Angular Displacement: ω=dtdQ
Angular acceleration :α=dtdω
α: constant
dtdω=α= constant
Integrate: ω = αt+c
at t=t0,ω=ω0⇒c=w0
ω=ω0+αt
t = αω−ω0
Linear Motion | Rotational motion |
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v=u+at | ω=w0+αt |
s=s0+ut+21a2 | θ=θ0+ω0t+21αt2 |
v2=u2+2as | ω2=ω02+2αθ |
dtdθ=ω0+αt
Integrate: θ=ω0t+2αt2 + c (c: a-constant)
θ(t=0)=θ0⇒c=θ0
θ=θ0+ω0t+2αt2
v = r ω
where: (v=ω×r)
v = dtdr=rdtdθ
PQ=a(A)
at=dtdv=dtd(rω)
=rdtdω=rα
(α×r=Ar)
ar=rv2 (centripetal acceleration)
=r(rω)2=rω2=r(dtdθ)2=rθ2
Ar=−rθ˙2e^r
a=A=Arer^+Ateθ^ :
a = ∣a∣ = at2+ar2 = (rα)2+(rω2)2
a = rα2+ω4
Ft=mat
The torque about the origin due to force Ft
τ=Ftr=(mat)r
at=rα
∴tau=(mr2)α=Iα
τ=Iα⇒τ∝α
I: Proportionality constant
dFt=(dm)at
Torque d τ due to the force dFT about the origin:
dτ=rdFt=(rdm)at
= (r dm) (rα)
τ=α∫r2dm
⇒τ=Iα
τ=r×F
[τ] = Dimension work or Energy
τ rotates the object dθ, then the work done for this infinitedecimal rotation:
dω=τdθ(Fdx)
dtdω=τdtdθ=τω
Rate at which the work is done is callas instantaneous power.
P = τω
("P = F v")
Work energy theorem in rotational motion
we have
τ=Iα=Idtdω=Idtdωdtdθ=Iωdθdω
So, τdθ=Iωdω=dW
Integrate,
w=∫θ0θτdθ=∫θ0θIωd ω=2I(ω2−ω02)
θ−θ0 : the change in angular displacement.
ω−ω0 : the corresponding change in angular speed.
W = 23(ω2−ω02)
W = 2m(v2−ω2)in linear motion
Rotational Motion
Linear Motion
Rotational Motion | Linear Motion |
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1. Angular velocity: ω=dtdθ=θ˙ | 1. Linear velocity: v= dtdx |
2. Angular acceleration: α=dtdω=θ¨ | 2. Linear acceleration: a = dtdv=x¨ |
3. Torque τ=Iα | 3. Force: F=Ma |
4. ω=ω0+αt | 4. v=u+at |
θ=θ0+ω0t+21αt2 | s=x0+ut+21at2 |
ω2=ω02+2a(θ−θ0) | v2=u2+2a(s−x0) |
5. Work done (W) = ∫θ0θτdθ | 5. Work done (W) =∫x0xFxdx |
6. KE =21Iω2 | 6. kE=21mv2 |
7. Power (P) = τω | 7. Power P = F v |
8. Angular momentum (L)= Iω | 8. Linear momentum p=mv |
9. τ=dtdL← newton's 2nd law | 9. F=dtdp |