Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Rotational motion about a fixed axis-angular momentum System of Particles and Rotational Motion
→ \rightarrow → → \rightarrow → Rotational motion about a fixed axis-angular momentum System of Particles and Rotational Motion → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Angular Momentum - Rotation About a Fixed Axis
(1) Expression for the orbital angular momentum.
(2) Principle of conservation of angular momentum
(3) Rotation, roling and slipping
(4) An example
→ \rightarrow → → \rightarrow → Angular Momentum - Rotation About a Fixed Axis → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Angular Momentum
l ⃗ = r ⃗ × p ⃗ p ⃗ = n v ⃗ \vec{l} =\vec{r} \times \vec{p} \quad \vec{p}=n \vec{v} l = r × p p = n v
r ⃗ = o p ⃗ = o c → + c p ⃗ \vec{r}=\vec{op} =\overrightarrow{o c}+\vec{cp} r = o p = oc + c p
l ⃗ = ( o c → + c p ⃗ ) × m v ⃗ \vec{l} =(\overrightarrow{o c}+\vec{cp}) \times m \vec{v} l = ( oc + c p ) × m v
=( o c ⃗ × m v ⃗ ) + ( c p ⃗ × m v ⃗ ) (\vec{oc} \times m \vec{v})+(\vec{cp} \times m \vec{v}) ( oc × m v ) + ( c p × m v )
l ⃗ = ( o c ⃗ × m v ⃗ ) + ( r ⊥ 2 m ω k ^ ) l ⃗ z \vec{l} =(\vec{oc} \times m \vec{v})+(r_{\perp}^2 m \omega \hat{k}) \vec{l}_z l = ( oc × m v ) + ( r ⊥ 2 mω k ^ ) l z
l ⃗ = l ⃗ z + o c → × m v ⃗ \vec{l} =\vec{l}_z+\overrightarrow{o c} \times m \vec{v} l = l z + oc × m v
l z ⃗ ∥ k ^ \vec{l_z} \parallel \hat{k} l z ∥ k ^
Rotational motion about a fixed axis-angular momentum System of Particles and Rotational Motion → \rightarrow → → \rightarrow → Angular Momentum → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Angular Momentum
In general ı ⃗ and ω ⃗ \vec{\imath} \text { and } \vec{\omega} and ω
L ⃗ = ∑ i l i ⃗ = ∑ l ⃗ i z + ∑ i o c → i × m v ⃗ i \vec{L}=\sum_i \vec{l_i}=\sum \vec{l}_{i_z}+\sum_i \overrightarrow{o c}_i \times m \vec{v}_i L = ∑ i l i = ∑ l i z + ∑ i oc i × m v i
L ⃗ = L z ⃗ + L ⊥ ⃗ , L z ⃗ \vec{L}=\vec{L_z}+\vec{L_\perp}, \vec{L_z} L = L z + L ⊥ , L z
= ∑ l i z ⃗ = ( ∑ i m i r ⊥ i 2 ) ω k ^ =\sum \vec{l_{i_z}}=\left(\sum_i m_i r_{\perp_i}^2\right) \omega \hat{k} = ∑ l i z = ( ∑ i m i r ⊥ i 2 ) ω k ^
L ⃗ z = I ω k ^ ( p = m v ) \vec{L}_z=I \omega \hat{k} \quad(p=m v) L z = I ω k ^ ( p = m v )
So, L ⊥ ⃗ \vec{L_{\perp}} L ⊥
L ⃗ = L ⃗ z = I ω k ^ \vec{L}=\vec{L}_z= I \omega \hat{k} L = L z = I ω k ^
Angular Momentum - Rotation About a Fixed Axis → \rightarrow → → \rightarrow → Angular Momentum → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Example
L ⃗ = 2 ω k ^ ; d L ⃗ d t = I d ω d t k ^ = ( I α ) k ^ = τ k ^ \vec{L}=2 \omega \hat{k} ; \frac{d \vec{L}}{d t}=I \frac{d \omega}{d t} \hat{k}=(I \alpha) \hat{k}=\tau \hat{k} L = 2 ω k ^ ; d t d L = I d t d ω k ^ = ( I α ) k ^ = τ k ^
similarly L ⃗ = L z ^ + L ⊥ ⃗ ; d L z ⃗ d t = τ k ^ ; d L ⊥ ⃗ d t \vec{L}=\hat{L_z}+\vec{L_{\perp}} ; \frac{d \vec{L_z}}{d t}=\tau \hat{k} ; \frac{d \vec{L_{\perp}}}{d t} L = L z ^ + L ⊥ ; d t d L z = τ k ^ ; d t d L ⊥
Angular Momentum → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Example
Angular Momentum → \rightarrow → → \rightarrow → Example → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Principle of Conservation of Angular Momentum (PCAM)
The total angular momentum of a system is constant, if the resultant external torque acting in the system
is zero.
I i ω i = I f ω f I_i \omega_i=I_f \omega_f I i ω i = I f ω f
Example → \rightarrow → → \rightarrow → Principle of Conservation of Angular Momentum (PCAM) → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Illustration
The line of motion of bullet perpendicular to the axis of the cylinder
Before collision: l i = m v o d l_i=m v_o d l i = m v o d
After the collision: l f = I ω l_f = I \omega l f = I ω
( I solid cylinder + I projectile ) ω (I_\text{{solid cylinder}} + I_\text{{ projectile}} ) \omega ( I solid cylinder + I projectile ) ω
( M R 2 2 + m R 2 ) ω = m v 0 d \left(\frac{M R^2}{2}+m R^2\right) \omega=m v_0 d ( 2 M R 2 + m R 2 ) ω = m v 0 d
ω = m v 0 d ( M R 2 2 + m R 2 ) \omega =\frac{mv_0d}{\left(\frac{M R^2}{2}+m R^2\right)} ω = ( 2 M R 2 + m R 2 ) m v 0 d
Example → \rightarrow → → \rightarrow → Illustration → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Illustration 2
Question :- ω c = ? \omega_c = ? ω c = ?
l = I ω l=I \omega l = I ω l i = l f l_i=l_f l i = l f
I i = I c p + I m = M R 2 2 + m R 2 I_i=I_{c p}+I_m=\frac{M R^2}{2}+m R^2 I i = I c p + I m = 2 M R 2 + m R 2
I f = I c p + I m = M R 2 2 + m x 2 I_f=I_{c p}+I_m \quad=\frac{M R^2}{2}+m x^2 I f = I c p + I m = 2 M R 2 + m x 2
PCAM : I i ω i = I f ω f I_i \omega_i = I_f \omega_f I i ω i = I f ω f
( M R 2 2 + m R 2 ) ω = ( M R 2 2 + m x 2 ) ω c (\frac{M R^2}{2}+m R^2) \omega = (\frac{M R^2}{2}+m x^2) \omega_c ( 2 M R 2 + m R 2 ) ω = ( 2 M R 2 + m x 2 ) ω c
ω c = ( M R 2 2 + m R 2 ) ( M R 2 2 + m x 2 ) ω \omega_c=\frac{\left(\frac{M R^2}{2}+m R^2\right)}{\left(\frac{M R^2}{2}+m x^2\right)} \omega ω c = ( 2 M R 2 + m x 2 ) ( 2 M R 2 + m R 2 ) ω ω c > ω \omega_c > \omega ω c > ω
Principle of Conservation of Angular Momentum (PCAM) → \rightarrow → → \rightarrow → Illustration 2 → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Rotation, Rolling and Slipping
The rotating disc is moves gently on the table frictionless.
v A = R ω 0 , v B = R ω 0 v_A=R \omega_0, v_B =R \omega_0 v A = R ω 0 , v B = R ω 0
v c = R 2 ω 0 v_c =\frac{R}{2} \omega_0 v c = 2 R ω 0
Disc only rotates ; it will not roll !
Illustration → \rightarrow → → \rightarrow → Rotation, Rolling and Slipping → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Rotation, Rolling and Slipping
At P 0 : ∣ v P 0 ⃗ ∣ = v C M ⃗ = R ω 0 P_0:\left|\vec{v_{P_0}}\right|=\vec{v_{C M}}_=R \omega_0 P 0 : ∣ v P 0 ∣ = v CM = R ω 0
P 0 P_0 P 0
v C M = R ω 0 v_{CM} = R\omega_0 v CM = R ω 0
Illustration 2 → \rightarrow → → \rightarrow → Rotation, Rolling and Slipping → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Kinetic Energy of Rolling motion
K = K E of a rolling body K =K E \text { of a rolling body} K = K E of a rolling body
= K E of translation + K E of rotation =K E \text { of translation }+K E \text { of rotation} = K E of translation + K E of rotation
KE of a system of particles \text { KE of a system of particles} KE of a system of particles
= K E ∣ + K E rotational motion about the CM =K E |+K E \text { rotational motion about the CM } = K E ∣ + K E rotational motion about the CM
k = 1 2 m v c m 2 + 1 2 I ω 2 k=\frac{1}{2} m v_{c_m}^2+\frac{1}{2} I \omega^2 k = 2 1 m v c m 2 + 2 1 I ω 2
Rotation, Rolling and Slipping → \rightarrow → → \rightarrow → Kinetic Energy of Rolling motion → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Kinetic Energy of a system of particles
= K E ∣ + K E =K E |+K E = K E ∣ + K E
k = 1 2 m v c m 2 + 1 2 I ω 2 k=\frac{1}{2} m v_{c_m}^2+\frac{1}{2} I \omega^2 k = 2 1 m v c m 2 + 2 1 I ω 2
v C M = R ω v_{CM} = R\omega v CM = R ω
I = m k 2 , ′ k ′ : I=mk^2, ' k ': I = m k 2 , ′ k ′ :
k = 1 2 m k 2 v C M 2 R 2 + 1 2 m v C M 2 k=\frac{1}{2} m k^2 \frac{v_{CM}^2}{R^2}+\frac{1}{2} m v_{C M}^2 k = 2 1 m k 2 R 2 v CM 2 + 2 1 m v CM 2
k = 1 2 m v c m 2 ( 1 + k 2 R 2 ) k=\frac{1}{2} m v_{c m}^2\left(1+\frac{k^2}{R^2}\right) k = 2 1 m v c m 2 ( 1 + R 2 k 2 )
The K E K E K E
=K E KE K E K E KE K E
Rotation, Rolling and Slipping → \rightarrow → → \rightarrow → Kinetic Energy of a system of particles → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Kinetic Energy of a system of particles
a ring, a solid cylinder, a sphere
mgh = m v 2 2 ( 1 + k 2 R 2 ) ⇒ v = ( 2 g h 1 + k 2 R 2 ) 1 / 2 \operatorname{mgh}=\frac{m v^2}{2}\left(1+\frac{k^2}{R^2}\right) \Rightarrow v=\left(\frac{2gh}{1+\frac{k^2}{R^2}}\right)^{1 / 2} mgh = 2 m v 2 ( 1 + R 2 k 2 ) ⇒ v = ( 1 + R 2 k 2 2 g h ) 1/2 Greatest velocity
Object
k k k v v v
Circular ring
R
g h \sqrt{gh} g h
0 r 0^r 0 r R 2 \frac{R}{\sqrt{2}} 2 R 4 3 g h \sqrt{\frac{4}{3}gh} 3 4 g h
Sphere (solid)
2 5 R \sqrt{\frac{2}{5}}R 5 2 R 10 7 g h \sqrt{\frac{10}{7}gh} 7 10 g h
Kinetic Energy of Rolling motion → \rightarrow → → \rightarrow → Kinetic Energy of a system of particles → \rightarrow → → \rightarrow →
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Thank You
Kinetic Energy of a system of particles → \rightarrow → → \rightarrow → Thank You → \rightarrow → → \rightarrow →
Resume presentation
Rotational Motion Abouta Fixed Axis Angular Momentum System L-8 Rotational motion about a fixed axis-angular momentum System of Particles and Rotational Motion $\rightarrow$ $\rightarrow$ Rotational motion about a fixed axis-angular momentum System of Particles and Rotational Motion $\rightarrow$ Angular Momentum - Rotation About a Fixed Axis $\rightarrow$ Angular Momentum
