Problem :
The angular position of a particle (in radian), on circular path of radius 0.5 m, is given by :
θ
=

0.2
t
2

0.04
θ
=

0.2
t
2

0.04
At t = 1 s, find (i) angular velocity (ii) linear speed (iii) angular acceleration (iv) magnitude of tangential acceleration (v) magnitude of centripetal acceleration and (vi) magnitude of total acceleration.
Solution :
Angular velocity is :
ω
=
đ
θ
đ
t
=
đ
đ
t
−
0.2
t
2
−
0.04
=
−
0.2
X
2
t
=
−
0.4
t
ω
=
đ
θ
đ
t
=
đ
đ
t
−
0.2
t
2
−
0.04
=
−
0.2
X
2
t
=
−
0.4
t
The angular speed, therefore, is dependent as it is a function in "t". At t = 1 s,
ω
=
−
0.4
rad
/
s
ω
=
−
0.4
rad
/
s
The magnitude of linear velocity, at t = 1 s, is :
v
=
ω
r
=
0.4
X
0.5
=
0.2
m
/
s
v
=
ω
r
=
0.4
X
0.5
=
0.2
m
/
s
Angular acceleration is :
α
=
đ
ω
đ
t
=
đ
đ
t

0.4
t
=
−
0.4
rad
/
s
2
α
=
đ
ω
đ
t
=
đ
đ
t

0.4
t
=
−
0.4
rad
/
s
2
Clearly, angular acceleration is constant and is independent of time.
The magnitude of tangential acceleration is :
⇒
a
T
=
α
r
=
0.4
X
0.5
=
0.2
m
/
s
2
⇒
a
T
=
α
r
=
0.4
X
0.5
=
0.2
m
/
s
2
Tangential acceleration is also constant and is independent of time.
The magnitude of centripetal acceleration is :
⇒
a
R
=
ω
v
=
0.4
t
X
0.2
t
=
0.08
t
2
⇒
a
R
=
ω
v
=
0.4
t
X
0.2
t
=
0.08
t
2
At t = 1 s,
⇒
a
R
=
0.08
m
/
s
2
⇒
a
R
=
0.08
m
/
s
2
The magnitude of total acceleration, at t =1 s, is :
a
=
a
T
2
+
a
R
2
a
=
a
T
2
+
a
R
2
a
=
{
0.2
2
+
0.08
2
}
=
0.215
m
/
s
2
a
=
{
0.2
2
+
0.08
2
}
=
0.215
m
/
s
2