170
4 Rotational Dynamics
4.38 The horizontal component of force at Q is mv 2 /R. The drop in height in coming down to Q is
(6R − R) = 5R
Gain in kinetic energy = loss in potential energy
7
10
mv
2
= (mg)(5R)
∴
mv 2
R
=
50
7
mg
4.39 Let the velocity on the top be v. Energy conservation gives
1
2
mv
2
0 = mgr cos θ 0 +
1
2
mv
2
(1)
where r cos θ 0 is the height to which the particle is raised. Angular momentum
conservation gives
mvr = mv 0 r sin θ 0
(2)
Eliminating v between (1) and (2) and simplifying
v 0 =
2gr
cos θ 0
4.40 Equation of motion is
ma = mg sin θ − T
(1)
Torque τ = TR = I α =
1
2
m R
2 a
R
∴ T =
1
2
ma
(2)
Using (2) in (1)
a =
2
3
g sin θ =
2
3
g sin 30
◦
=
g
3
4.41 < ω > =
ωdt
dt
(1)
τ = I α = C
√
ω
where C is a constant.
4 Rotational Dynamics
4.38 The horizontal component of force at Q is mv 2 /R. The drop in height in coming down to Q is
(6R − R) = 5R
Gain in kinetic energy = loss in potential energy
7
10
mv
2
= (mg)(5R)
∴
mv 2
R
=
50
7
mg
4.39 Let the velocity on the top be v. Energy conservation gives
1
2
mv
2
0 = mgr cos θ 0 +
1
2
mv
2
(1)
where r cos θ 0 is the height to which the particle is raised. Angular momentum
conservation gives
mvr = mv 0 r sin θ 0
(2)
Eliminating v between (1) and (2) and simplifying
v 0 =
2gr
cos θ 0
4.40 Equation of motion is
ma = mg sin θ − T
(1)
Torque τ = TR = I α =
1
2
m R
2 a
R
∴ T =
1
2
ma
(2)
Using (2) in (1)
a =
2
3
g sin θ =
2
3
g sin 30
◦
=
g
3
4.41 < ω > =
ωdt
dt
(1)
τ = I α = C
√
ω
where C is a constant.