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The time-dependent position of a point object which moves counterclockwise along the circumference of a circle (radius $R )$ in the $x y$ plane with constant speed $v$ is given by $\vec{\mathbf{r}}=\hat{\mathbf{i}} R \cos \omega t+\hat{\mathbf{j}} R \sin \omega t$ where the constant $\omega=v / R$ . Determine the velocity $\nabla$ and angular velocity $\vec{\omega}$ of this object and then show that these three vectors obey the relation $\vec{\mathbf{v}}=\vec{\omega} \times \vec{\mathbf{r}}$

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03:52

Averell Hause

Physics 101 Mechanics

Chapter 11

Angular Momentum; General Rotation

Moment, Impulse, and Collisions

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

Lectures

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

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for this problem on the topic of angular momentum and general rotation were given the time dependent position function often object which moves counterclockwise along this company Off the circle which has radius are in the X Y plane. The object moves with constant speed v and we had told that the angular velocity omega is able to be over our We want to determine the velocity and angular velocity of the object and ensure that these vectors of a relation B is equal to omega across our So the velocity we know is derivative off the position. So to find the velocity vector we This is simply d bye DT off the position Vector are and this is D by d t off our times the call sign off amega t the direction plus our have the sign off omega t along deejay direction So performing this derivative we get this to B minus Omega are sign Omega t I plus Omega are co sign Omega t times J Now we know that omega tents are Tzvi so we can write this as V into minus sign omega t along the direction plus the co sign off omega t in the J direction. So V here is the magnitude off the velocity or speed off the object motion. Now, from the right hand rule, a counterclockwise rotation of the X Y plane produces an angular velocity in the positive direction. So therefore, Victor Angular philosophy Omega is equal to the over our oriented in the direction so we can take the cross product. Omega Cross are as follows good. So the cross product. We'll use the method of co factors here. Omega Cross Our is equal to the over our, which are Skelos along the K direction, and that's crossed with the R. Victor, which were given as our co sign Omega T I. That's our sine omega T people along the J direction. So if we use the method of co factors, that's I J. K. And the first Vector only has a key component, which is the over our The second vector has no que component and has I am J components. I component off our call sign Omega T and a J component of our sine omega t. He came from 0.20 So if we perform this cross product, we get the result to be minus B claims the sign off omega T in the direction plus the plans The co sign off omega T in the J direction. And this is familiar because this is the velocity vector we found above. And so we can see therefore, that the velocity vector V is indeed the angular velocity Omega Cross with the position vector are Thank you. Yes.

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