Question

Angular position of a particle performing circular motion is given by $\theta =\left(t^2-2t\right)$. Consider the ads circular track to be 2m. Angular distance of the particle at t=3s.

Answer 1

Hint:Angular distance of a particle at a particular time can be calculated by substituting the value of t in the given equation of angular distance of the particle.

Formula Used:The given equation for the angular distance of the particle is given as:
$d=r\theta$
Where r is the radius of the circular track and $\theta$is the angular position.

Step by step answer:
The angular distance of the particle performing circular motion is given by the equation:
$\theta =(t^2-2t)$
According to the question, the radius of the circular path $r=2m$
Now substituting the value of $\theta$and $r$in the formula of Angular distance we have:
$d=2(t^2-2t)=(2t^2-4t)$
In this question it has been asked for us to find the value of angular distance of the particle at time, $t=3s$.
Thus, we have to now substitute the value of $t$into the equation obtained above. Doing this, we get:
$d=2(3^2)-4(3)=18-12$
Thus, solving the above equation we get the angular distance:
$d=6m$
Thus, a particle travels an angular distance of 6 metres.

Additional Information:Angular distance might sound a bit synonymous to the word angle but it is actually suggestive of the linear distance between two points or objects.

Note:This numerical problem can also be solved by first putting the value of t in the formula for angular position and then substituting the value of the radius of the circular path covered by the body but it is not advised as this could further complicate the equation and this might lead to calculation errors. It is always better to take an approach with the least hurdles so as to arrive in the correct answer.