### 剛体の回転運動

```(angular acceleration)
Δω dω d2θ
β = lim
= dt = dt2
Δt
(β=
ω
ω = ω0 + β t
1
θ = ω0 t +
β t2
2
ω
θ
)
½βt2
ω2 − ω02 = 2 β θ
(angular velocity)
dθ
Δθ
ω = lim
=
dt
Δt→0 Δt
(ω=
ω
(
’ v = v0 + a t
)
θ=ωt
v
θ
’ x = v0 t +
x=vt
’
v0 :
x
t
ω0t
ω0 :
v2
−
v02
a=
t
)
v
1 2
at
2
=2ax
½at2
v0t
t
t
1
d2θ
dt2 = β (
)
t
dθ
= β t + ω0
dt
(
ω0
)
t=
ω – ω0
β
θ=
ω – ω0 2
1
ω – ω0
β(
) + ω0 (
)
β
2
β
ω 2 − ω 02 = 2 β θ
ω
ω = ω0 + β t
(
)
t
1
θ = 2 β t2 + ω0 t + θ0
θ0
θ0 = 0
2
x=rθ
β:
r=
)
t
dx
dθ
=r
dt
dt
a=rβ
a:
(
v=rω
r:
t
β
θ
x=rθ
r
dv
dω
=
r
dt
dt
a=rβ
□
x
a
3
m
F
a
N
N:
I:
β:
(
N=Iβ
)
(
N=Fr
F=ma
)
N=mar
a=rβ
N = m r2 β
(
)
I = m r2
r
r
m
□
I
β
F
r
N=Iβ
I = m r2
a
m
4
(ii)
ρ
m
(i)
ρ
m
r
r
L
dL
x
dx
(
(
)
ρdx)
di = ρdx x2
L
di
(I
I = ∫ di = ∫0 ρ x2 dx = ρ [
=mr2)
x3 L ρL3
] =
3
3 0
(
di
(I
=mr2)
ρ r2 dL = ρ r2
dL
ρdx)
di = ρdL r2
I = ∫ di =
)
dL = rdθ
2π
(
(θ
)
2π
dL = ∫0 r dθ = r [θ] 0 = 2 π r
m = ρL
1
I = 3 m L2
I = ρ r2 2 π r
m=ρ 2πr
I = m r2
5
(iii)
ρ
dr
m
R
r
(
dr)
di
(I =mr2)
di = ρ 2πr3 dr
R
R
I = ∫ di = ∫0 ρ 2πr3 dr = 2πρ ∫0 r3 dr = 2πρ
R4
4
m = ρ πR2
I=
1
m R2
2
6
(
(
mA
)
A
B
TB
A
mAa = TA – mAg
B
mBa = mBg – TB
TA = TB = T (
)
+
(mA + mB)a = (mB − mA)g
mB − mA
a=
g
mA + mB
)
A
mB
B
g
A B
TA
r
I
0
M
a
mB
mA
I=
1
M R2
2
a
7
M(≠0)
B
A
TA < TB
TB
TA
+
**
*
**
{(mB − mA)g − (mA + mB)a}r2
a=
I
1
I=
M R2
2
(mB − mA)g
a=
1
M + mA + mB
2
β
r
mAa = TA – mAg
mBa = mBg – TB
(mA + mB)a = (mB − mA)g + TA – TB
TB – TA = (mB − mA)g − (mA + mB)a
I β = (TB – TA) r
TA
A
B
TB
M=0
a
a=rβ
mB
mA
a
(TB – TA) r2
a=
I
*
8
a
r
2
m
f
g
θ
a
ma = mgsinθ – f
β
θ
I = ½ m r2 a = r β
ma = mgsinθ – f
f
a = gsinθ –
m
ma = mgsinθ – ½ ma
Iβ=fr
½ ma = f
a=
2
gsinθ
3
9
```