# Dynamics of Rotational Motion

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Dynamics of Rotational Motion

Cross Product

Cross Product The Cross Product (or Vector Product) of two vectors A and B is a multiplication of vectors where the result is a vector quantity C with a direction perpendicular to both vectors A and B, and the magnitude equal to ABsin : Magnitude of the cross (vector) product of two vectors A and B A is the magnitude of the first vector, B is the magnitude of the second vector and f is the angle between the two vectors. The direction of the cross product is perpendicular to the plane formed by the two vectors in the product. This leaves two possible choices which are resolved by using the Right Hand Rule.

Cross Product Properties 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.

Cross Product The above formula for the cross product is useful when the magnitudes of the two vectors and the angle between them are known. If you only know the components of the two vectors: Components of cross product vector

Cross Product Right-handed coordinate system, in which:
Left-handed coordinate system, in which:

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 or turning action. The quantative measure of the tendency of a force to cause or change the rotational motion of a body is called torque. This body can rotate about axis through O,  to the plane. It is acted by three forces (in the plane of figure). The tendency of any force to cause the rotation depends on its magnitude and on the perpendicular distance (lever arm) between the line of action of the force and point O.

Torque

Torque Torque is a vector quantity that measures the tendency of a force to rotate an object about an axis. The magnitude of the torque produced by a force is defined as where r = distance between the pivot point and the point of application of the force. F = the magnitude of the force. f = the angle between the force and a line extending thru the pivot and the point of application. Ftan = F sin(f) = the component of the force perpendicular to the line connecting the pivot and the point of application. L = r sin(f) = moment arm or lever arm = distance from the pivot to the line of action of the force.

Torque

Torque Some important points about torque
Torque has units of N·m. Despite the fact that this unit is the same as a Joule it is customary to leave torque expressed in N·m (or foot·pounds). Engineers will often use the term "moment" to describe what physicists call a "torque". We will adopt a convention that defines torques that tend to cause clockwise 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 simply say the "torque of F". Instead you must say the "torque of F relative to point X". More general definition for the torque is given by the vector (or cross product). When a force acting at a point which has position vector r relative to an origin O the torque exerted by the force about the origin is defined as

Torque

Torque and Angular Acceleration for a Rigid Body

Torque and Angular Acceleration for a Rigid Body

Torque and Angular Acceleration for a Rigid Body
This expression is the rotational form of Newton's Second Law for rigid body motion (for a fixed axis of rotation): N2L for a rigid body in rotational form Valid only for rigid bodies! If the body is not rigid like a rotating tank of 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 the rigid body exert on each other are equal and opposite. If these forces act along the line joining the two particles, their lever arms with respect to any axis are also equal. So the torques for each pair are equal and opposite and add to ZERO.

Torque and Angular Acceleration for a Rigid Body
ONLY external torques affect the rigid body’s rotation!

N2L in Rotational Form Rotational Dynamics for Rigid Bodies
Problem-Solving Strategy IDENTIFY 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 to change 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 using equation z=Iz is almost always the most efficient one.

N2L in Rotational Form Problem-Solving Strategy
SET UP the problem using the following steps: Draw a sketch of the situation and select the body or bodies to be analyzed. For each body, draw a free-body diagram isolating the body and including 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 body accurately, including all dimensions and angles you will need for torque calculations. Choose coordinate axes for each body and indicate a positive sense of rotation for each rotating body. If there is a linear acceleration, it’s usually simplest to pick a positive axis in its direction. If you know the sense of z in advance, picking it as the positive sense of rotation simplifies the calculations. When you represent a force in terms of its components, cross out the original force to avoid including it twice.

N2L in Rotational Form Problem-Solving Strategy
EXECUTE the solution as follows: For each body in the problem, decide whether it undergoes translational motion, rotational motion, or both. Depending on the behavior of the body in question, apply F=ma, z=Iz, or both to the body. Be careful to write separate equations of motion for each body. There may be geometrical relations between the motions of two or more bodies, as with a string that unwinds from a pulley while turning it or a wheel that rolls without slipping. Express these relations in algebraic form, usually as relations between two linear accelerations or between a linear acceleration and an angular acceleration. Check that the number of equations matches the number of unknown quantities. Then solve the equations to find the target variable(s).

Check that the algebraic signs of your results make sense. As an example, suppose the problem is about a spool of thread. If you are pulling thread off the spool, your answers should not tell you that the spool is turning in the direction that rolls the thread back on the spool! Whenever possible, check the results for special cases or extreme values of quantities and compare them with your intuitive expectations. Ask yourself: “Does this result make sense?”

Rigid-Body Rotation about a Moving Axis

Rigid-Body Rotation about a Moving Axis
Let’s extend analysis of rotational motion to cases in which the axis of rotation moves: the motion of a body is combined translation and rotation. Every possible motion of a rigid body can be represented as a combination of translational motion of the center of mass and rotation about an axis through the center of mass. It is applicable even when the center of mass accelerates (so that is not at rest in any inertial frame). The translation of the center of mass and the rotation about the center of mass can be treated as separate but related problems. The prove of all that is beyond of the scope of this course. We will learn concept only.

Rigid-Body Rotation about a Moving Axis
If a round object of cross-sectional radius R rolls without slipping then the distance along the surface that the object covers will be the same as the arc length along the edge of the circular object that has been in contact with the surface (i.e. s = Rq). Differentiating this expression with respect to time shows that the speed of the center of mass of the object will be given by 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 object with the surface. Superposition of translation of the center of mass plus rotation about the center of mass.

Rigid-Body Rotation about a Moving Axis
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. In order not to slip, the point of contact (where the wheel contacts the ground) is instantaneously at rest as well. Hence the velocity of the point of contact relative to the 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.

Rigid-Body Rotation about a Moving Axis
The velocity of a point on the wheel is the vector sum of the velocity of CM and the velocity 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 the wheel is moving forward twice as fast as the center of mass; points 2 and 4 at the sides have velocities at 45 degrees to the horizontal.

Rigid-Body Rotation about a Moving Axis
The kinetic energy of an object that is rolling without slipping is given by the sum of the rotational kinetic energy about the center of mass plus the translational kinetic energy of the center of mass: Rigid body with both translation and rotation If a rigid body changes height as it moves, you must also consider gravitational potential energy The gravitational potential energy associated with any extended body of mass M, rigid or not, is the same as if you replace the body by a particle of mass M located at the body’s center of mass:

Rigid-Body Rotation about a Moving Axis
Dynamics of Combined Translation and Rotation The combined translational and rotational motion of an object can also be analyzed from the standpoint of dynamics. In this case the object must obey both of the following forms of Newton's Second Law: Two following conditions should be met: The axis through the center of mass must be an axis of symmetry The axis must not change direction

Rolling Friction

Work and Power in Rotational Motion
The work done by a torque on an object that undergoes an angular displacement from q1 to q2 is given by Work done by a torque If the torque is constant then the work done is given by Work done by a constant torque Note: similarity between these expressions and the equations for work done by a force (W=FS).

Work and Power in Rotational Motion
The rotational analog to the Work - Energy Theorem is The change in rotational kinetic energy of a rigid body equals the work done by forces exerted from outside the body. The rate at which work is performed is the power

Angular Momentum

Angular Momentum of a Particle. Definition
The angular momentum L of a particle relative to a point O is the cross product of the particle's position r relative to O with the linear momentum p of the particle. Mass m is moving in XY plane Angular momentum of a particle The value of the angular momentum depends on the choice of the origin O, since it involves the position vector relative to the origin The units of angular momentum: kg·m2/s “Lever arm” Right-hand rule

Angular Momentum of a Particle
When a net force F acts on a particle, its velocity and linear momentum change. Thus, angular momentum may also change. Vector product of vector by itself = 0 For a particle acted on by net force F Rate of change of angular momentum L of a particle equals the torque of the net force acting on it.

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 Each particle in the slice moves in a circle centered in the origin O, and its velocity vi at each instant  to its position vector ri Thus, =90°, and particle of the mass mi at distance ri from O has speed vi=ri The direction of angular momentum Li is by right-hand rule and the magnitude: The total angular momentum of the slice is the sum of Li of particles:

Angular Momentum of a Rigid Body
For points not lying in XY plane, the position vectors have components in Z-direction as well as in X- and Y-directions. This gives the angular momentum of each particle a 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 rotates about an axis of symmetry, its angular momentum vector L lies along the symmetry axis, and its magnitude is L=I The angular velocity vector lies also along the rotation axis. Hence for a rigid body rotating around axis of symmetry, L and  have the same direction for a rigid body rotating around a symmetry axis

Angular Momentum of a Rigid Body

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 of the torques of all forces acting on all the particles Torques of the internal forces add to zero if these forces act along the line from one particle to another, so the sum of torques includes only the torques of external forces: for any system of particles If the system of particles is a rigid body rotating about its axis of symmetry (Z-axis), then Lz=Iz and I is constant. If this axis is fixed in space, then the vectors L and  change only in magnitude, not in direction If body is not rigid, I may change, and L changes even if  is constant

Angular Momentum of a Rigid Body
If the axis of rotation is not a symmetry axis, L does not in general lie along the rotation axis. Even if  is constant, the direction of L changes and a net torque is required to maintain rotation If the body is an unbalanced wheel of your car, this torque is provided by friction in the bearings, which causes the bearings to wear out Balancing a wheel means distributing the mass so that the rotation axis is an axis of symmetry, then L points along the rotation axis, and no net torque is required to keep the wheel turning

Conservation of Angular Momentum

Conservation of Angular Momentum
When the net external force torque acting on a system is zero, the total angular momentum of the system is constant (or conserved) Angular momentum conservation This principle is universal conservation law, valid at all scales from atomic 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 and rotating counterclockwise about her center of mass. When she pulls her arms and legs in, her moment of inertia Icm with respect to her center of mass changes from a large value I1 to much smaller value I2. The only external force is her weight, which has no torque (goes through center of mass). So angular momentum remains constant, and angular speed changes:

Physics of Falling Cats
How does a cat land on its legs when dropped? … Moment of inertia is important ... To understand how a cat can land on it's feet, you must first know some concepts of rotational motion, since the cat rotates as it falls. Reminder: The moment of inertia of an object is determined by the distance it's mass is distributed from the rotational axis. Relating this to the cat, if the cat stretches out it's legs and tail, it increases it's moment of inertia; conversely, it can decrease it's moment of inertia by curling up. Remember how it was proved by extending your professor’s arms while spinning around on a swivel chair? Just as a more massive object requires more force to move, an object with a greater moment of inertia requires more torque to spin. Therefore by manipulating it's moment of inertia, by extending and retracting its legs and rotating its tail, the cat can change the speed at which it rotates, giving it control over which part of it's body comes in contact with the ground.

Physics of Falling Cats
... and the conservation of angular momentum ... If a cat is dropped they almost always tend to land on their feet because they use the conservation of angular momentum to change their orientation When a cat falls, as you would expect, its centre of mass follows a parabolic path. The cat falls with a definite angular momentum about an axis through the cat’s centre of mass. When the cat is in the air, no net external torque acts on it about its centre of mass, so the angular momentum about the cat’s centre of mass cannot change. By pulling in its legs, cat can considerably reduce it rotational inertia about the same axis and thus considerably increase its angular speed. Stretching out its legs increases its rotational inertia and thus slows the cat’s angular speed. Conservation of angular momentum allows cat to rotate its body and slow its rate of rotation enough so that it lands on its feet

Conservation of Angular Momentum
Falling cat twists different parts of its body in different directions so that it lands feet first At all times during this process the angular momentum of the cat as a whole is zero A free-falling cat cannot alter its total angular momentum. Nonetheless, by swinging its tail and twisting its body to alter its moment of inertia, the cat can manage to alter its orientation