CLASS 11 SYSTEM OF PARTICLES AND ROTATIONAL MOTION
SYSTEM OF PARTICLES & ROTATIONAL MOTION
✤ CENTRE OF MASS :- It is a point inside a body at which total mass of the body is concentrated on suppose to be concentrated.
Coordinate of C.M:-
X = m₁x₁ + m₂x₂ + m₃x₃ / m₁ + m₂ + m₃
Y = m₁y₁ + m₂y₂ + m₃y₃ / m₁ + m₂ + m₃
① Rcm = m₁r₁ + m₂r₂ / m₁ + m₂
② If Rcm = 0, CM at origin
m₁r₁ + m₂r₂ = 0
m₂r₂ = -m₁r₁
r₂ = -m₁r₁ / m₂
If m₁ > m₂ r₂ > r₁
i.e CM always app. the heavier body.
③ Equation of CM
* F = (m₁ + m₂) * d²R / dt²
* F = M dVcm / dt
If F = 0 i.e no external force act on the body
[ Vcm = Constant
(4) Vcm = m₁v₁ + m₂v₂ / m₁+m₂ ← in same direction
Vcm = m₁v₁ - m₂v₂ / m₁+m₂ → in opposite direction
✤ TORQUE:-
T = r f sinθ It is defined as the product of force & perpendicular distance b/w line of action of force & axis of rotation
SI unit - Newton meter
Dimensions: ML2T-2
Direction of torque is always ⊥ to r & f
T = r X f
Some Special Cases :-
① If θ = 0° ② If θ = 90°
T = r f sinθ T = r f sin90°
T = 0 Tmax = fr
Torque in Vector form
Let r = r₁î+r₂ĵ+r₃k̂
F = f₁î+f₂ĵ+f₃k̂
T = r x f
T = [■(î&ĵ&k̂@r₁&r₂&r₃@f₁&f₂&f₃)]
= î [■(r2&r3@f2&f3)] - ĵ [■(r1&r3@f1&f3)] + k̂ [■(r1&r2@f2&f2)]
= î(r₂f₃-r₃f₂) - ĵ(r₁f₃-r₃f₁) + k̂ (r₁f₂-f₁r₂)
* Relation b/w torque & work:
Let dw to torque let angular displacement is dθ
∴ dw = T-dθ
✤ Relation blw Power & torque :-
We have
dw = Tdt
Divide by dt both side
Dω/dt = τ.dω/dt
[ P = τ.ω ]→ omega
❈ Angular Momentum: It is defined as the product of linear momentum & ⊥ distance blw line of action of momentum & ⊥ axis of rotation
L = rPsino
In vector form 
∴ τ =r×p Unit is kgm²/s
Dimensions [ML²T-1]
NOTE :-
* If θ = 90° We know If θ = 0°
L= rP r = rw L = 0
If p = mv L = mr²ω
L = mvr
We know, If θ = 0° L = 0
✤ Relation blw angular momentum (L) & torque (τ) :-
We have, -
L=Pr
differentiating dL/dt = r.dp/dt
dL = r.f
dl/dt = τ
Note: If T=0 then dl/dt =0
=> L = Constant
Conservation of angular momentum
If T = θ
then L1 + L2 + L3 = Constant
No of revolution = θ / 2π
Angular Speed = V°/r
w = V°/r
θ = 2πn
✤ Torque:-
It is defined as the produce of force & ⊥ distance b/w line of action of force & axis of rotation
Torque = fx r sin θ
T = fr sinθ
Slunt: Nm
Dimensions: ML²T-2
Direction of torque us always ⊥ to r & f
T = r × f
✤ SOME SPECIAL CASES:-
1) If θ=0° 2) If θ=90°
τ=fr Sinθ τ= fr Sin90°
τ=0 τmax = fr
Torque in Vector form:-
Let r = r₁î+r₂ĵ+r₃k
F = f₁î+f₂ĵ+f₃k
T = r × F
T = [■(î&ĵ&k̂@r₁&r₂&r₃@f₁&f₂&f₃)]
= î [■(r2&r3@f2&f3)] - ĵ [■(r1&r3@f1&f3)] + k̂ [■(r1&r2@f1&f2)]
= î(r₂f₃-r₃f₂) - ĵ(r₁f₃-r₃f₁) + k̂ (r₁f₂-f₁r₂)
RELATION B/W TORQUE & WORK:-
Let dw to torque let angular displacement is dθ
∴ [ dw = τ.dθ ]
RELATION B/W POWER & TORQUE:-
We have, dw=τdt
divide by dt both side.
Dw/dt = τ.dω/dt
[ P=τ.ω ]
ANGULAR MOMENTUM :-
It is defined as the product of linear momentum & ⊥ distance blur line of angular momentum & axis rotation.
L = rpsino
In Vector from
:. L = r x P
Units K gm²/s
Dim [ML²T-¹]
Note :- If θ = 90° We have, P=mv We know, V=rω
L = rp L=mvr L=mr²ω
If θ = 0°
L=0
RELATION Btw ANGULAR MOMENTUM (L) & TORQUE (τ):-
We have,
L = pr
diff. dL/dT = r. dp/dt
dL/dt = rf
[ dL/dt = τ ]
NOTE :- If τ = 0 then dL = 0 => L = Constant
✤ Conservation of Angular momentum
Angular Momentum is constant
Therefore
d/dt(L) = 0
Then Σ τ = 0 = constant
✤ INERTIA:-
It is defined as the inherent property by which a body opposes the change of state against direction
I = mr² SI Unit : Kgm²
-> Inertia depend upon the mass and position of mass
❈ Relation (Torque) (T) Inertia(I) :-
We know:- τ = fr
= m(a)r
= m(ar)r
= mr²α
[ τ= Iα ]
✤ Relation Between Angular Momentum & Inertia:-
We know:
L = fr = mvr = m(ωr)r = mr²ω
L= mr²ω
[ L = Iω ]
✤ Relation b/w K.E & Inertia:-
We Know
K.E = ½ mv²
= ½ m(ωr)²
= ½ mω²r²
[ l.e= 1/2 Iω² ]
✤ CONSERVATION OF ANGULAR MOMENTUM:-
If torque is absent then
I₁ω₁ = I₂ω₂
❈ THEOREM OF PARALLEL AXIS:-
(1) Ixy = IAB+ Mh²
(ii) Ixy = IAB + M(l/2)²
= 1/12 Ml² + 1/4 Ml²
= Ml²+3Ml²
= 4 Ml² = 1/3 Ml²
(iii) Ixy = IAB+ MR²
= 2MR²
❊ THEOREM OF ⊥ AXIS:-
Iy = Ix+ Iz
✤ EQUILIBRIUM OF FORCE
Anticlockwise torque = f₁ X OA
Clockwise torque = f₂ X OB
❈ IN EQUILIBRIUM:
=> f₁xOA - f₂xOB = 0
=>[ f₁xOA - f₂xOB ]
* * * * * * * * * * * * * * * *
