Question
A particle moves so that its position vector is given by $$r = \cos \omega t\hat x + \sin \omega t\hat y,$$ where $$\omega $$ is a constant.
A particle moves so that its position vector is given by $$r = \cos \omega t\hat x + \sin \omega t\hat y,$$ 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 form the origin
D.
Velocity and acceleration both are perpendicular to $$r.$$
Answer :
Velocity is perpendicular to $$r$$ and acceleration is directed towards to origin
Solution :
Position vector of the particle is given by $$r = \cos \omega t\hat x + \sin \omega t\hat y$$
where $$\omega $$ is a constant.
Velocity of the particle is
$$\eqalign{ & v = \frac{{dr}}{{dt}} = \frac{d}{{dt}}\left( {\cos \omega t\,\hat x + \sin \omega t\,\hat y} \right) \cr & = \left( { - \sin \omega t} \right)\omega \hat x + \left( {\cos \omega t} \right)\omega \hat y \cr & = - \omega \left( {\sin \omega t\,\hat x - \cos \omega t\hat y} \right) \cr} $$
Acceleration of the particles
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left[ { - \omega \sin \omega t\,\hat x + \omega \cos \omega t\,\hat y} \right] \cr & = - {\omega ^2}\cos \omega t\,\hat x - {\omega ^2}\sin \omega t\,\hat y \cr & = - {\omega ^2}\left( {\cos \omega t\,\hat x + \sin \omega t\widehat y} \right) \cr & \Rightarrow a = - {\omega ^2}r = {\omega ^2}\left( { - r} \right) \cr} $$
Assuming the particle as $$P,$$ then its position vector is directed as shown in the diagram.
Therefore, acceleration is directed towards $$- r$$ that is towards $$"O"$$ (origin).
$$\eqalign{ & {\text{Now, we have}}\,v \cdot r = - \omega \left( {\sin \omega t\,\hat x - \cos \omega t\,\hat y} \right)\left( {\cos \omega t\,\hat x + \sin \omega t\,\hat y} \right) \cr & = - \omega \left[ {\sin \omega t \cdot \cos \omega t + 0 + 0 - \sin \omega t \cdot \cos \omega t} \right] \cr & = - \omega \left[ 0 \right] = 0\,\,\,\left[ {\because \hat x \bot \hat y} \right] \cr & v \bot r \cr} $$
Thus, velocity is perpendicular to $$r.$$
Position vector of the particle is given by $$r = \cos \omega t\hat x + \sin \omega t\hat y$$
where $$\omega $$ is a constant.
Velocity of the particle is
$$\eqalign{ & v = \frac{{dr}}{{dt}} = \frac{d}{{dt}}\left( {\cos \omega t\,\hat x + \sin \omega t\,\hat y} \right) \cr & = \left( { - \sin \omega t} \right)\omega \hat x + \left( {\cos \omega t} \right)\omega \hat y \cr & = - \omega \left( {\sin \omega t\,\hat x - \cos \omega t\hat y} \right) \cr} $$
Acceleration of the particles
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left[ { - \omega \sin \omega t\,\hat x + \omega \cos \omega t\,\hat y} \right] \cr & = - {\omega ^2}\cos \omega t\,\hat x - {\omega ^2}\sin \omega t\,\hat y \cr & = - {\omega ^2}\left( {\cos \omega t\,\hat x + \sin \omega t\widehat y} \right) \cr & \Rightarrow a = - {\omega ^2}r = {\omega ^2}\left( { - r} \right) \cr} $$
Assuming the particle as $$P,$$ then its position vector is directed as shown in the diagram.
Therefore, acceleration is directed towards $$- r$$ that is towards $$"O"$$ (origin).
$$\eqalign{ & {\text{Now, we have}}\,v \cdot r = - \omega \left( {\sin \omega t\,\hat x - \cos \omega t\,\hat y} \right)\left( {\cos \omega t\,\hat x + \sin \omega t\,\hat y} \right) \cr & = - \omega \left[ {\sin \omega t \cdot \cos \omega t + 0 + 0 - \sin \omega t \cdot \cos \omega t} \right] \cr & = - \omega \left[ 0 \right] = 0\,\,\,\left[ {\because \hat x \bot \hat y} \right] \cr & v \bot r \cr} $$
Thus, velocity is perpendicular to $$r.$$
