How to Prepare Rotational Motion (Physics) for JEE Main

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Important tips for physics for JEE Main by Prof. Ashish Dubey.

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How to Prepare Rotational Motion (Physics) for JEE Main

  1. 1. Dynamics of RotationalMotion
  2. 2. Cross Product
  3. 3. Cross Product The Cross Product (or Vector Product) of two vectors A and B isa multiplication of vectors where the result is a vector quantity Cwith a direction perpendicular to both vectors A and B, and themagnitude equal to ABsin :sinABC  Magnitude of the cross (vector)product of two vectors A and BBAC A is the magnitude of the first vector, B is the magnitude of thesecond vector and  is the angle between the two vectors. The direction of the cross product is perpendicular to the planeformed by the two vectors in the product. This leaves two possiblechoices which are resolved by using the Right Hand Rule.
  4. 4. Cross ProductProperties of the Cross Product Cross Product is Anti-Commutative. Parallel Vectors have Cross Product of zero. Cross Product obeys the Distributive Law. Product Rule for Derivative of a Cross Product.ABBACABACBA )(BdtAddtBdABAdtd  )(
  5. 5. Cross Product The above formula for the cross product is useful when the magnitudesof the two vectors and the angle between them are known. If you only know the components of the two vectors:kBABAjBABAiBABABA xyyxzxxzyzzyˆ)(ˆ)(ˆ)( sinABC yzzyx BABAC zxxzy BABAC xyyxz BABAC BACComponents of crossproduct vector
  6. 6. Cross Product Right-handed coordinatesystem, in which:kji ˆˆˆ ikj ˆˆˆ jik ˆˆˆ  Left-handed coordinatesystem, in which:kji ˆˆˆ ikj ˆˆˆ jik ˆˆˆ 
  7. 7. Torque Net force applied to a body gives that body an acceleration. What does it take to give a body angular acceleration? Force is required! It must be applied in a way that gives a twisting orturning action.The quantative measureof the tendency of a forceto cause or change therotational motion of abody is called torque. This body can rotate about axisthrough O,  to the plane. It is actedby three forces (in the plane offigure). The tendency of any force tocause the rotation depends on itsmagnitude and on the perpendiculardistance (lever arm) between theline of action of the force and point O.
  8. 8. Torque
  9. 9. Torque Torque is a vector quantity that measures the tendency of a force torotate an object about an axis. The magnitude of the torque producedby a force is defined asrFrFFl tansin   wherer = distance between the pivot point and the point of application of theforce.F = the magnitude of the force. = the angle between the force and a line extending thru the pivot andthe point of application.Ftan = F sin() = the component of the force perpendicular to the lineconnecting the pivot and the point of application.L = r sin() = moment arm or lever arm = distance from the pivot tothe line of action of the force.
  10. 10. Torque
  11. 11. TorqueFrSome important points about torque Torque has units of N·m. Despite the fact that this unit is the same as aJoule it is customary to leave torque expressed in N·m (or foot·pounds). Engineers will often use the term "moment" to describe what physicistscall a "torque". We will adopt a convention that defines torques that tend to causeclockwise rotation as negative and torques that tend to cause counter-clockwise rotation as positive. Torques are always defined relative to a point. It is incorrect to simplysay the "torque of F". Instead you must say the "torque of F relative topoint X". More general definition for the torque is given by the vector (or crossproduct). When a force acting at a point which has position vector rrelative to an origin O the torque exerted by the force about the origin isdefined as
  12. 12. Torque
  13. 13. Torque and Angular Acceleration for aRigid Body
  14. 14. Torque and Angular Acceleration for a RigidBody If we consider a rigid body rotating about a fixed axis as made up of a collectionof individual point particles, all of which have to obey Newtons Second Law fora particle, then we can show that the net torque acting on the body about thegiven axis of rotation will equal the moment of inertia of the body about thataxis times the angular acceleration.tan,11tan,1 amF  Z-axis is the axis of rotation; the first particle hasmass m1 and distance r1 from the axis of rotation. The net force acting on this article has a componentF1,rad along the radial direction, a component F1,tan thatis tangent to the circle of radius r1 in which particlemoves, and component F1z along axis of rotation. N2L for tangential component iszra 1tan,1  zrmrF 2111tan,1  zz I  11 F1,rad and F1z do not contribute to the torque about z-axis.ziiziiiiiz Irm    2For all particles
  15. 15. Torque and Angular Acceleration for a RigidBody This expression is the rotational form of Newtons Second Lawfor rigid body motion (for a fixed axis of rotation):  zz I N2L for a rigid body in rotational form Valid only for rigid bodies! If the body is not rigid like a rotating tankof water, the angular acceleration is different for different particles. z must be measured in rad/s2 (we used atan=rz in derivation) The torque on each particle is due to the net force on that particle,which is the vector sum of external and internal forces. According to N3L, the internal forces that any pair of particles in therigid body exert on each other are equal and opposite. If these forcesact along the line joining the two particles, their lever arms withrespect to any axis are also equal. So the torques for each pair areequal and opposite and add to ZERO.
  16. 16. Torque and Angular Acceleration for a RigidBodyONLY externaltorques affect therigid body’s rotation!
  17. 17. N2L in Rotational FormIDENTIFY the relevant concepts: The equation z=Iz is useful whenever torques act on a rigid body- that is, whenever forces act on a rigid body in such a way as tochange the state of the body’s rotation. In some cases you may be able to use an energy approach instead.However, if the target variable is a force, a torque, an acceleration,an angular acceleration, or an elapsed time, the approach usingequation z=Iz is almost always the most efficient one.Problem-Solving Strategy Rotational Dynamicsfor Rigid Bodies
  18. 18. N2L in Rotational FormSET UP the problem using the following steps:1. Draw a sketch of the situation and select the body or bodies to beanalyzed.2. For each body, draw a free-body diagram isolating the body andincluding all the forces (and only those forces) that act on the body,including its weight. Label unknown quantities with algebraic symbols.A new consideration is that you must show the shape of the bodyaccurately, including all dimensions and angles you will need fortorque calculations.3. Choose coordinate axes for each body and indicate a positive senseof rotation for each rotating body. If there is a linear acceleration, it’susually simplest to pick a positive axis in its direction. If you know thesense of z in advance, picking it as the positive sense of rotationsimplifies the calculations. When you represent a force in terms of itscomponents, cross out the original force to avoid including it twice.Problem-Solving Strategy
  19. 19. N2L in Rotational FormEXECUTE the solution as follows: For each body in the problem, decide whether it undergoes translationalmotion, rotational motion, or both. Depending on the behavior of thebody in question, apply F=ma, z=Iz, or both to the body. Becareful to write separate equations of motion for each body. There may be geometrical relations between the motions of two ormore bodies, as with a string that unwinds from a pulley while turning itor a wheel that rolls without slipping. Express these relations in algebraicform, usually as relations between two linear accelerations or between alinear acceleration and an angular acceleration. Check that the number of equations matches the number of unknownquantities. Then solve the equations to find the target variable(s).Problem-Solving Strategy
  20. 20. N2L in Rotational FormEVALUATE your answer: Check that the algebraic signs of your results make sense. As an example, suppose the problem is about a spool of thread. Ifyou are pulling thread off the spool, your answers should not tellyou that the spool is turning in the direction that rolls the threadback on the spool! Whenever possible, check the results for special cases or extremevalues of quantities and compare them with your intuitiveexpectations. Ask yourself: “Does this result make sense?”Problem-Solving Strategy
  21. 21. Rigid-Body Rotation about aMoving Axis
  22. 22. Rigid-Body Rotation about a MovingAxis Let’s extend analysis of rotational motion tocases in which the axis of rotation moves: themotion of a body is combined translation androtation. Every possible motion of a rigid body can berepresented as a combination of translationalmotion of the center of mass and rotationabout an axis through the center of mass. It is applicable even when the center of massaccelerates (so that is not at rest in anyinertial frame). The translation of the center of mass and therotation about the center of mass can betreated as separate but related problems. The prove of all that is beyond of the scopeof this course. We will learn concept only.
  23. 23. Rigid-Body Rotation about a MovingAxis If a round object of cross-sectional radius R rolls without slipping thenthe distance along the surface that the object covers will be the same asthe arc length along the edge of the circular object that has been incontact with the surface (i.e. s = Rq). Differentiating this expression with respect to time shows that the speedof the center of mass of the object will be given byRvcm  Condition for rolling without slipping This condition must be satisfied if an object is rolling without slipping.Rolling motion can be thought of in two different ways: Pure rotation about the instantaneous point of contact (P) of the objectwith the surface. Superposition of translation of the center of mass plus rotation aboutthe center of mass.
  24. 24. Rigid-Body Rotation about a MovingAxis The wheel is symmetrical, so its CM is at its geometric center. We view the motion in inertial frame of reference in which the surface is at rest. Inorder not to slip, the point of contact (where the wheel contacts the ground) isinstantaneously at rest as well. Hence the velocity of the point of contact relative tothe CM must have the same magnitude but opposite direction as the CM velocity. If the radius of the wheel is R and its angular speed about CM is : vcm=R.
  25. 25. Rigid-Body Rotation about a MovingAxis The velocity of a point on the wheel is the vector sum of the velocity of CM and thevelocity of the point relative to the center of mass. Thus, the point of the contact is instantaneously at rest, point 3 at the top of thewheel is moving forward twice as fast as the center of mass; points 2 and 4 at thesides have velocities at 45 degrees to the horizontal.
  26. 26. Rigid-Body Rotation about a MovingAxis The kinetic energy of an object that is rolling without slipping is givenby the sum of the rotational kinetic energy about the center of massplus the translational kinetic energy of the center of mass:222121cmcm IMvK  Rigid body with bothtranslation and rotation If a rigid body changes height as it moves, you must also considergravitational potential energy The gravitational potential energy associated with any extended body ofmass M, rigid or not, is the same as if you replace the body by a particleof mass M located at the body’s center of mass:cmMgyU 
  27. 27. Rigid-Body Rotation about a MovingAxisDynamics of Combined Translation and Rotation The combined translational and rotational motion of an object can alsobe analyzed from the standpoint of dynamics. In this case the objectmust obey both of the following forms of Newtons Second Law:cmext aMFzcmz I  Two following conditions should be met:1. The axis through the center of massmust be an axis of symmetry2. The axis must not change direction
  28. 28. Rolling Friction
  29. 29. Work and Power in Rotational Motion The work done by a torque on an object that undergoes an angulardisplacement from q1 to q2 is given byqqqdW z21Work done by a torque If the torque is constant then the work done is given byqqq  zzW )( 12 Work done by a constant torque Note: similarity between these expressions and the equations for workdone by a force (W=FS).
  30. 30. Work and Power in Rotational Motion The rotational analog to the Work - Energy Theorem is2122212121IIdIW zztot   The change in rotational kinetic energy of a rigid body equals thework done by forces exerted from outside the body. The rate at which work is performed is the powerzzP 
  31. 31. Angular Momentum
  32. 32. Angular Momentum of a Particle.Definition The angular momentum L of a particle relative to a point O is thecross product of the particles position r relative to O with the linearmomentum p of the particle.vmrprLAngular momentum of a particle The value of the angular momentumdepends on the choice of the origin O,since it involves the position vectorrelative to the origin The units of angular momentum: kg·m2/sMass m is moving in XY planeRight-hand rule“Leverarm”
  33. 33. Angular Momentum of a Particle When a net force F acts on a particle, its velocity and linear momentumchange. Thus, angular momentum may also change. FramrdtLd For a particle acted on by net force F)()()(amrvmvdtvmdrvmdtrddtvmrddtLd  Rate of change of angular momentum L of a particle equals thetorque of the net force acting on it.Vector product ofvector by itself = 0
  34. 34. Angular Momentum of a Rigid Body Rigid body rotating about Z-axis with angular speed  Consider a thin slice of the body lying in XY plane 2)( iiiiii rmrrmL  Each particle in the slice moves in a circlecentered in the origin O, and its velocity viat each instant  to its position vector ri Thus, =90°, and particle of the mass mi atdistance ri from O has speed vi=ri The direction of angular momentum Li is byright-hand rule and the magnitude: The total angular momentum of the slice is thesum of Li of particles:   IrmLL iii   2
  35. 35. Angular Momentum of a Rigid Body For points not lying in XY plane, the position vectors have components in Z-directionas well as in X- and Y-directions. This gives the angular momentum of each particlea component perpendicular to Z-axis. But if Z-axis is the axis of symmetry, -components for particles on opposite sides of this axis add up to ZERO. Thus, when a rigid body rotatesabout an axis of symmetry, itsangular momentum vector L liesalong the symmetry axis, andits magnitude is L=I The angular velocity vector liesalso along the rotation axis.Hence for a rigid body rotatingaround axis of symmetry, L and have the same directionIL for a rigid body rotatingaround a symmetry axis
  36. 36. Angular Momentum of a Rigid Body
  37. 37. Angular Momentum and Torque For any system of particles (including both rigid and non-rigid bodies),the rate of change of the total angular momentum equals the sum ofthe torques of all forces acting on all the particles Torques of the internal forces add to zero if these forces act along theline from one particle to another, so the sum of torques includes only thetorques of external forces:dtLdfor any system of particles If the system of particles is a rigid body rotating about its axis ofsymmetry (Z-axis), then Lz=Iz and I is constant. If this axis is fixed inspace, then the vectors L and  change only in magnitude, not in directionzzzIdtIddtLd  • If body is not rigid, I maychange, and L changes even if is constant
  38. 38. Angular Momentum of a Rigid Body If the axis of rotation is not asymmetry axis, L does not ingeneral lie along the rotation axis. Even if  is constant, the directionof L changes and a net torque isrequired to maintain rotation• If the body is an unbalanced wheel ofyour car, this torque is provided byfriction in the bearings, which causes thebearings to wear out• Balancing a wheel means distributing themass so that the rotation axis is an axisof symmetry, then L points along therotation axis, and no net torque isrequired to keep the wheel turning
  39. 39. Conservation of AngularMomentum
  40. 40. Conservation of AngularMomentum When the net external force torque acting on a system iszero, the total angular momentum of the system is constant(or conserved)Angular momentum conservation This principle is universal conservation law, valid at all scales fromatomic and nuclear systems to the motions of galaxies Circus acrobat, diver, ice skater use this principle: Suppose acrobat has just left a swing with arms and legs extended androtating counterclockwise about her center of mass. When she pulls herarms and legs in, her moment of inertia Icm with respect to her center ofmass changes from a large value I1 to much smaller value I2. The onlyexternal force is her weight, which has no torque (goes through center ofmass). So angular momentum remains constant, and angular speedchanges:constLdtLdz ,0zz II 2211  
  41. 41. Physics of Falling CatsHow does a cat land on its legs when dropped?… Moment of inertia is important ... To understand how a cat can land on its feet, you must first knowsome concepts of rotational motion, since the cat rotates as itfalls. Reminder: The moment of inertia of an object is determined bythe distance its mass is distributed from the rotational axis. Relating this to the cat, if the cat stretches out its legs and tail, itincreases its moment of inertia; conversely, it can decrease itsmoment of inertia by curling up. Remember how it was proved by extending yourprofessor’s arms while spinning around on a swivel chair? Just as a more massive object requires more force to move, anobject with a greater moment of inertia requires more torque tospin. Therefore by manipulating its moment of inertia, byextending and retracting its legs and rotating its tail, the cat canchange the speed at which it rotates, giving it control over whichpart of its body comes in contact with the ground.
  42. 42. Physics of Falling Cats... and the conservation of angular momentum ... If a cat is dropped they almost always tend to land on their feetbecause they use the conservation of angular momentum tochange their orientation When a cat falls, as you would expect, its centre of mass followsa parabolic path. The cat falls with a definite angular momentumabout an axis through the cat’s centre of mass. When the cat is in the air, no net external torque acts on it aboutits centre of mass, so the angular momentum about the cat’scentre of mass cannot change. By pulling in its legs, cat can considerably reduce it rotationalinertia about the same axis and thus considerably increase itsangular speed. Stretching out its legs increases its rotational inertia and thusslows the cat’s angular speed. Conservation of angular momentum allows cat to rotate its bodyand slow its rate of rotation enough so that it lands on its feet
  43. 43. Conservation of AngularMomentum Falling cat twists different parts of its body indifferent directions so that it lands feet first At all times during this process the angularmomentum of the cat as a whole is zero A free-falling cat cannot alter its total angularmomentum. Nonetheless, by swinging its tail andtwisting its body to alter its moment of inertia, thecat can manage to alter its orientation
  44. 44. Falling Cats: More InformationHow does a cat land on its legs when dropped? Cats have the seemingly unique ability to orient themselves in a fall allowing them toavoid many injuries. This ability is attributed to two significant feline characteristics:“righting reflex” and unique skeletal structure. The “righting reflex” is the cat’s ability to first, know up from down, and thenthe innate nature to rotate in mid air to orient the body so its feet facedownward. Animal experts say that this instinct is observable in kittens as young as threeto four weeks, and is fully developed by the age of seven weeks. A cat’s “righting reflex” is augmented by an unusually flexible backbone and theabsence of a collarbone in the skeleton. Combined, these factors allow for amazingflexibility and upper body rotation. By turning the head and forefeet, the rest of thebody naturally follows and cat is able reorient itself. Like many small animals, cats are said to have a non-fatal terminal falling velocity.That is, because of their very low body volume-to-weight ratio these animals areable to slow their decent by spreading out (flying squirrel style). Animals with thesecharacteristics are fluffy and have a high drag coefficient giving them a greaterchance of surviving these falls.

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