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SBBS 112: GENERAL PHYSICS

SBBS 112: GENERAL PHYSICS. University of Health and Allied Sciences SECOND SEMESTER 2013/2014. Schedule. Topics for yesterday and today Vectors Newton’s laws of motion Linear momentum Collisions and impulse Equilibrium Projectiles Circular motion. Reading: Y&F Chs . 1-5, 8, 11.

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SBBS 112: GENERAL PHYSICS

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  1. SBBS 112: GENERAL PHYSICS University of Health and Allied Sciences SECOND SEMESTER 2013/2014

  2. Schedule • Topics for yesterday and today • Vectors • Newton’s laws of motion • Linear momentum • Collisions and impulse • Equilibrium • Projectiles • Circular motion Reading: Y&F Chs. 1-5, 8, 11

  3. Schedule • Topics for week 3 • Rotational motion • Moment of inertia • Kinetic energy of rotational motion • Work of varying forces • Work-energy theorem • Conservation of energy • Conservative and non-conservative forces • Vibrations and oscillatory motion Reading: Y&F Chs. 6, 7, 9, 10, 12, 13

  4. Schedule (cont.) • Topics for week 5 • Heat • Temperature and temperature scales • Thermal equilibrium; zerothlaw of thermodynamics • Heat capacity • Heat transfer mechanisms • Equation of state; ideal gas law • Internal energy • First law of thermodynamics and applications • Kinetic theory • Entropy

  5. Cosine rule C=A+B A q B

  6. Example

  7. Coefficient of restitution

  8. Circular motion • Circular motion designates motion on a circle Circular motion Rectilinear motion a v(t1) v(t2) v(t1) a R a v(t2) v(t2) v(t1) a is parallel to vand can be zero a has a radial component and is nonzero NB: Velocity is always tangential to path

  9. Types of circular motion Y&F, Sect. 3.2 • There are two types of circular motion: uniform and non-uniform • Resolve acceleration into tangential (atan) and radial (arad) components • Tangential accel. changes speed • Radial accel. changes direction (- sign implies toward O) • Uniform circular motion: atan = 0 hence constant speed E.g. • Vehicle rounding a constant-radius curve at constant speed • Satellite in circular orbit • Non-uniform circular motion: atan ≠ 0 E.g. • Skater on a vertical circular track • In general, magnitude of acceleration is Non-uniform circular motion Uniform circular motion O v(t1) v(t1) arad arad a atan=0 atan v(t2) v(t2) O Constant speed, zero tangential acceleration Non-zero speed and tangential acceleration

  10. Acceleration in Uniform Circular Motion 180o - Df In uniform circular motion acceleration is radial atan = 0 and arad depends on speed and radius P1 O' v(t)=v1 Ds R Df Df P2 R O v(t+Dt)=v2

  11. Period and Angular Speed • Time for one complete revolution is the period T • The number of revolutions in unit time is frequency f = 1/T • The angular frequency w= 2pf =2p/T • w is also called angular speed (unit: radians per second) • Angular velocity is a vector pointing perpendicular to the plane of the circle, in a right-handed sense • In one revolution, distance travelled is the circumference 2pR of the circle • Speed v = 2pR/T = 2pRf = Rw • Centripetal acceleration arad = v2/R = Rw2 = 4p2R/T2 = 4p2f2R arad = v2/Ralso gives radial acceleration when speed is v at a point of radius R on any curvilinear path v arad R

  12. Example Problem • A particle moves in a circular orbit of radius 1.5 m with constant orbital speed 3 m s-1. Determine the • acceleration • period and • angular frequency of the motion.

  13. Example Problem Solution Since orbital speed is constant, acceleration is radial: arad = v2/R = (32/1.5) m s-2 = 6 m s-2. The period is T = 2pR/v = 3.14 s. The angular frequency is w = 2p/T = 2 rad s-1. Note that frequency can also be expressed as number of revolutions in unit time (usually 1 minute). In this example, f = 0.318s-1 can be given as 19.08rpm. w = 2pRPM/60 • A particle moves in a circular orbit of radius 1.5 m with constant orbital speed 3 m s-1. Determine the • acceleration • period and • angular frequency of the motion.

  14. Non-uniform circular motion • Recall: In uniform circular motion tangential acceleration is zero and speed is constant • In non-uniform circular motion • Tangential acceleration is not zero • Speed is not constant • Tangential acceleration changes speed • Tangential acceleration is not constant because speed is not constant • Centripetal acceleration is still given byarad = v2/R

  15. Example Problem • A body moves in a circle of fixed radius R. If its speed increases uniformly from v1 to v2 in time t, find the • change in centripetal acceleration and • average tangential acceleration.

  16. Example Problem Solution Initial centripetal accelerationarad1 = v12/RFinal centripetal accelerationarad2 = v22/RChange in centripetal accelerationDarad= arad2 – arad1 = (v22 – v12)/R Average tangential acceleration <atan> = (v2 – v1)/tSince speed changes uniformly, the average also represents the instantaneous tangential acceleration atan. Note radial acceleration is not uniform even though tangential acceleration is uniform and speed changes uniformly. This is a manifestation of the quadratic dependence of centripetal acceleration on speed. • A body moves in a circle of fixed radius R. If its speed increases uniformly from v1 to v2 in time t, find the • change in centripetal acceleration and • average tangential acceleration.

  17. Dynamics of circular motion Reading: Y&F Sect. 5.4 • The force needed for the radial component of acceleration is known as centripetal force • Centripetal force is always directed towards the centre of the circle • By Newton’s Second Law, a body of mass m in circular motion is subject to a centripetal force Frad = marad = mv2/R • If a system of forces act on the body, Frad equals the radial component of the resultant • In uniform circular motion, the resultant is radial (no tangential component) Frad is provided by the tension in the string

  18. Example Problem (Y&F) • A sled with mass 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.0 m rope to a post set in the ice. Once given a push the sled revolves uniformly in a circle around the post. If the sled makes 5 complete revolutions every minute, find the force exerted on it by the rope.

  19. Example Problem (Y&F) Solution Period T = time for 1 rev. = time for 5 revs. /5 = 12 s Angular frequency w = 2p/T= 2p/12 rad/s = 0.52 rad/sarad = Rw2Centripetal force Frad = marad = mRw2 = 33.8 N =34 N See alternative solution in Y&F • A sled with mass 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.0 m rope to a post set in the ice. Once given a push the sled revolves uniformly in a circle around the post. If the sled makes 5 complete revolutions every minute, find the force exerted on it by the rope.

  20. Motion in Horizontal Circle y Conical pendulum x b • The conical pendulum comprises a ball of mass m moving in a horizontal circle at the end of a string inclined at an angle b to the vertical • Apply Newton’s Second Law • Absence of vertical motion implies vertical component of resultant of all forces is zero • Horizontal component of resultant equals centripetal force

  21. Examples Governor Centrifuge http://www.euronuclear.org/ http://www.gunt.de

  22. Vehicle rounding a curve Rounding a flat curve • Friction provides centripetal force needed to keep vehicle in circular path • Friction force depends on normal force and coefficient of friction (Section 5.3) • For motion without skidding Frad < fs • Maximum speed

  23. Vehicle rounding a curve (cont.) Rounding a frictionless banked curve • If the curve is banked, the radial component of the normal force can provide the needed centripetal force

  24. Vehicle rounding a curve (cont.) Rounding a banked curve with friction β β Fs ncosβ n • In addition to the radial component of the normal force, friction contributes to centripetal force nsinβ Fs cosβ Fs sin β Fs w = mg

  25. Motion in a vertical circle The Ferris Wheel (Y&F) • A passenger on a carnival Ferris wheel moves in a vertical circle of radius R with constant speed v. The seat remains upright during the motion. Find expressions for the force the seat exerts on the passenger at the top of the circle and at the bottom. http://norman.walsh.name/

  26. Motion in a vertical circle (cont.) Reading: Y&F Sect. 5.4, 7.1 A ball moving in a vertical circle Newton’s Second Law Conservation of energy Work done by tension is zero because it is perpendicular to displacement E = Gravitational P.E. + K.E. is conserved T q

  27. Vertical Circle (cont.) E > 2mgR E < 2mgR q =180o -qmax qmax q E = 2mgR v = 0

  28. Example Problem (Y&F, Ex. 7.4) Solution At top, total energy E = mgR At bottom v = √(2E/m) = √(2gR) = 7.67 m s-1 Frad = n – w = mv2/Rn = mv2/R + w = m(2gR)/R + w = 2mg + mg = 735 N • Your cousin Throckmorton skateboards down a curved playground ramp. If we treat Throcky and his skateboard as a particle, he moves through a quarter-circle with radius R = 3.00 m (Fig. 7.9). The total mass of Throcky and his skateboard is 25.0 kg. He starts from rest and there is no friction. (a) Find his speed at the bottom of the ramp. (b) Find the normal force that acts on him at the bottom of the curve.

  29. Satellite motion Elliptic orbit • Satellite is body orbiting a planet • Moon (natural satellite) • Space station (artificial) • Satellite orbit can be circular or elliptic • General orbit is elliptical • Other orbit types are unbound: parabolic, hyperbolic • Centripetal force is provided by gravity Earth at focus R Satellite http://media.pearsoncmg.com/bc/aw_young_physics_11/pt1a/Media/CircularMotion/SatellitesOrbit/Main.html

  30. http://media.pearsoncmg.com/bc/aw_young_physics_11/pt1a/Media/CircularMotion/SatellitesOrbit/Main.htmlhttp://media.pearsoncmg.com/bc/aw_young_physics_11/pt1a/Media/CircularMotion/SatellitesOrbit/Main.html

  31. Satellite Motion (cont.): Circular orbits • Uniform circ. motion • Radial acceleration • Orbital speed • Orbital period

  32. Examples Problems Solutions • Determine the distance between a geostationary satellite and: a) the centre of the earth; b) the surface of the earth. • Determine the orbital speed of a geostationary satellite

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