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Safko Authors: Rockets are propelled by the momentum reaction of the exhaust gases expelled from the tail. The total frictional force is: Homework 5: It is called the weak law of action and reaction. Jordan Jeong.

Actions Shares. Embeds 0 No embeds. No notes for slide. Elementary Principles1. Angular Momentum: Torque is the time derivative of angular momentum: A force is considered conservative if the work is the same for any physicallypossible path. Independence of W12 on the particular path implies that thework done around a closed ciruit is zero: Potential Energy: The capacity to do work that a body or system has by viture of is positionis called its potential energy.

V above is the potential energy. To express workin a way that is independent of the path taken, a change in a quantity thatdepends on only the end points is needed. This quantity is potential energy. The change is -V. Energy Conservation Theorem for a Particle: The Conservation Theorem for the Linear Momentum of a Particle statesthat linear momentum, p, is conserved if the total force F, is zero.

The Conservation Theorem for the Angular Momentum of a Particle statesthat angular momentum, L, is conserved if the total torque T, is zero. It is called the weak law of action and reaction. Center of mass: This is how rockets work in space. Total linear momentum: If the total external force is zero, the total linear momentum is conserved. The strong law of action and reaction is the condition that the internal forcesbetween two particles, in addition to being equal and opposite, also lie alongthe line joining the particles.

Then the time derivative of angular momentumis the total external torque: Conservation Theorem for Total Angular Momentum: L is constant in timeif the applied torque is zero.

Linear Momentum Conservation requires weak law of action and reaction. Angular Momentum Conservation requires strong law of action and reaction. Total Angular Momentum: Total angular momentum about a point O is the angular momentum of mo-tion concentrated at the center of mass, plus the angular momentum of motionabout the center of mass.

If the center of mass is at rest wrt the origin then theangular momentum is independent of the point of reference. Total Work: Total kinetic energy: Total potential energy: The term on the right is called the internal potential energy. For rigid bodiesthe internal potential energy will be constant. For a rigid body the internalforces do no work and the internal potential energy remains constant.

Equations of motion are not all independent, because coordinates are no longer all independent 2. Forces are not known beforehand, and must be obtained from solution. For holonomic constraints introduce generalized coordinates.

Degrees offreedom are reduced. Use independent variables, eliminate dependent coordi-nates. This is called a transformation, going from one set of dependent variablesto another set of independent variables. Generalized coordinates are worthwhilein problems even without constraints.

Examples of generalized coordinates: Two angles expressing position on the sphere that a particle is constrained to move on. Two angles for a double pendulum moving in a plane. Amplitudes in a Fourier expansion of rj. Quanities with with dimensions of energy or angular momentum. For nonholonomic constraints equations expressing the constraint cannot beused to eliminate the dependent coordinates.

This is the only restric-tion on the nature of the constraints: This is great news, but itis not yet in a form that is useful for deriving equations of motion. Transformthis equation into an expression involving virtual displacements of the gener-alized coordinates. The generalized coordinates are independent of each otherfor holonomic constraints.

Theresult is: The equation of motion can be dervied for the x-dirction, and notice theyare identical component wise: The rate of energy dissipation due to friction is 2Fdis and the component ofthe generalized force resulting from the force of friction is: Scalar functions T and V are much easier to deal withinstead of vector forces and accelerations.

Write T and V in generalized coordinates. Form L from them. Solve for the equations of motion. Simple examples are: Forces of contstraint, do not appear in the Lagrangian formulation.

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You can download our homework help app on iOS or Android to access solutions manuals on your mobile device. Asking a study question in a snap - just take a pic. Textbook Solutions. Looking for the textbook? We have solutions for your book! Show that for a single particle with constant mass the equation of motion implies the following differential equation for the kinetic energy: Step-by-step solution:.

JavaScript Not Detected. Comment 0. The kinetic energy of the particle of mass is given as follows: From the commutative law of scalar product, the scalar product is given as follows: Therefore, the result is proved. Treat the mass of the particle varies with time and multiply the kinetic energy of particle with mass as follows: The result is: If you remember the indi- vidual coefficients vanish, and allow the forces derivable from a scaler potential function, and forgive me for skipping some steps, the result is:.

The velocity dependent potential is important for the electromagnetic forces on moving charges, the electromagnetic field. For a charge mvoing in an electric and magnetic field, the Lorentz force.

The equation of motion can be dervied for the x-dirction, and notice they are identical component wise:. The rate of energy dissipation due to friction is 2Fdis and the component of the generalized force resulting from the force of friction is:. The Lagrangian method allows us to eliminate the forces of constraint from the equations of motion.

Scalar functions T and V are much easier to deal with instead of vector forces and accelerations. Forces of contstraint, do not appear in the Lagrangian formulation. They also cannot be directly derived. Show that for a single particle with constant mass the equation of motion implies the follwing differential equation for the kinetic energy:.

Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation:. From the equations of the motion of the individual particles show that the in- ternal forces between particles satisfy both the weak and the strong laws of ac- tion and reaction. The argument may be generalized to a system with arbitrary number of particles, thus proving the converse of the arguments leading to the equations above.

First, if the particles satisfy the strong law of action and reaction then they will automatically satisfy the weak law.

The weak law demands that only the forces be equal and opposite. The strong law demands they be equal and oppo- site and lie along the line joining the particles. The first equation of motion tells us that internal forces have no effect. The equations governing the individual particles are. If the particles obey.

For two particles, the internal torque contribution is. A constraint condition of this type is holonomic only if an integrating function f x1,. Clearly the function must be such that. Show that no such integrating factor can be found for either of the equations of constraint for the rolling disk. The only way for f to satisfy this equation is if f is constant and thus appar- ently there is no integrating function to make these equations exact. Performing the same procedure on the second equation you can find.

If this question was confusing to you, it was confusing to me too. That makes me feel better. Two wheels of radius a are mounted on the ends of a common axle of length b such that the wheels rotate independently.

The whole combination rolls with- out slipping on a palne. Show that there are two nonholonomic equations of constraint,. The trick to this problem is carefully looking at the angles and getting the signs right. I think the fastest way to solve this is to follow the same procedure that was used for the single disk in the book, that is, find the speed of the disk, find the point of contact, and take the derivative of the x component, and y component of position, and solve for the equations of motion.

Once you have the equations of motion, from there its just slightly tricky algebra. Here goes:.

The contact points come from the length of the axis being b as well as x and y being the center of the axis. The components of the distance are cos and sin for x and y repectively.

This will give us the components of the velocity. Make sure you get the angles right, they were tricky for me. The plus sign is there because of the derivative of cos multiplied with the negative for the primed wheel distance from the center of the axis. For the y parts:. It is negative because I decided to have axis in the first quadrent heading south-east.

I also have the primed wheel south-west of the non-primed wheel. So just think about it. Now we are done with the physics. The rest is manipulation of these equa- tions of motion to come up with the constraints.

For the holonomic equation use 1 - 2. A particle moves in the xy plane under the constraint that its velocity vector is always directed towards a point on the x axis whose abscissa is some given function of time f t.

Show that for f t differentiable, but otherwise arbitrary,. The abscissa is the x-axis distance from the origin to the point on the x-axis that the velocity vector is aimed at. It has the distance f t. I claim that the ratio of the velocity vector components must be equal to the ratio of the vector components of the vector that connects the particle to the point on the x-axis. The directions are the same. The velocity vector components are:. That will show that they can be written as displayed above.

Now to show the terms with F vanish. The electromagnetic field is invariant under a gauge transformation of the scalar and vector potential given by.

What effect does this gauge trans- formation have on the Lagrangian of a particle moving in the electromagnetic field? Is the motion affected? This is all that you need to show that the Lagrangian is changed but the motion is not. This problem is now in the same form as before:. Suppose we transform to another set of independent coordinates s1, Such a transformatin is called a point transformation.

Consider a uniform thin disk that rolls without slipping on a horizontal plane. A horizontal force is applied to the center of the disk and in a direction parallel to the plane of the disk.