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

Step-by-step 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 * (180)/(\pi )`

`\theta =76.36^(\circ)`

(c)velocity is
`v=\omgea * r`

`v=1.55* 2.55=3.95 m/s`