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A particle moves so that its position vector is given by r=cos⁡(ωt)x^+sin⁡(ωt)y^\mathbf{r} = \cos(\omega t)\hat{x} + \sin(\omega t)\hat{y} where ω\omega is a constant. Based on the information given, which of the following is true?

A

The velocity and acceleration, both are parallel to r\mathbf{r}.

B

The velocity is perpendicular to r\mathbf{r} and acceleration is directed towards the origin.

C

The velocity is not perpendicular to r\mathbf{r} and acceleration is directed away from the origin.

D

The velocity and acceleration, both are perpendicular to r\mathbf{r}.

Step-by-Step Solution

  1. Velocity Vector (v\mathbf{v}): Velocity is the time derivative of the position vector r\mathbf{r}. v=drdt=ddt[cos⁡(ωt)x^+sin⁡(ωt)y^]=−ωsin⁡(ωt)x^+ωcos⁡(ωt)y^\mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}[\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}] = -\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y}
  2. Check Perpendicularity (v⋅r\mathbf{v} \cdot \mathbf{r}): Calculate the dot product of velocity and position vectors. v⋅r=[−ωsin⁡(ωt)][cos⁡(ωt)]+[ωcos⁡(ωt)][sin⁡(ωt)]\mathbf{v} \cdot \mathbf{r} = [-\omega \sin(\omega t)] [\cos(\omega t)] + [\omega \cos(\omega t)] [\sin(\omega t)] v⋅r=−ωsin⁡(ωt)cos⁡(ωt)+ωsin⁡(ωt)cos⁡(ωt)=0\mathbf{v} \cdot \mathbf{r} = -\omega \sin(\omega t)\cos(\omega t) + \omega \sin(\omega t)\cos(\omega t) = 0 Since the dot product is zero, the velocity is perpendicular to the position vector r\mathbf{r}.
  3. Acceleration Vector (a\mathbf{a}): Acceleration is the time derivative of velocity. a=dvdt=ddt[−ωsin⁡(ωt)x^+ωcos⁡(ωt)y^]\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}[-\omega \sin(\omega t)\hat{x} + \omega \cos(\omega t)\hat{y}] a=−ω2cos⁡(ωt)x^−ω2sin⁡(ωt)y^=−ω2[cos⁡(ωt)x^+sin⁡(ωt)y^]\mathbf{a} = -\omega^2 \cos(\omega t)\hat{x} - \omega^2 \sin(\omega t)\hat{y} = -\omega^2 [\cos(\omega t)\hat{x} + \sin(\omega t)\hat{y}] a=−ω2r\mathbf{a} = -\omega^2 \mathbf{r}
  4. Direction of Acceleration: The relation a=−ω2r\mathbf{a} = -\omega^2 \mathbf{r} shows that acceleration is antiparallel to the position vector r\mathbf{r}. Since r\mathbf{r} is directed outward from the origin, acceleration is directed towards the origin (Centripetal Acceleration) .
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