A particle moves so that its position vector is given by `vec r = cos omega t hat x + sin omega t hat y`, where `omega` is a constant which of the following is true ?
A particle moves so that its position vector is given by `vec r = cos omega t hat x + sin omega t hat y`, where `omega` is a constant which of the following is true ?
A
Velocity and acceleration both are parallel to r
B
velocity is perpendicular to r and acceleration is directed towards to origin
C
Velocity is perpendicular to r and acceleration is directed away from the origin
D
Velocity and acceleration both are perpendicular to r
Text Solution
Verified by Experts
The correct Answer is:
B
Position vector of the particle is given by
`r = cos omega t x +sin omega ty`
where, `omega` is a constant
Velocity of the particle is
`v = (dr)/(Dt) = (d)/(dt) (cos omega t x +sin omega ty)`
`= (-sin omega t)omega x +(cos omegat) omega y`
`=- omega (sin omega t x- cos omega ty)`
Acceleration of the particles
`a = (dv)/(dt) = (d)/(dt)[-omega sin omega t x +omega cos omega ty]`
`=- omega^(2) cos omega t x - omega^(2) sin omega t y`
`rArr =- omega^(2)r = omega^(2)(-r)`
Assuming the particle as P, then its position vector is directed as shown in the diagram.
Therefore, acceleraton is directed towards -r that is towards O (origin) `v.r =- omega (sin omega t x - cos omega t y) (cos omega tx +sin omega t y)`
`=- omega [sin omega t cos omega t +0 +0 - sin omega t. cos omegat t]`
`=- omega (0) = 0`
`rArr v _|_ r [ :' x _|_ y]`
Thus, velocity is perpendicular to r.
`r = cos omega t x +sin omega ty`
where, `omega` is a constant
Velocity of the particle is
`v = (dr)/(Dt) = (d)/(dt) (cos omega t x +sin omega ty)`
`= (-sin omega t)omega x +(cos omegat) omega y`
`=- omega (sin omega t x- cos omega ty)`
Acceleration of the particles
`a = (dv)/(dt) = (d)/(dt)[-omega sin omega t x +omega cos omega ty]`
`=- omega^(2) cos omega t x - omega^(2) sin omega t y`
`rArr =- omega^(2)r = omega^(2)(-r)`
Assuming the particle as P, then its position vector is directed as shown in the diagram.
Therefore, acceleraton is directed towards -r that is towards O (origin) `v.r =- omega (sin omega t x - cos omega t y) (cos omega tx +sin omega t y)`
`=- omega [sin omega t cos omega t +0 +0 - sin omega t. cos omegat t]`
`=- omega (0) = 0`
`rArr v _|_ r [ :' x _|_ y]`
Thus, velocity is perpendicular to r.
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