Question

A particle starts from the point (0m,8m) and moves with a uniform velocity of 3i^ m/s. After 5 seconds , its angular velocity about the origin will be:
A)8/289 rad/s
B)3/8 rad/s
C)24/289 rad/s
D)8/17 rad/s

Answer 1

Dear student

Particle starts from (0m , 8m) from rest with uniform velocity 3 m/s in the x-direction so after 5 sec the particle will traverse a distance along x-axis is 3*5=15 m so new position of the particle is ​(15m , 8m). So the distance of this new position to the origin is $\sqrt{{\left(15-0\right)}^2+{\left(8-0\right)}^2}=\sqrt{225+64}=\sqrt{289}=17m$. Thus the angular velocity about origin is given by
$$\begin{gathered} \omega =\frac{v}{r} \\ \omega =\frac{3}{17}rad/s \end{gathered}$$