PHY1012F ROTATION II Gregor Leigh gregor.leigh@uct.ac.za

Learning outcomes: At the end of this chapter you should be able to… PHY1012F ROTATIONAL ENERGY ROTATIONAL ENERGY Learning outcomes: At the end of this chapter you should be able to… Use the laws of conservation of mechanical energy and angular momentum to solve rotational problems, including those involving rolling motion. Use vector mathematics to describe and solve problems involving rotational problems.

PHY1012F ROTATIONAL ENERGY ROTATIONAL ENERGY Each particle in a rigid rotating body has kinetic energy. m2 r3 m3 r2 The sum of all the individual kinetic energies of each of the particles is the rotational kinetic energy of the body: r1 axle  m1 Krot = ½ m1v12 + ½ m2v22 + …  Krot = ½ m1r122 + ½ m2r222 + …  Krot = ½ (m1r12)2  Krot = ½ I2

CONSERVATION OF ENERGY PHY1012F ROTATIONAL ENERGY CONSERVATION OF ENERGY As usual, energy is conserved (in frictionless systems). If, however, a horizontal axis of rotation does not coincide with the centre of mass, the object’s potential energy will vary. axle So we write: Emech = Krot + Ug = ½ I2 + MgyCM K U Emech = K + U Emech = 0 4 4

ROTATION ABOUT A FIXED AXIS PHY1012F ROTATION OF A RIGID BODY ROTATION ABOUT A FIXED AXIS A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position? pivot  i = 0 rad i = 0 rad/s f2 = i2 + 2 You canNOT use rotational kinematics to solve this problem! Why not?  f2 = 0 + 2(–15)(–0.5 )  = –15 rad/s2 (Not constant) vt =  r  f = –0.5 rad f = ?  vt = –6.8  1 = 6.8 m/s Use rotational kinematics to find angular positions and velocities. (Not this time!)

PHY1012F ROTATIONAL ENERGY A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position? x y L = 1 m pivot M = 0.07 kg yCM = 0 m 0 = 0 rad/s Before: ½ I02 + MgyCM 0 = ½ I12 + MgyCM 1 yCM = –0.5 m 1 = ? vtip = ? After:

PHY1012F ROTATIONAL ENERGY ROLLING MOTION Rolling is a combination of rotation and translation. (We shall consider only objects which roll without slipping.) As a wheel (or sphere) rolls along a flat surface… 2R each point on the rim describes a cycloid; the axle (the centre of mass) moves in a straight line, covering a distance of 2R each revolution; the speed of the wheel is given by

FUNNY THING ABOUT THE CYCLOID… PHY1012F ROTATIONAL ENERGY FUNNY THING ABOUT THE CYCLOID… If a farmer’s road surface were rutted into a cycloid form, the smoothest way to get his sheep to market would be to use a truck with… SQUARE wheels!

PHY1012F ROTATIONAL ENERGY ROLLING MOTION The velocity of a particle on a wheel consists of two parts: TRANSLATION + ROTATION = ROLLING vCM R v = 2vCM = 2R vCM v = R + = v = 0 vCM –R P So the point, P, at the bottom of an object which rolls (without slipping) is instantaneously at rest…

KINETIC ENERGY OF A ROLLING OBJECT PHY1012F ROTATIONAL ENERGY KINETIC ENERGY OF A ROLLING OBJECT If we regard P as an instantaneous axis of rotation, the object’s motion simplifies to one of pure rotation, and thus its kinetic energy is given by:  v = 2R K = Krot about P = ½ IP2 v = R Using the parallel axis theorem, IP = (ICM + MR2) v = 0 K = ½ ICM2 + ½ M(R)2 P  K = ½ ICM2 + ½ M(vCM)2 I.e. K = Krot + KCM

THE GREAT DOWNHILL RACE PHY1012F ROTATIONAL ENERGY THE GREAT DOWNHILL RACE Kf = Ui  ½ ICM2 + ½ M(vCM)2 = Mgh h ICM = cMR2 and  I.e. The actual values of M and R do not feature, but where the mass is situated is of critical importance.

THE GREAT DOWNHILL RACE PHY1012F ROTATIONAL ENERGY THE GREAT DOWNHILL RACE vCM2 = 0 + 2ax h x where  I.e. The acceleration of a rolling body is less than that of a particle by a factor which depends on the body’s moment of inertia.

VECTOR DESCRIPTION OF ROTATIONAL MOTION PHY1012F ROTATIONAL ENERGY VECTOR DESCRIPTION OF ROTATIONAL MOTION Using only “clockwise” and “counterclockwise” is the rotational analogue of using “backwards” and “forwards” in rectilinear kinematics. A more general handling of rotational motion requires vector quantities. The vector associated with a rotational quantity… has magnitude equal to the magnitude of that quantity; has direction as given by the right-hand rule. E.g. The angular velocity vector, , of this anticlockwise-turning disc points… in the positive z-direction.

PHY1012F ROTATIONAL ENERGY THE CROSS PRODUCT The magnitude of the torque exerted by force applied at displacement from the turning point is:   rFsin Once again, the quantity rF sin is the product of two vectors, and , at an angle  to each other. This time, however, we use the orthogonal components to determine the cross product of the vectors: . y 1 In RH system: x 1 z 1

THE CROSS PRODUCT Notes: PHY1012F ROTATIONAL ENERGY THE CROSS PRODUCT Notes: The more orthogonal the vectors, the greater the cross product; the more parallel, the smaller… Since it is a vector quantity, the cross product is also known as the vector product. . Derivative of a cross product:

PHY1012F ROTATIONAL ENERGY ANGULAR MOMENTUM We have shown that in circular motion (where vt and r are perpendicular) a particle has angular momentum L = mrvt. mvt = p z  L = rp More generally (allowing for and to be at any angle )…  = (mrv sin, direction from RH rule) I.e. (Cf. in linear motion: )

ROTATIONAL MOMENTUM & ENERGY PHY1012F ROTATIONAL ENERGY ROTATIONAL MOMENTUM & ENERGY Summary of corresponding quantities and relationships: Linear Rotational KCM = ½ MvCM2 Krot = ½ I2 (around an axis of symmetry) Linear momentum, , is con-served if there is no net force Angular momentum, , is con-served if there is no net torque

Learning outcomes: At the end of this chapter you should be able to… PHY1012F ROTATIONAL ENERGY ROTATIONAL ENERGY Learning outcomes: At the end of this chapter you should be able to… Use the laws of conservation of mechanical energy and angular momentum to solve rotational problems, including those involving rolling motion. Use vector mathematics to describe and solve problems involving rotational problems.