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The position of a particle with mass $m$ travelin…

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Problem 69 Hard Difficulty

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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Video Transcript

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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Physics for Scientists and Engineers with Modern Physics

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Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Liev Birman

Meghan Miholics

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

Liev Birman

Meghan Miholics

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Christina Krawiec

Andy Chen

Liev Birman

Meghan Miholics

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics