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UT Arlington PHYS 1441 - PHYS 1441 Lecture Notes

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PHYS 1441 Section 002 Lecture 23 Monday Dec 6 2010 Dr Jaehoon Yu Monday Dec 6 2010 Similarities Between Linear and Rotational Quantities Conditions for Equilibrium How to Solve Equilibrium Problems A Few Examples of Mechanical Equilibrium Elastic Properties of Solids Density and Specific Gravity PHYS 1441 002 Fall 2010 Dr Jaehoon Yu 1 Announcements The Final Exam Date and time 11am 1 30pm Monday Dec 13 Place SH103 Comprehensive exam Covers from CH1 1 what we finish Wednesday Dec 8 Plus appendices A 1 A 8 Combination of multiple choice and free response problems Bring your Planetarium extra credit sheet to the class next Wednesday Dec 8 with your name clearly marked on the sheet Reading assignments Ch9 3 9 7 Monday Dec 6 2010 PHYS 1441 002 Fall 2010 Dr Jaehoon Yu 2 Similarity Between Linear and Rotational Motions All physical quantities in linear and rotational motions show striking similarity Quantities Mass Length of motion Speed Acceleration Force Work Linear Mass Distance Dr Dt Dv a urDt I Angle Radian L Dq Dt Dw a Dt v r Force F ma r r Work W F d ur r P F v u r r p mv Momentum Monday Dec 6 2010 Moment of Inertia M Power Kinetic Energy Rotational Kinetic 1 K mv 2 2 w r ur Torque I Work W tq P u r ur L I Rotational PHYS 1441 002 Fall 2010 Dr Jaehoon Yu 1 K R I 2 2 3 Conditions for Equilibrium What do you think the term An object is at its equilibrium means The object is either at rest Static Equilibrium or its center of mass is moving at a constant velocity Dynamic Equilibrium When do you think an object is at its equilibrium ur Translational Equilibrium Equilibrium in linear motion F 0 Is this it The above condition is sufficient for a point like object to be at its translational equilibrium However for an object with size this is not sufficient One more condition is needed What is it Let s consider two forces equal in magnitude but in opposite direction acting on a rigid object as shown in the figure What do you think will happen F d d CM F Monday Dec 6 2010 r The object will rotate about the CM Thus the net t 0 torque acting on the object about any axis must be 0 For an object to be at its static equilibrium the object should not have linear or angular speed vCM 0 w 0 PHYS 1441 002 Fall 2010 Dr Jaehoon Yu 4 More on Conditions for Equilibrium To simplify the problem we will only deal with forces acting on x y plane giving torque only along z axis What do you think the conditions for equilibrium be in this case The six possible equations from the two vector equations turns to three equations F F ur F 0 x y 0 0 r AND t 0 t z 0 What happens if there are many forces exerting on an object 2 F F 1 r5 Or O 5 F F3 F4 Monday Dec 6 2010 If an object is at its translational static equilibrium and if the net torque acting on the object is 0 about one axis the net torque must be 0 about any arbitrary axis Why is this true Because the object is not moving moving no matter what the rotational axis is there should not be any motion It is simply a matter of mathematical PHYS 1441 002 Fall 2010 Dr Jaehoon 5 manipulation Yu 1 2 3 4 5 6 7 How do we solve static equilibrium problems Select the object to which the equations for equilibrium are to be applied Identify all the forces and draw a free body diagram with them indicated on it with their directions and locations properly indicated Choose a convenient set of x and y axes and write down force equation for each x and y component with correct signs Apply the equations that specify the balance of forces at equilibrium Set the net force in the x and y directions equal to 0 Select the most optimal rotational axis for torque calculations Selecting the axis such that the torque of one of the unknown forces become 0 makes the problem easier to solve Write down the torque equation with proper signs Solve the force and torque equations for the desired unknown quantities Monday Dec 6 2010 PHYS 1441 002 Fall 2010 Dr Jaehoon Yu 6 Example for Mechanical Equilibrium A uniform 40 0 N board supports the father and the daughter each weighing 800 N and 350 N respectively and is not moving If the support or fulcrum is under the center of gravity of the board and the father is 1 00 m from CoG what is the magnitude of the normal force n exerted on the board by the support 1m F M Fg x n M Bg Since there is no linear motion this system is in its translational equilibrium D F 0 x MDg F y Therefore the magnitude of the normal force n M B g M F g M D g 0 n 40 0 800 350 1190 N Determine where the child should sit to balance the system The net torque about the fulcrum by the three forces are Therefore to balance the system the daughter must sit Monday Dec 6 2010 M B g 0 n 0 M F g 1 00 M D g x 0 x MFg 800 1 00m 1 00m 2 29m MDg 350 PHYS 1441 002 Fall 2010 Dr Jaehoon Yu 7 Example for Mech Equilibrium Cont d Determine the position of the child to balance the system for different position of axis of rotation Rotational axis 1m F M Fg x n x 2 D M Fg M Bg The net torque about the axis of rotation by all the forces are M B g x 2 M F g 1 00 x 2 n x 2 M D g x 2 0 n M B g M F g M D g M B g x 2 M F g 1 00 x 2 M B g M F g M D g x 2 M D g x 2 Since the normal force is The net torque can be rewritten M F g 1 00 M D g x 0 Therefore x Monday Dec 6 2010 MFg 800 1 00m 1 00m 2 29m MDg 350 PHYS 1441 002 Fall 2010 Dr Jaehoon Yu What do we learn No matter where the rotation axis is net effect of the torque is identical 8 Example 9 7 A 5 0 m long ladder leans against a wall at a point 4 0m above the ground The ladder is uniform and has mass 12 0kg Assuming the wall is frictionless but ground is not determine the forces exerted on the ladder by the ground and the wall FW FBD FGy O FGx m g First the translational equilibrium using components F F F 0 F mg F 0 x Gx y W Gy …


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UT Arlington PHYS 1441 - PHYS 1441 Lecture Notes

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