Question

A small object with mass 3.80 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.70 m centered at the origin. It starts at the point with position vector 2.70 m. Then it undergoes an angular displacement of 8.70 rad.
(a) What is its new position vector?
in meters
(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?
in m/s
(d) In what direction is it moving?
_____° from the +x direction.
(e) What is its acceleration?
in m/s2
(f) What total force is exerted on the object?
in N

Answer 1

Answer:

Explanation:

angular velocity

ω = 1.65 rad /s

radius R = 2.70 m

angular displacement = 8.70 rad

a )

New position vector in vector form

= R cos8.7 i + R sin8.7 j

= 2.7 cos8.7 i + 2.7 sin8.7 j

= 2.7 x .748 i + 2.7 x .663 j

= 2.01 i + 1.79 j

b )

8.7 radian = 180/π x 8.7 degree

= 498.72 degree

= 498.72 - 360

= 138.72 degree

It will be in second quadrant .

angle made with positive x - axis

= 138.72 degree .

c )

velocity

v = ω R

= 1.65 x 2.7

= 4.455 m /s

d )

It is moving in a direction making 138.72° with positive x direction .

e )

acceleration will be centripetal acceleration

= v²/ R  

= 4.455² / 2.7

= 7.35 m /s²

f ) force = mass x acceleration

= 3.8 x 7.35

= 27.93 N .