Question

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: r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^.The particle's motion can be described by ____________.(A) an ellipse starting at time t=0 on the positive x axis(B) an ellipse starting at time t=0 on the positive y axis(C) a circle starting at time t=0 on the positive x axis(D) a circle starting at time t=0 on the positive y axis

Answer 1

Answer:

(C) a circle starting at time t=0 on the positive x axis

Explanation:

particle's position is

r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^

this is a parametric equation of a circle, because the axis at x and y are the same = R.

for t=0:

r=Ri^

so: circle starting at time t=0 on the positive x axis

On the other hand:

$v=\frac{dx}{dt}= Rw[-sin(wt)i+cos(wt)j]\\a=\frac{dv}{dt}= Rw^{2}[-cos(wt)i-sin(wt)j]$

The value of the magnitude of the acceleration is:

$a=Rw^{2}(cos^{2}(wt)+sin^{2}(wt))=Rw^{2}$

we can recognise that this represent the centripetal acceleration.