116
3 Rotational Kinematics
3.7 a =
a 2
N + a 2
T
(Fig. 3.9)
a N
a T
= tan 30
◦
=
1
√
3
a N =
a T
√
3
(1)
a T = α R
(2)
ω = αt
(3)
∴ a N = ω
2 R =
a T
√
3
=
α R
√
3
ω
2
=
α
√
3
α
2 t
2
=
α
√
3
α =
1
√
3t 2
=
1
√
3.1 2
= 0.577 rad/s
2
Fig. 3.9
3.8 a =
a 2
N + a 2
T
12
√
10 =
(ω 2 R) 2 + α 2 R 2 =
(α 2 t 2 R) 2 + α 2 R 2
= α R
α 2 t 4 + 1 = 3R
3 2 × 1 2 + 1
R = 4 cm
3.9 (i) Equating the centripetal force to the frictional force
mv 2
max
R
= μ mg
∴ v max =
μ g R
3 Rotational Kinematics
3.7 a =
a 2
N + a 2
T
(Fig. 3.9)
a N
a T
= tan 30
◦
=
1
√
3
a N =
a T
√
3
(1)
a T = α R
(2)
ω = αt
(3)
∴ a N = ω
2 R =
a T
√
3
=
α R
√
3
ω
2
=
α
√
3
α
2 t
2
=
α
√
3
α =
1
√
3t 2
=
1
√
3.1 2
= 0.577 rad/s
2
Fig. 3.9
3.8 a =
a 2
N + a 2
T
12
√
10 =
(ω 2 R) 2 + α 2 R 2 =
(α 2 t 2 R) 2 + α 2 R 2
= α R
α 2 t 4 + 1 = 3R
3 2 × 1 2 + 1
R = 4 cm
3.9 (i) Equating the centripetal force to the frictional force
mv 2
max
R
= μ mg
∴ v max =
μ g R