
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.
Answer
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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.
Complete step by step answer:
\[\Rightarrow \overrightarrow{v}={{t}^{2}}\widehat{i}+2t\widehat{j}\]
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
Additional information:
When the body moves in a circular motion it has both linear velocity and angular velocity. If a particle moves with a constant speed then its speed remains constant but its velocity changes because velocity has both magnitude and direction. When the body moves there is a constant force which acts on the particle towards the centre of the path. This is called centripetal force and if during the moving particle is allowed to move freely it flies away tangentially.
Note:We are not given with the value of radius of the circle. The angular velocity is linear velocity divided by the radius. This is measured in radians per second and is defined as the rate of change of angle projected by the particle moving in the circle.
Complete step by step answer:
\[\Rightarrow \overrightarrow{v}={{t}^{2}}\widehat{i}+2t\widehat{j}\]
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
Additional information:
When the body moves in a circular motion it has both linear velocity and angular velocity. If a particle moves with a constant speed then its speed remains constant but its velocity changes because velocity has both magnitude and direction. When the body moves there is a constant force which acts on the particle towards the centre of the path. This is called centripetal force and if during the moving particle is allowed to move freely it flies away tangentially.
Note:We are not given with the value of radius of the circle. The angular velocity is linear velocity divided by the radius. This is measured in radians per second and is defined as the rate of change of angle projected by the particle moving in the circle.
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