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A particle of mass $m$ uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius $R :$$\vec{\mathbf{r}}=\hat{\mathbf{i}} R \cos \theta+\hat{\mathbf{j}} R \sin \theta$with $\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2},$ where the constants $\omega_{0}$ and $\alpha$ are the initial angular velocity and angular acceleration, respectively. Determine the object's tangential acceleration $\vec{a}_{\text { tan }}$ and determine the torque acting on the object using $(a) \overline{\tau}=\vec{\mathbf{r}} \times \vec{\mathbf{F}}$ . $(b) \vec{\tau}=I \vec{\alpha} .$

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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

University of Washington

University of Winnipeg

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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from Problem 25. We know that the tangential acceleration vector is equaling the cross product between the angular acceleration vector and the position vector are for this object specifically rotating counter clockwise and gaining angular speed. We can then say that the angular acceleration Alfa here is going going to be equal to the magnitude of the angular acceleration in the V K hat direction. And so here the tangential acceleration would be again equaling Alfa Cross product. With the position this would be called the to the determinant of the following three by three matrix. I had J hat que hat and then we have 00 Alfa and then we have our co sign of data are sign of data and then zero. This is giving us negative. Alfa are sign of data I had and then plus also our co sign of data J and so for party. Excuse me. We need the acceleration in order to calculate the torque vector which we know to be the cross product between the position vector and the four specter. And so we know the force consists of two components the radial component and the tangential component. And there's no torque associated with the radio component because we can say that, uh, angle the angle between the position specter and the centripetal force factor, it's gonna be 180 degrees. Therefore, they're essentially parallel to one another, but in the opposite directions. So that means that there is no force associated with that radio component. And so we can say that. Then the torque factor is gonna be equaling again. The cross product between the position vector and the force factor. And this would be equaling the cross product between the position director and the tangential component of the force factor. This is gonna be equaling the position vector, um, cross product with the mass times, the tangential acceleration specter. And this would be equaling the mass times the position vector cross product with the tangential acceleration Dr. And so the torque is gonna be equaling the mass. And then this would be multiplied by the determinant of the following three by three matrix. We have I hat J Hat K hat and then our co sign of data. Our sign of theta zero and then negative Alfa are sign of failure. Let me have Alfa are co sign of Fada and then again, zero. And so the torque vector would be equaling. The mass times are squared Alfa Co sign squared of theta I hat, brother, My apologies. Just coasting spread of Alfa and then plus R squared Alfa sine squared of theta And then the direction It would actually be on Lee in the K direction We know that co sign squared of theta and sign screwed If data due to the trigonometry identity would just simply equal one And so the torque factor is then going to be equaling Um R squared Alfa Times K hat And so this would be our final answer. Four part eh, For part B, we know that the moment of inertia of a particle is I equals m are square. So the torque factor would be equaling the moment of inertia times the angular acceleration vector. And we know this to be equaling em r squared times the magnitude of the acceleration angular acceleration times K hat, which is the exact same answer that we had found in port, eh? But that's how we would prove, uh, party. That is the end of the solution. Thank you for

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