Question

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?

Answer 1

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

Answer 2

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}`