Question

(a) A small object with mass 3.75 kg moves counterclockwise with constant speed 1.55 rad/s in a circle of radius 2.55 m centered at the origin. It starts at the point with position vector 2.55 i m. Then it undergoes an angular displacement of 8.95 rad.
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity?

Answer 1

Answer:3.95 m/s

Explanation:

Given

mass of object $m=3.75 kg$

$\omega =1.55 rad/s$

radius of circle $=2.55 m$

initial Position $r=2.55 \hat{i}$

angular displacement $\theta _0=8.95 rad$

8.95 radian can be written as

$1.42 (2\pi )$

i.e. Particle is at first quadrant with $\theta =0.4242\pi \times \frac{180}{\pi }$

$\theta =76.36^{\circ}$

(c)velocity is $v=\omgea \times r$

$v=1.55\times 2.55=3.95 m/s$