A particle moves so that its position vector | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the position vector: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$.$1$. Velocity $\vec{v}$ is the derivative of position with

Vedclass · Accepted Answer

Velocity is perpendicular to $\vec{r}$ and acceleration is directed towards the origin.

Vedclass · Answer

(B) Given the position vector: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$.$1$. Velocity $\vec{v}$ is the derivative of position with respect to time:$\vec{v} = \frac{d\vec{r}}{dt} = -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j}$.$2$. Acceleration $\vec{a}$ is the derivative of velocity with respect to time:$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2 \cos(\omega t) \hat{i} - \omega^2 \sin(\omega t) \hat{j} = -\omega^2 \vec{r}$.$3$. Since $\vec{a} = -\omega^2 \vec{r}$,the acceleration is directed towards the origin (centripetal acceleration).$4$. To check the relationship between $\vec{r}$ and $\vec{v}$,calculate the dot product:$\vec{r} \cdot \vec{v} = (\cos(\omega t))(-\omega \sin(\omega t)) + (\sin(\omega t))(\omega \cos(\omega t)) = -\omega \sin(\omega t) \cos(\omega t) + \omega \sin(\omega t) \cos(\omega t) = 0$.Since the dot product is $0$,the velocity is perpendicular to the position vector $\vec{r}$.

$A$ particle moves so that its position vector is given by $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$,where $\omega$ is a constant. Which of the following is true?

Similar Questions

$A$ fan is making $600$ revolutions per minute. If after some time it makes $1200$ revolutions per minute,then the increase in its angular velocity is

$A$ particle performing uniform circular motion of radius $\frac{\pi}{2} \,m$ makes $x$ revolutions in time $t$. Its tangential velocity is

$A$ body moves along a circular path of diameter $30 \text{ cm}$. It starts from one end of a diameter, moves along the circular path, and reaches the other end of the diameter in $3 \text{ seconds}$. The angular speed of the body in radians per second is:

$A$ particle is moving in uniform circular motion. About which point in the plane of the circle will the angular momentum of the particle always remain conserved?

Two bodies of equal masses revolve in circular orbits of radii $R_1$ and $R_2$ with the same period. Their centripetal forces are in the ratio:

Vedclass Test Series

Exam Paper Generator

Online Exam Module