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A particle is in uniform circular motion about the origin of an xy coordinate system, moving counter-clockwise with a period of 6.50 s. At one instant, its position vector (from the origin) is r with arrow = (6.00 m)i hat − (6.00 m)j. At that instant, what is its velocity in unit-vector notation?

User Jwezorek
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2 Answers

4 votes

Final answer:

To find the velocity in unit-vector notation, differentiate the position vector with respect to time. Use the given position vector and the equations for the x and y components of velocity in uniform circular motion to find the velocity in unit-vector notation.

Step-by-step explanation:

To find the velocity in unit-vector notation, we can differentiate the position vector with respect to time.

Given, r(t) = (6.00 m)i - (6.00 m)j

Diffrentiating r(t) gives, v(t) = (dx/dt)i + (dy/dt)j

Since the particle is in uniform circular motion, the x and y components of the velocity are:

dx/dt = -6.00 m * (2π/6.50 s) * sin(2πt/6.50 s)

dy/dt = 6.00 m * (2π/6.50 s) * cos(2πt/6.50 s)

Therefore, the velocity in unit-vector notation is:

v(t) = (-6.00 m * (2π/6.50 s) * sin(2πt/6.50 s))i + (6.00 m * (2π/6.50 s) * cos(2πt/6.50 s))j

User BlogueroConnor
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7.0k points
4 votes

Answer:
\hat{v}=(1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}

Step-by-step explanation:

Given

Time Period
T=6.5 s

Position Vector
\vec{r}=6\hat{i}-6\hat{j}

Since it is moving in a circular path therefore magnitude of position vector will give the radius of circular path


|r|=√(6^2+6^2)


|r|=6√(2)


distance=velocity* time


6√(2)=velocity* 6.5


velocity=(6√(2))/(6.5)


v=1.305

this velocity is at angle of
\tan \theta =(-6)/(6)


\theta =-45^(\circ)

velocity
\vec{v}=v\cos \theta +v\sin \theta


\vec{v}=1.305\cos (-45)+1.305\sin (-45)

Therefore unit velocity vector


\hat{v}=(1)/(√(2))\hat{i}-(1)/(√(2))\hat{j}

User Mudassar Shaheen
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6.6k points