XII Physics Rotational Motion

(1)

Rotational Motion

(2)

CAUSE & CONSEQUENCE

in Rotational Motion

(12)

Force produces translation i.e. linear acceleration, ‘a’

Couple Moment produces rotation i.e. angular acceleration, ‘’

F



(13)



 • Rigid body subjected to

torque 

• Rotating about a fixed axis with angular acceleration 

(14)

  1 n 2 R₁ R₂ R_n

• Consider ‘n’ particles of the body in circular motion with masses m₁, m₂, … , m_n.

(15)

  R₁ R₂ R_n a₁ a_n a₂

• Linear tangential accelerations of constituents; a₁, a₂, … , a_n • Using a_T = R a₁ = R₁  a₂ = R₂  … … … … … … … … a_n = R_n  (1)

(16)

  R₁ R₂ R_n a₁ a_n a₂ F₁ F₂ F_n a₁ = R₁ , a₂ = R₂ , … , a_n = R_n  _(1) • Using Newton’s II Law; F = ma

F₁ = m₁a₁ = m₁R₁  F₂ = m₂a₂ = m₂R₂  … … … … … … F_n = m_na_n = m_nR_n  (2)

(17)

Prof. Sameer Sawarkar 17 F₁ F₂ F_n   R₁ R₂ R_n a₁ a_n a₂ ₁ _n F₁ F₂ F_n F₁ = m₁R₁ , F₂ = m₂R₂ , … … … F_n = m_nR_n  _(2) • Using definition of torque;  = d*F

₁ = R₁F₁ = R₁(m₁R₁) ₁ = m₁R₁2 ₂ = m₂R₂2 … … … … … … _n = m_nR_n2 (3)

(18)

F₁ F₂ F_n   R₁ R₂ R_n a₁ a_n a₂ ₁ _n F₁ F₂ F_n ₁ = m₁R₁2_, 2 = m2R22, … … … _n = m_nR_n2_{_(3)}

• Sum of all individual constituent torques must be equal to the

externally applied original torque.

 = ₁ + ₂ + … + _n

 = m₁R₁2_{+ m}

2R22 + … + mnRn2



= (



m

_i

R

_i2

₎



i = 1, 2, … , n.

(19)

Prof. Sameer Sawarkar 19 Translational Motion Rotational Motion



= (



m

_i

R

_i2

_)*



F = m*a

F   a  

m





m

_i

R

_i2

Quantity m_iR_i2 _{is called as Moment of Inertia of rotating}

(20)

Moment of Inertia (m_iR_i2 _{) about a given axis of rotation is}

defined as the sum of product of mass of each constituent and

square of its distance from the axis of rotation.

Moment of Inertia (abbreviated as MI, denoted by I) represents inertia in rotational motion i.e. reluctance of a rigid body to

(21)

Prof. Sameer Sawarkar 21 With regular geometric boundaries,

where division in discrete shapes is possible, MI is expressed as;

I = m_iR_i2

With irregular geometric boundaries, where division in elemental shapes is necessary, MI is expressed as;

(22)

• I = (m_i, R_i2₎

• MI represents mass distribution of the rotating rigid body. • Rotational motion depends not just upon total mass but upon

(23)

(24)

Moment of Inertia

I =



m

_i

R

_i2

_{or I =}



_R

2

_dm

Unit: kg-m

2

(25)

KINETIC ENERGY

in Rotational Motion

(26)

 • Rigid body rotating about a

fixed axis with angular velocity 

(27)

 1 n 2 R₁ R₂ R_n

(28)

 R₁ R₂ R_n V₁ V_n V₂

• Linear tangential velocities of constituents; V₁, V₂, … , V_n • Using V = R V₁ = R₁  V₂ = R₂  … … … … … … … … V_n = R_n  (1)

(29)

 R₁ R₂ R_n V₁ V_n V₂ V₁ = R₁ , V₂ = R₂ , … , V_n = R_n  _(1) • KE = ½ mV2_{= ½ m(R}22_{) of each} constituent. U₁ = ½ m₁R₁22 U₂ = ½ m₂R₂22 … … … … … … U_n = ½ m_nR_n22 (2)

(30)

 R₁ R₂ R_n V₁ V_n V₂ U₁ = ½ m₁R₁22_{, U} 2 = ½ m2R222, … … U_n = ½ m_nR_n22 _{_(2)}

• Total KE of the rotating rigid body; U = U₁ + U₂ + … + U_n U = ½ m₁R₁22_{+ ½ m} 2R222 + … + ½ m_nR_n22 U = ½ (m_iR_i2₎2

U = ½ I



2

(31)

ANGULAR MOMENTUM

in Rotational Motion

(32)

V, mV

L

R

m

Angular Momentum: Property possessed by a rotating body by virtue of its

angular velocity.

Defined as; moment of linear momentum. i.e. L = R*P = R*(mV) Like linear momentum, angular momentum is a vector.

(33)

Prof. Sameer Sawarkar 33 P

L

R

m

Vector relation between linear momentum and angular

momentum:

From scalar relation; L = R*P and using Right-hand rule;

P

R

L





(34)

 • Rigid body rotating about a

fixed axis with angular velocity 

(35)

 1 n 2 R₁ R₂ R_n

(36)

 R₁ R₂ R_n V₁ V_n V₂

• Linear tangential velocities of constituents; V₁, V₂, … , V_n • Using V = R V₁ = R₁  V₂ = R₂  … … … … … … … … V_n = R_n  (1)

(37)

, L R₁ R₂ R_n V₁ V_n V₂ V₁ = R₁ , V₂ = R₂ , … , V_n = R_n  _(1) • Linear momentum P = mV for each

constituent.

• Angular momentum for each constituent; L = R*P = RmV = Rm(R) = mR2 L₁ = m₁R₁2 L₂ = m₂R₂2 … … … … … … L_n = m_nR_n2 (2)

(38)

, L R₁ R₂ R_n V₁ V_n V₂ L₁ = m₁R₁2_{, L} 2 = m2R22, … … L_n = m_nR_n2_{_(2)}

• Total angular momentum of the rotating rigid body;

L = L_i, i = 1, 2, … , n. L = m₁R₁2 _{+ m}

2R22 + … + mnRn2

L = (m_iR_i2₎

(39)

PRINCIPLE OF CONSERVATION OF

(40)

 

L

dt

d

I

dt

d

dt

d

I

















,

0 



_L

If then is constant.

In absence of an external torque, the angular momentum of the system remains constant

(41)

(42)

PARALLEL AXES THEOREM

(43)

Prof. Sameer Sawarkar 43 • Rigid body with mass M • Purely rotating about an

axis through C.M. • MI = I_G (known) I_G

(44)

I_P

• It is desired that MI

about a parallel axis at a distance ‘h’ through P i.e. I_P be found. I_G G P h

(45)

Prof. Sameer Sawarkar 45 I_P

G P

• Assume elemental mass dm at an arbitrary point Q. I_G Q G P h

(46)

I_P • Construction I_G G P Q (dm) S h

(47)

Prof. Sameer Sawarkar 47 I_P _I G G P Q (dm) S h I_G = QG2_dm I_P = QP2_dm QP2_{= PS}2_{+ SQ}2 = (PG + GS)2_{+ SQ}2 = PG2_{+ 2PG*GS + (GS}2₊ SQ2₎ QP2 _{= PG}2_{+ 2PG*GS + QG}2

(48)

I_P _I G G P Q (dm) S h QP2 _{= PG}2_{+ 2PG*GS + QG}2

Multiplying throughout by dm and integrating; QP2 _{dm =}_PG2_{dm + 2PG}_GSdm +  QG2_dm QP2 _{dm = I} P  QG2_{dm = I} G  PG2_{dm = PG}2_{dm = Mh}2

(49)

Prof. Sameer Sawarkar 49 I_P Substituting; I_P = I_G + Mh2 I_G G P Q (dm) S h

MI of a rigid body about any axis is equal to sum of its MI about a parallel axis through center of mass and product of mass of body and square of the distance between two parallel axes.

(50)

• Rigid with mass M

• Laminar body (thickness very small compared to surface area)

(51)

Prof. Sameer Sawarkar 51 • System of 3 mutually

perpendicular axes through any point O. • X and Y in the plane of

the lamina, Z being perpendicular to the plane. O X Y Z

(52)

• Imagine elemental mass dm at a distance ‘r’ from Z axis. O X Y Z r dm

(53)

Prof. Sameer Sawarkar 53 • Moment of inertia of

the lamina @ Z axis; I_Z =  r2_dm O X Y Z r dm I_Z

(54)

• Construction –

perpendiculars on X and Y axes from elemental mass. O X Y Z r dm I_Z x y

(55)

Prof. Sameer Sawarkar 55 • MI of lamina about X axis; I_X =  y2_dm • MI of lamina about Y axis; I_X =  x2_dm O X Y Z r dm I_Z x y _I Y I_X

(56)

r2_{= x}2_{+ y}2 Multiplying throughout by dm and integrating;  r2_{dm =}_x2_{dm +}_y2_dm Substituting; O X Y Z r dm I_Z x y _I Y I_X I_Z = I_X + I_Y Moment of inertia of a lamina about an axis

perpendicular to its plane is equal to sum of its moments of inertia about two mutually perpendicular axes in the plane of lamina and

(57)

RADIUS OF GYRATION

(58)

Radius of Gyration (K) w.r.t. the given axis of rotation is the

theoretical distance at which, when entire mass of the body is

assumed to be concentrated, gives same MI (of idealized point mass system) as that of the original rigid body. If MK2₌_R2_{dm, then K is}

the radius of gyration.

I = R2_dm _{I = MK}2

M

(59)

Prof. Sameer Sawarkar 59 I_G = ½MR2 _I

G = MK2

K M

REAL SYSTEMS IDEALIZED SYSTEMS

I_G = 2MR2_/5 I_G = MK2 K M MK2_{= ½MR}2  K = R/2 MK2_{= 2MR}2_/5  K = R*(2/5)

(60)

XII Physics Rotational Motion

Rotational Motion

Contents

CAUSE & CONSEQUENCE

in Rotational Motion



= (



m

R

)





= (



m

R

)*



F = m*a

m





m

R

Moment of Inertia

I =



m

R

or I =



R

dm

Unit: kg-m

KINETIC ENERGY

in Rotational Motion

U = ½ I



ANGULAR MOMENTUM

in Rotational Motion

P

R

L





PRINCIPLE OF CONSERVATION OF

 

 

L

dt

d

I

dt

d

dt

d

I

I





















,

0





L

PARALLEL AXES THEOREM

RADIUS OF GYRATION

₎

_)*

_{or I =}

_R

_dm

_L