To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration. Suppose that a particle's position is given by the following expression: r⃗ (t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^.
Choose the answer that best completes the following sentence:
The particle's motion can be described by ____________.
an ellipse starting at time t=0 on the positive x axis
an ellipse starting at time t=0 on the positive y axis
a circle starting at time t=0 on the positive x axis
a circle starting at time t=0 on the positive y axis
Part B
When does the particle first cross the negative x axis?
Express your answer in terms of some or all of the variables ? (Greek letter omega), R, and ?.
Part C
Find the particle's velocity as a function of time.
Express your answer using unit vectors (e.g., A i^+ B j^, where A and B are functions of ?, R, t, and ?).
Part D
Find the speed of the particle at time t.
Express your answer in terms of some or all of the variables ?, R, and ?.
Part E
Now find the acceleration of the particle.
Express your answer using unit vectors (e.g., A i^+ B j^, where A and B are functions of ?, R, t, and ?).
Part F
Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r? (t).
Express your answer in terms of some or all of the variables r? (t) and ?.
Part H
Finally, express the magnitude of the particle's acceleration in terms of R and v using the expression you obtained for the speed of the particle.
Express your answer in terms of one or both of the variables R and v.

The particle's motion is a circle starting at t=0 on the positive x axis since r(0)=R[cos(0)i^+sin(0)j^]=R[i^]. The particle first cross the negative x axis when r(t)=-R[i^], which means cos(ωt)=-1, or [tex]\omega t=\pi[/tex], so we have [tex]t=\pi/\omega[/tex]. The particle's velocity is the derivative of its position, so v(t)=Rω[-sin(ωt)i^+cos(ωt)j^], while its speed is the magnitude of that vector, v(t)=Rω (since the magnitude of the vector -sin(ωt)i^+cos(ωt)j^ is 1). The particle's acceleration is the derivative of its velocity, so a(t)=-R[tex]\omega^2[/tex][cos(ωt)i^+sin(ωt)j^], or in terms of its position a(t)=-[tex]\omega^2[/tex]r(t), and its magnitude using the expression obtained for the speed of the particle, a=R[tex]\omega^2[/tex]=R[tex]v^2[/tex]/[tex]R^2[/tex]=[tex]v^2[/tex]/R.

kasliwalalvi24 kasliwalalvi24 · Answer 2 · 2022-01-31T08:39:17+01:00

Motion-The particle's motion is a circle starting at t=0 on the positive x-axis since r(0)=R[cos(0)i^+sin(0)j^]=R[i^].

A) A circle starting at time t=0 on the positive x-axis.

B) t=π/ωC)

v(t)=Rω[-sin(ωt)i^+cos(ωt)j^]D)
v(t)=RωE)
a(t)=-Rω²[cos(ωt)i^+sin(ωt)j^]F)
a(t)=ω²r(t)G)

(There is no Part G)H) a=∨²/R

What is the particle's velocity?

The particle first crosses the negative x-axis when r(t)=-R[i^], which means cos(ωt)=-1, or ωt=π, so we have t=π/ω.

Velocity-The velocity is the derivative of its position, so v(t)=Rω[-sin(ωt)i^+cos(ωt)j^], while its speed is the magnitude of that vector, v(t)=Rω (since the magnitude of the vector -sin(ωt)i^+cos(ωt)j^ is 1).

What is the particle's acceleration?

Acceleration-The acceleration is the derivative of its velocity, so a(t)=-Rω²[cos(ωt)i^+sin(ωt)j^], or in terms of its position a(t)=ω²r(t), and its magnitude using the expression obtained for the speed of the particle, a=Rω²=R∨²/R²=∨²/R.

Therefore, these are the correct answers.

Learn about the velocity and acceleration here:

https://brainly.com/question/2239252