Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #5Monday, Sept. 18, 2002Dr. Jaehoon Yu1. Newton’s Laws of Motion2. Application of Newton’s Laws3. Friction4. Newton’s laws and its use in uniform and non-uniform circular motionToday’s homework is homework #6, due 1am, next Wednesday!!Remember the first term exam on Monday, Sept. 30!!Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Uniform Circular Motion• A motion with a constant speed on a circular path.– The velocity of the object changes, because the direction changes– Therefore, there is an accelerationvivf∆vrvivf∆θ∆θ∆rr2r1The acceleration pulls the object inward: Centripetal Accelerationtvttvvaifif∆∆=−−=Average Accelerationrrvv ∆=∆=∆ θAngle is ∆θrvrvvrvtraattr200limlim =×=∆∆==→∆→∆Instantaneous AccelerationIs this correct in dimension?What story is this expression telling you?rrvv∆=∆rrtva∆∆=Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Non-uniform Circular Motion• Motion not on a circle but through a curved path– Requires both tangential (at) and radial acceleration (ar) arataaratadtvdat =rvar2=arataTangential Acceleration:Radial Acceleration:∧r∧θθxyOr∧∧−=+= rrvdtvdaaa tr2θTotal Acceleration:Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Relative Velocity and AccelerationThe velocity and acceleration in two different frames of references can be denoted, using the formula in the previous slide:OFrame Sr’O’Frame S’v0v0trtvrr0' −=Galilean transformation equation 0' vvv −=What does this tell you?The accelerations measured in two frames are the same when the frames move at a constant velocity with respect to each other!!!The earth’s gravitational acceleration is the same in a frame moving at a constant velocity wrt the earth.0'vdtrddtrd−=0' vvv −=dtvddtvddtvd0'−=constant is when ,'0vaa =Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Newton’s First Law and Inertial FramesGalileo’s statement on natural states of matter: Any velocity once imparted to a moving body will be rigidly maintained as long as the external causes of retardation are removed!!This statement is formulated by Newton into the 1stlaw of motion (Law of Inertia):In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity. What does this statement tell us? 1. When no force is exerted on an object, the acceleration of the object is 0. 2. Any isolated object, the object that do not interact with its surrounding, is either at rest or moving at a constant velocity.3. Objects would like to keep its current state of motion, as long as there is no force that interferes with the motion. This tendency is called the Inertia.A frame of reference that is moving at constant velocity is called an Inertial FrameMonday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Newton’s Second Law of MotionamFii=∑The acceleration of an object is directly proportional to the net force exerted on it and inversely proportional to the object’s mass. How do we write the above statement in a mathematical expression?xiixmaF =∑Since it’s a vector expression, each component should also satisfy:From the above vector expression, what do you conclude the dimension and unit of force are?]][[]][[2−= LTMam22/]][[]][[][ smkgLTMamForce ⋅===−The dimension of force is The unit of force in SI is See Table 5.1 for lbs to kgm/s2 conversion. yiiymaF =∑ziizmaF =∑lbssmkgN41/112≈⋅≡Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Free Body Diagrams• Diagrams of vector forces acting on an object ⇒ A great tool to solve a problem using forces or using dynamics1. Select a point on an object and w/ information given2. Identify all the forces acting only on the selected object3. Define a reference frame with positive and negative axes specified4. Draw arrows to represent the force vectors on the selected point5. Write down net force vector equation6. Write down the forces in components to solve the problems⇒ No matter which one we choose to draw the diagram on, the results should be the same, as long as they are from the same motionMWhich one would you like to select to draw FBD?What do you think are the forces acting on this object?Gravitational forcegMFG=A force supporting the object exerted by the floorMeWhich one would you like to select to draw FBD?What do you think are the forces acting on this elevator?NFgMFG=Gravitational force The force pulling the elevator (Tension)mWhat about the box in the elevator?Gravitational forceNormal forceNFTFgMFG =gmFGB=TFgMFG=NFgmFBG=NFMonday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Acceleration Vector aF2F1Example 5.1Determine the magnitude and direction of acceleration of the puck whose mass is 0.30kg and is being pulled by two forces, F1 and F2, as shown in the picture, whose magnitudes of the forces are 8.0 N and 5.0 N, respectively.()NFFx0.460cos0.8cos11=×==1oθθ1=60οθ2=−20οxxxxmaNFFF==+=+=7.87.40.421Components of F1Components of F2Components oftotal force F2/293.07.8smmFaxx===Magnitude and direction of acceleration a()NFFy9.660sin0.8sin11=×==1oθ()NFFx7.420cos0.5cos222=−×==oθ()NFFy7.120sin0.5sin222−=−×==oθyyyymaNFFF ==−=+= 2.57.19.6212/173.02.5 smmFayy===( ) ( )222/341729 sma =+=o302917tantan 11===−−xyaaθ2/1729 smjijaiaayx+=+=∧∧∧∧Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Gravitational Force and WeightSince weight depends on the magnitude of gravitational acceleration, g, it varies depending on geographical location.The attractive force exerted on an object by the Earth Gravitational Force, FggmamFG==Weight of an object with mass M isMggMFWG===By measuring the forces one can determine masses. This is why you can measure mass using spring scale.Actual unit of weight is in the unit of force but the unit of mass is commonly used in place of force.Monday, Sept. 18, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu10Newton’s Third Law (Law of Action and Reaction)If two objects interact, the force, F12, exerted on object 1 by object 2 is equal in magnitude and opposite in direction to the force, F21, exerted on object 1 by object 2. F12F212112FF −=The action force is equal in magnitude to the reaction force but in opposite
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