a)New position vector in vector form= r = 0.233404 î + 2.94056j m
b) Lies in second quadrant at 161.391°
c)Velocity =4.8675 m/s
d)It is moving in a direction making 161.391° with positive x-direction.
e)Acceleration will be centripetal acceleration (8.031 m/s²).
Given:
Mass of the object m = 3.95 kg
ω=1.65 rad/s
Radius of the circle = 2.95 m
a)
new position vector in vector form
=R cos1.65 î + R sin 1.65 j
= 2.95 cos1.65 î +2.95 sin1.65 j
= 2.95 x 0.07912 î + 2.95 x 0.9968 j
r = 0.233404 î + 2.94056j
b)
Angular Displacement = θ₀ = 9.10 rad
9.10 radian = 180/π× 9/10 degree
= 521.391°
=521.391°- 360°
=161.391°
This will lie in second quadrant.
Angle made with positive x-axis
=161.391°
c)
Velocity
v = ω R
= 1.65 x 2.95
=4.8675 m/s
d)
It is moving in a direction making 161.391° with positive x-direction.
e)
Acceleration will be centripetal acceleration.
= v²/R
=(4.8675)² / 2.95
=23.6925562 / 2.95
=8.031 m/s²
f) Position, Velocity and Acceleration graph:
Learn more about Angular displacement here:
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