7. Rotational motion In pure rotation every point of an object moves in a circle whose center lies on the axis of rotation (in translational motion the movement occurs along a straight or curved line). In rotational motion one introduces the kinematic angular variables: angular position θ, angular velocity ω and angular acceleration a (7.1) (7.2) ω1 ω2 t1 t2 7.1. Newton’s law for rotation Second Newton’s law for translational motion (7.3) can be transformed to the shape useful for the analysis of rotational motion. Multiplying both sides of (7.3) by the position vector one obtains

Newton’s law for rotation, cont. (7.4) The left side of Eq.(7.4) can be transfomed to Eq.(7.4) can be then written as or (Newton’second law for rotation) (7.5) where the angular momentum of a particle is and is the torque. The magnitude of torque is what indicates that in rotation not only the magnitude of force is important but also how far from the axis it is applied and in what direction (d-moment arm of F).

7.2. Flat rotational motion with constant radius In the flat rotation with constant r and the reference frame origin in the center of a circular path the velocity vector is perpendicular to the position vector. In this case we have The magnitude of an angular momentum is then (7.6) The quantity mr2 in Eq.(7.6) is called the rotational inertia (moment of inertia) of a particle of mass m rotating about an axis with a distance r. (7.7) The angular momentum (7.6) in a vector form can then be written as (7.8) Eq.(7.8) is formally similar to the equation defining linear momentum corresponds to I corresponds to m

Flat rotational motion with constant radius, cont. If the force is placed in a plain of motion, all vectors , , are perpendicular to this plane and the scalar notation can be used. In this case Eq.(7.5) can be written as follows or (7.9) For I = const Eq.(7.9) transforms to (7.10) Eq.(7.10) is the rotational analog of Newton’s second law (F = ma): the net torque acting on a particle is equal to the product of the rotational inertia about the rotation axis and the resulting angular acceleration about that axis.

7.3. Rotational motion of a rigid body The angular momentum of a small particle mi being a part of a rigid body, with respect to an origin O is equal where , The torque acting on the mass element mi vs. origin O is For the whole rigid body all angular momenta must be summarized (7.11) The net torque is equal (7.12) And finally one obtains (7.13) The net external torque acting on a rigid body is equal to the time rate of change of the total angular momentum of a body.

Rotational motion of a rigid body, cont. From (7.13) it follows that for (7.15) This illustrates the law of conservation of angular momentum: If the net external torque acting on a system attains zero, the angular momentum of the system remains constant. Example 1: the spinning student Fig(a) A student sitting on a stool and holding dumbells in his outsstretched hands, rotates freely about a vertical axis with an angular velocity wi. Fig(b) The student reduces his rotational inertia by pulling in his hands. As no net external torque acts on the student-stool system, the angular momentum of the system remains unchanged and the angular speed of the student increases to wf. wf > wi (figure from HRW)

Conservation of angular momentum, cont. Example 2: a student with a heavy bicycle wheel Fig(a) A student initially at rest is sitting on a stool and is holding a bicycle wheel (with rotational inertia Iwh) which rotates counterclockwise about its central axis with wwh.. Fig(b) The student inverts the wheel which now rotates clockwise. Fig.(c) The net angular momentum of the system before and after the inversion remains constant. Question: With what angular speed wb and in what direction does the system (the student, the stool and the wheel) rotate after the inversion? (figure from HRW)

7.4. Rotation of a rigid body about a fixed symmetrical axis Equation (7.13) is general but can be simplified if the rotation takes place about a fixed axis. In this case the vectors of torque , angular momentum and angular velocity are collinear and the vector notation can be dropped (7.16) An example is the rotation of a hoop about a fixed axis z passing through its center. The magnitude of the angular momntum is given by (7.11) The summation in the expression for a moment of inertia can be replaced by integration, what gives (7.12) where M is the total mass of a hoop and R its radius. If the acting forces form a couple, i.e. the torque exerted by a couple equals where d is the perpendicular distance between the two forces (in the figure d = 2r). Hoop (obręcz)

7.5. Kinetic energy of rotation about a fixed symmetrical axis The kinetic energy of the ith mass point of a rotating rigid body equals The total kinetic energy of a body is then (7.13) The kinetic energy of a rotating body equals one half the moment of inertia multiplied by the square of the angular velocity. Eq.(7.13) is the angular equivalent of the expression for the kinetic energy in translational motion. In this case I corresponds to m ω corresponds to v

7.6. Rotational inertia of selected bodies The moment of inertia of a rotating rigid body is calculated from the expression (7.14) r – the distance of mass dm from the axis of rotation Introducing density ρ one can write dm = ρ dV and when the body has uniform density (ρ = const) integral (7.14) transforms to (7.15) The integration in (7.15) is over volume V but for the symmetrical objects the calculation in many cases can be reduced to, e.g. integration over one dimension only. Examples of calculations Annular cylinder (ring) about central axis where is the volume of a thin ring of radius r, thickness dr and height h. The limits of integration are R1 and R2 .

Annular cylinder, cont. Finally one obtains (7.16) From (7.16) it follows that for a solid cylinder (R1 = 0) we obtain (the height L is not important so it also holds for a flat disk) Again, for the case (cylinder) one obtains from (7.16) what is in accordance with expression (7.12) for the moment of inertia of a hoop.

For this case one can introduce the linear density λ. Thus 2. Thin rod about perpendicular to lenght axis through center (7.17) For this case one can introduce the linear density λ. Thus The moment of inertia is then equal Making use of the introduced linear density one obtains the final result (7.18)

(twierdzenie Steinera) 7.7. Parallel Axis Theorem (twierdzenie Steinera) The question is how to determine the moment of inertia about a given axis, which does not pass through the center of mass of a body. The parallel-axis theorem states that: If the moment of inertia about an axis passing thorough the center of mass IC is known, then the moment of inertia about any other parallel axis I is given by (7.19) where a is the distance between the two parallel axes and M is the mass of a body. Sample problem What is the moment of inertia of a rod about an axis perpendicular to the rod and passing through its end? From the parallel-axis theorem one gets