Question

A A particle undergoes uniform circular motion.The velocity and angular of particle at an instant time is V=3I+4J m/s and omega(w)= XI+6J.Find X in rad/sec & radius and acceleration.

Answer 1

Acceleration x = -8

Explanation:

Given,

$\vec{v} = 3\times \hat{i} + 4\times \hat{j}$

$\vec{\omega} = x\times \hat{i} + 6\times \hat{j}$

We know that,

$\vec{v} = \vec{\omega} \times \vec{r}$

$\vec{v}\perp\vec{\omega}$

⇒ $\vec{v}\times \vec{\omega} = 0$

⇒ $3\times x + 24 = 0$

or,

$x = -8$

Answer 2

Given that,

Velocity $v=(3i+4j)\ m/s$

Angular velocity $\omega=(xi+6j)\ rad/s$

A particle undergoes uniform circular motion.

It means velocity is perpendicular to angular velocity.

We need to calculate the value of x

Using formula of velocity

$v=\omega\times r$

The velocity is perpendicular to angular velocity

So, $v\cdot\omega=0$

Put the value into the formula

$(3i+4j)\cdot(xi+6j)=0$

$3x+24=0$

$3x=-24$

$x=-8\ rad/s$

So, the angular velocity is

$\omega=(-8i+6j)\ rad/s$

The magnitude of the velocity is

$|v|=\sqrt{3^2+4^2}$

$|v|=5\ m/s$

The magnitude of the angular velocity is

$|\omega|=\sqrt{(-8)^2+6^2}$

$|\omega|=10\ rad/s$

We need to calculate the radius

Using formula of angular velocity

$v=r\omega$

$r=\dfrac{v}{\omega}$

Put the value into the formula

$r=\dfrac{5}{10}$

$r=0.5\ m$

We need to calculate the radial acceleration

Using formula of radial acceleration

$a_{c}=\dfrac{v^2}{r}$

Put the value into the formula

$a_{c}=\dfrac{10^2}{5}$

$a_{c}=20\ m/s^2$

Hence, The radius of particle is 5 m/s.

The radial acceleration of the particle is 20 m/s².