Question

A particle is moving in a circular path in the x-y plane. The center of the circle is at the origin and the rotation is counterclockwise at a rate (angular speed) of ω = 6.41 rad/s. At time t = 0, the particle is at y = 0 and x = 5.94 m. a. What is the x coordinate of the particle, in meters, at t = 16.7 s? b. What is the y coordinate of the particle, in meters, at t = 16.7 s? c. What is the x component of the particle's velocity, in m/s, at t = 16.7 s? d. What is the y component of the particle's velocity, in m/s, at t = 16.7 s? e. What is the x component of the particle's acceleration, in m/s2, at t = 16.7 s? f. What is the y component of the particle's acceleration, in m/s2, at t = 16.7 s? 

          A particle is moving in a circular path in
the x-y plane. The center of the circle is at
the origin and the rotation is counterclockwise at a rate (angular
speed) of ω = 6.41 rad/s. At time t =
0, the particle is at y = 0 and x = 5.94 m.
a. What is the x coordinate of the
particle, in meters,
at t = 16.7 s? 
b. What is the y coordinate of the
particle, in meters,
at t = 16.7 s? 
c. What is the x component of the
particle's velocity, in m/s,
at t = 16.7 s? 
d. What is the y component of the
particle's velocity, in m/s,
at t = 16.7 s? 
e. What is the x component of the
particle's acceleration, in m/s2,
at t = 16.7 s? 
f.  What is the y component of the
particle's acceleration, in m/s2,
at t = 16.7 s?
Show more…

Added by Steve J.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
A particle is moving in a circular path in the x-y plane. The center of the circle is at the origin and the rotation is counterclockwise at a rate (angular speed) of ω = 6.41 rad/s. At time t = 0, the particle is at y = 0 and x = 5.94 m. a. What is the x coordinate of the particle, in meters, at t = 16.7 s? b. What is the y coordinate of the particle, in meters, at t = 16.7 s? c. What is the x component of the particle's velocity, in m/s, at t = 16.7 s? d. What is the y component of the particle's velocity, in m/s, at t = 16.7 s? e. What is the x component of the particle's acceleration, in m/s2, at t = 16.7 s? f. What is the y component of the particle's acceleration, in m/s2, at t = 16.7 s?
Close icon
Play audio
Feedback
Powered by NumerAI
David Collins Ivan Kochetkov
Danielle Fairburn verified

Aarya B and 86 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question
*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs
*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.
*

Recommended Videos

-
a-particle-moves-horizontally-in-uniform-circular-motion-over-a-horizontal-xy-plane-at-one-instant-it-moves-through-the-point-at-coordinates-490-m-470-m-with-a-velocity-of-730-i-ms-and-an-ac-73741

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (4.90 m, 4.70 m) with a velocity of -7.30 i m/s and an acceleration of +12.0 j m/s^2. What are the (a) x and (b) y coordinates of the center of the circular path?

Mahendra K.
a-particle-moves-along-a-circular-path-over-a-horizontal-x-y-coordinate-system-at-constant-speed-at-

A particle moves along a circular path over a horizontal $x y$ coordinate system, at constant speed. At time $t_{1}=4.00$ s, it is at point $(5.00 \mathrm{m}, 6.00 \mathrm{m})$ with velocity $(3.00 \mathrm{m} / \mathrm{s}) \hat{\mathrm{j}}$ and acceleration in the positive $x$ direction. At time $t_{2}=10.0 \mathrm{s}$ , it has velocity $(-3.00 \mathrm{m} / \mathrm{s}) \hat{\mathrm{i}}$ and acceleration in the positive $y$ direction. What are the $(a) x$ and (b) $y$ coordinates of the center of the circular path if $t_{2}-t_{1}$ is less than one period?

Fundamentals of Physics

a-particle-moves-along-a-circular-path-over-a-horizontal-x-y-coordinate-system-at-constant-speed-at-

A particle moves along a circular path over a horizontal $x y$ coordinate system, at constant speed. At time $t_{1}=4.00$ s, it is at point $(5.00 \mathrm{m}, 6.00 \mathrm{m})$ with velocity $(3.00 \mathrm{m} / \mathrm{s}) \hat{\mathrm{j}}$ and acceleration in the positive $x$ direction. At time $t_{2}=10.0 \mathrm{s}$ , it has velocity $(-3.00 \mathrm{m} / \mathrm{s}) \hat{\mathrm{i}}$ and acceleration in the positive $y$ direction. What are the $(a) x$ and (b) $y$ coordinates of the center of the circular path if $t_{2}-t_{1}$ is less than one period?

Fundamentals of Physics

*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,421 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,914 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,810 solutions
*

Transcript

-
00:01 Hi, in this question it is given that a particle is moving in a circular path with an angular speed of 6 .41 radius per second and the center of this circle is at origin and initially at time equal to 0 the particle is at x equal to 5 .94 meter and the y coordinate is 0 so since the particle is moving along the circular path then equation of circle is given as x squared plus y squared equal to r squared here r is the radius of this circle so in polar form this x is equal to r cos omega t and the y coordinate is r sine omega t since it is given that at t equal to at t equal to 0 second this x is equal to 5 .94 so here we can write r c c0 equal to 5 .9 4 so here we can write r coss 0 equal to 5.
01:02 94 so from here we can write this r value is equal to 5 .94 meter so radius of circle is 5 .94.
01:10 So from here we can write x is equal to 5 .94 cos omega t and the y coordinate is 5 .94 sine omega t so in the first part of the question we need to find at t is equal to 16.
01:34 7 second the horizontal coordinate of the particle that is x at t is given as 5 .94 cost and this omega value is given as 6 .41 so this is 6 .41 multiplied by 16 .7 and we will get this x of 16 .7 equal to so if we solve this then we will get minus 1 .74 meter.
02:10 So this is the answer for the first part of the question.
02:16 Now, in the second part of the question, we need to find same that is p equal to 16 .7 second y coordinate of the particle.
02:25 So y of t is given as 5 .94 sine.
02:31 Omega value is 6 .41, multiplied by 16 .7 and the unit is meter.
02:40 So again if you saw this y of 16 .7 is equal to 5 .68 meter and this is the answer for the second part of the question.
02:53 Now in the third part of the question we need to find at t equal to 16 .7 second velocity of the particle.
03:00 So in horizontal direction that is vx t will be equal to d x over dd so if you differentiate this position vector so this we will get us 5 .94 omega and cos omega t if you differentiate this so this will become negative of 5 .94 omega sine omega t so now if you find the value of bx at 16 .7 so this will be minus 5 .94 multiplied by 6 .41 multiplied by sine 6 .41 multiplied by 16 .7.
03:47 So the horizontal component of velocity will be equal to so if we solve this then we will get minus 36 .4 meter per second and this is the answer for the third part of the question...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever