity between any two point objects of mass m1 an m2 is attractive and of magnitudeF-gravity forms action-reaction pair.where G is the universal gravitational constant, G = 6.67 x 10-11 Nm2/kg2Newtons Law of Universal Gravitation

*FAP 0015 PHYSICS I*Dependence of the Gravitational Force on Separation Distance, rThe force diminishes rapidly with the distance, but never completely vanishes. Thus, gravity is a force of infinite range.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*WeightPreviously we defined the weight of a body as the attractive gravitational force exerted on it by the earth. Now, we can broaden the definition as: the weight of the body is the total gravitational force exerted on the body by all other bodies in the universe.When the body near the earth, we can neglect all other gravitational forces and consider the weight as just the earths gravitational attraction.At the surface of the moon we can neglect all others forces and consider the bodys weight to be gravitational attraction of the moon, and so on.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*So, the weight of a body of mass, m, near the earth surface,where Me and re are the mass and radius of the earth respectively.For a body of mass, m, at a distance h from the earth surface,Me= 5.98 1024 kgre= 6.38 106 m

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*FAP 0015 PHYSICS I*Mass in Circular Orbit

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Exercise*EPF 0014 PHYSICS I*Determine the average radius of orbit of the moon around the earth based on its period of orbit. T = 28 days = 28 24 3600 s = 2.4 106 s G = 6.67 x 10-11 Nm2/kg2me= 5.98 1024 kgr = 387.5 106 m= 58.2 1024 m3

EPF 0014 PHYSICS I

*FAP 0015 PHYSICS I*Gravitational and Inertial MassWe have had two definitions of mass: The property of an object that resists change in state of motion. Appears as the constant in Newtons second law F = ma. It is called inertial mass.The property of an object that determines the strength of the gravitational force F = mg. It is called gravitational mass.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*ExampleFind the acceleration of gravity on the surface of the moon.The lunar rover has a mass of 225 kg. What is its weight on the earth and on the moon? [note, the mass of the moon is Mm = 7.35 x 1022 kg and its radius is Rm = 1.74 x 106 m.]

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*FAP 0015 PHYSICS I*Conceptual Question Other things being equal, would it be easier to drive at high speed around unbanked horizontal curve on the moon than to drive around the same curve on the earth? Explain.

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*FAP 0015 PHYSICS I*REASONING AND SOLUTIONThe maximum safe speed with which a car can round an unbanked horizontal curve of radius r is given by . Since the acceleration due to gravity on the moon is roughly one sixth that on earth, the safe speed for the same curve on the moon would be less than that on earth. In other words, other things being equal, it would be more difficult to drive at high speed around an unbanked curve on the moon as compared to driving around the same curve on the earth.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Conceptual QuestionA stone is tied to a string and whirled around in a circular path at a constant speed. Is string more likely to break when the circle is horizontal or when it vertical? Account for to your answer assuming the constant speed is the same in each case.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*REASONING AND SOLUTIONWhen the string is whirled in a horizontal circle, the tension in the string, FT, provides the centripetal force which causes the stone to move in a circle. Since the speed of the stone is constant, and the tension in the string is constant.When the string is whirled in a vertical circle, the tension in the string and the weight of the stone both contribute to the centripetal force, depending on where the stone is on the circle.

FAP 0015 PHYSICS I

*EPF 0014 PHYSICS I*Free body diagramsHorizontal circleVertical circleAt the top of circleAt the bottom of circle

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*FAP 0015 PHYSICS I*Now, however, the tension increases and decreases as the stone traverses the vertical circle. When the stone is at the lowest point in its swing, the tension in the string pulls the stone upward, while the weight of the stone acts downward. Therefore, the centripetal force is .

Thus This tension is larger than in the horizontal case. Therefore, the string has a greater chance of breaking when the stone is whirled in a vertical circle.

FAP 0015 PHYSICS I

*FAP 0015 PHYSICS I*Loop the LoopThe rider who perform the loop-the loop trick know that he must have a minimum speed at the top of the circle to remain on the track.

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*FAP 0015 PHYSICS I*At point 3,

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The rollercoaster**At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers will not fall out? Assume a radius of curvature of 7.4 m.FAP 0015 PHYSICS I

FAP 0015 PHYSICS I