- 1. • Francis Marlon Cabredo • Mary Minette Geñorga • Mary Judith Verdejo
- 2. VCircular motion: Motion of a body along a circular path is known as circular motion. Uniform circular motion: Motion of a body long a circular path with the constant speed is called uniform circular motion.
- 3. ▪Uniform Circular Motion is the motion of an object in a circle with a constant or uniform speed
- 4. SPEED: ▪ In UCM, distance= the circumference of the circle ▪ T(period) is the time (or number of seconds) to make one revolution
- 5. PROBLEM 1: Tire Balancing Machine • The wheel of a car has a radius of 0.29m and is being rotated at 0.072 sec/rev on a tire-balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving. Use 3.14 as the value for .
- 7. VELOCITY ▪ The velocity of an object in UCM has a constant magnitude and a constant change in direction ▪ The object does not have constant velocity since its direction changes at every point along the circle. ▪ Also, The velocity is always tangent to the path of the object.
- 8. In UCM, the velocity is always tangent to the path of the object. The tangent specifies the direction of the motion. VA VD VB VC At a time t0, the car is located at point A with a velocity of VA, this is tangent to the circle at point A. This means that… The direction of the velocity of the Car at point A is due west.
- 9. CENTRIPETAL ACCELERATION ▪ in UCM, an object has no constant velocity. Therefore, If there is a change in velocity, then that must mean that an object has acceleration. Centripetal Acceleration. This acceleration has constant magnitude but changing direction and is directed radially inwards.
- 10. CENTRIPETAL ACCELERATION • Symbol: ac • Vector • It is the rate of change of Tangential velocity • Always perpendicular to the path of the motion. • Points toward the center of the circle. VA VD VB VC A B C D
- 12. PROBLEM 2: Centripetal Acceleration The bobsled track at the 1994 Olympics in Norway, contained turns with radii of 33 m and 24 m, as the figure illustrates. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was achieved in the two-man event..
- 13. From ac=v2/r Radius=33m Radius=24m 2 2 /35 33 )/34( sm m sm ac 2 2 /48 24 )/34( sm m sm ac
- 15. NEWTON’S SECOND LAW OF MOTION STATES THAT ALL ACCELERATIONS ARE CAUSED BY A NET FORCE ACTING ON AN OBJECT. IN THE CASE OF UCM, THE NET FORCE IS A SPECIAL FORCE CALLED THE CENTRIPETAL FORCE .
- 16. CENTRIPETAL IS LATIN FOR "CENTER SEEKING". IT IS THE INWARD NET FORCE WHICH KEEPS AN OBJECT MOVING WITH A UNIFORM VELOCITY ALONG A CIRCULAR PATH. THIS FORCE IS DIRECTED ALONG THE RADIUS TOWARDS THE CENTER. CENTRIPETAL FORCE
- 17. F ma F mv r 2 •APPLYING NEWTON’S SECOND LAW OF MOTION CENTRIPETAL FORCE Centripetal Force ac=v2/r F=ma mv2 ___ r
- 19. PROBLEM 3A: Centripetal Force • A 400-g rock attached to a 1.0-m string is whirled in a horizontal circle at a constant speed of 10.0m/s. Neglecting the effects of gravity, what is the centripetal force acting on the rock?
- 20. SOLUTION: Given: V=10.0 m/s R=1.0 m M=400g= 0.4 kg Find: Fc (0.4 kg)(10 m/s)2___________ 1.0 m mv2___ r 40 N 40 kg m/s2
- 21. • FORCES SUCH AS THE GRAVITATIONAL FORCE (w=mg), TENSION FORCE (tied to a string; pushing/ pulling), FRICTIONAL FORCE (Ex. a car turning) and Normal force (on a surface) can be the centripetal forces • GRAVITATIONAL FORCE. For satellites in orbit around a planet, the centripetal force is supplied by gravity. • TENSILE FORCE. For an object swinging around on the end of a rope in a horizontal plane, the centripetal force on the object is supplied by the tension of the rope.
- 22. THE CENTRIFUGAL FORCE ACTS AWAY FROM THE CENTER. THE WORD ITSELF MEANS “FLEEING FROM THE CENTER” . THIS FORCE IS A FICTITIOUS FORCE. IT DOESN’T ACT ON A BODY IN MOTION, BUT ONLY ON NON-INERTIAL COORDINATE SYSTEMS SUCH AS A ROTATING ONE. CENTRIFUGAL FORCE
- 23. PROBLEM 5: Which way will the object go? • An object on a guideline is in uniform circular motion. The object is symbolized by a dot, and at point O it is release suddenly from its circular path. • If the guideline is cut suddenly, will the object move along OA or OP ?
- 24. ANSWER: • the object would move along the straight line between points O and A, not on the circular arc between points O and P.
- 25. SOLUTION: • NEWTON’S LAW OF MOTION: “An object continues in a state of rest/ motion at a constant speed unless compelled to changes to its net force.” When the object was suddenly released from its path, there was no longer a net force (i.e. centripetal force) being applied to the object. In the absence of a net force, the object will continue to move at a constant speed but, along a straight line in the direction it had at the time of release.
- 26. CENTRIFUGAL FORCE vs. INERTIA If you let go of the rope (or the rope breaks) the object will no longer be kept in that circular path and it will be free to fly off on a tangent.
- 27. 27 Uniform circular motion emphasizes that 1.The speed, or the magnitude of the velocity vector, is constant. 2. Direction of the vector is not constant. 3. Change in direction, means acceleration 4. “Centripetal acceleration” , points toward the center of the circle. 5. “Centripetal Force” is the net force that causes centripetal acceleration
- 28. Assignment: • Solve the problems in the Practice Math in yourbook, page 19. (1/2 crosswise pad paper) • Study for the Long Test on, July 18. Topics: Types of Forces, Friction, Laws of Motion, Uniform Circular Motion
- 29. AP PHYSICS Circular Motion Mrs. Coyle Practical ad Explorational Physics Padua et al You and The Natural world Physics Navaza, Valdes INTERNET YOUTUBE GOOGLE…….. SOURCES: • Francis Marlon Cabredo • Mary Minette Geñorga • Mary Judith Verdejo APPLICATIONS PROBLEMS