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Problem 75 Hard Difficulty

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

Physics for Scientists and Engineers with Modern Physics

Chapter 11

Angular Momentum; General Rotation

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Video Transcript

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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Physics for Scientists and Engineers with Modern Physics

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Problem

A projectile with mass $m$ is launched from the g…

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Related Courses

Physics for Scientists and Engineers with Modern Physics

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Watch More Solved Questions in Chapter 11

Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Marshall Styczinski

Jared Enns

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Marshall Styczinski

Jared Enns

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Marshall Styczinski

Jared Enns

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics