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# A particle moves in a circle of radius r with a velocity given by $\overrightarrow{v}={{t}^{2}}\widehat{i}+2t\widehat{j}$. Find the angular velocity for time 1 second.

Last updated date: 13th Jun 2024
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Hint:We are given with a velocity vector in terms of time and we are given that the particle is moving in a circle of radius r, we need to find the angular velocity at t=1 s. we can find the value of linear velocity after 1 sec and then can use the relationship between linear velocity and angular velocity to arrive at the solution.

$\Rightarrow \overrightarrow{v}={{t}^{2}}\widehat{i}+2t\widehat{j}$
At t=1 s,
$\Rightarrow \overrightarrow{v}={{(1)}^{2}}\widehat{i}+2(1)\widehat{j}$
$\Rightarrow \overrightarrow{v}=\widehat{i}+2\widehat{j}$
Finding its magnitude by taking its dot product with itself,
\begin{align} &\Rightarrow \overrightarrow{v.}\overrightarrow{v}=\widehat{(i}+2\widehat{j}).\widehat{(i}+2\widehat{j}) \\ &\Rightarrow v=\sqrt{1+4} \\ &\Rightarrow v=\sqrt{5} \\ \end{align}
So, the value of velocity after 1s is $\sqrt{5}$units. Now we know for a particle in a circular motion the relationship between linear velocity and angular velocity is given by, $\Rightarrow v=r\omega$
Putting the values, we get $\dfrac{\sqrt{5}}{r}=\omega$
This is the angular velocity after 1 sec