Rotational DynamicsBasics

1

FORMULA

Torqueτ=Iα=dL→dt

2

FORMULA

Angular MomentumL=Iω

3

FORMULA

Kinetic Energy in Pure RotationK.E.=12Iω2

4

FORMULA

Angular Impulse in Pure RotationAngular Impulse  = ΔL→

5

FORMULA

Torque and Force in Pure Rotationτ=r→ ×F→

6

FORMULA

L→ in Rotation + TranslationL→=Icomω+Mr→×Vcom→

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FORMULA

τ in Rotation + Translationτ=Icomα F=Macom

8

FORMULA

K.E. in Rotation + TranslationK.E.=12Iω2+ 12MV2com

9

FORMULA

Linear velocity in Rotation + Translation

v→=ωr+vcom→

10

FORMULA

Friction in Rolling - Force above center

fs=[MRr−IcomIo]F

11

FORMULA

Friction in Rolling - Force on Top most point

fs=[MR2−IcomIo]F

12

FORMULA

Friction in Rolling - Force at Center

fs=[−IcomIo]F

13

FORMULA

Friction in Rolling - Down an incline

fs=[−IcomIo]Mgcosθ

14

FORMULA

Friction in Rolling - Force below center

fs=−[MRr+IcomIo]F

15

FORMULA

Moment of Inertia - Basic defintionM.I=∫r2dm

16

FORMULA

Total K.E of a rolling body :K.Etotal=K.Etranslatory+K.Erotational

=12mV2c+12Iω2   =12mV2c(1+K2R2)

WhereVc= Velocity of C.MK= radius of gyrationR= radius

KEtrans.:KErot.:KEtot.=1:K2R2:(1+K2R2)

17

FORMULA

Velocity of Center of Mass:Velocity of centre of mass is

VCM=m1v1+m2v2+.....m1+m2+.....

=total momentumtotal mass

18

FORMULA

Acceleration of centre of mass:Acceleration of centre of mass is 

aCM=m1a1+m2a2+......m1+m2+.....

=net external forcetotal mass

19

FORMULA

Torque or Moment of Force (τ):It is the turning effect of a force.Magnitude of torque = Force acting on the body x perpendicular distance between line of action of force and axis of rotation.

τ→=r→×F→    ∴ The magnitude of Torque τ is

 τ=rF sinθwhere θ= angle between r→,F→Direction of torque is parallel to the axis of rotation or perpendicular to the plane of rotation.Magnitude of torque is maximum when r→,F→ are perpendicular to each other.

Couple :

 Two equal and opposite non-collinear forces simultaneously acting on a body constitute couple. Couple always produces turning effect. Turning effect of couple is called moment of couple. Magnitude of moment of the couple = magnitude of one of the two forces of couple × perpendicular distance between two forces.

Force in translatory motion is analogous to torque in rotational motion. 

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FORMULA

Moment of Inertia:Moment of inertia of a particle of mass m is

I=mr2where r = perpendicular distance of particle from axis of rotationMoment of inertial of a group or system of particles is

I=m1r21+m2r22+......+mnr2nI=S mr2where m1,m2.......mn are masses of particles and r1,r2,...rn are their perpendicular distances from axis of rotationMoment of inertia of a rigid body of mass M is

I=MK2where K = radius of gyrationK=r21+r22+......+r2nn−−−−−−−−−−−−−−−−√

where n is total number of particles in the body and r1,r2......rn are their perpendicular distances from axis of rotation

Moment of inertia in rotational motion is analogous (similar) to mass in translatory motion.

21

FORMULA

Parallel Axes Theory:I=IC+Md2

WhereI = Moment of inertia of a body about an arbitrary axisIC=Moment of inertia of same body about a parallel axis to the first axis and passing through centre of mass of body. 

M = Mass of bodyd = perpendicular distance between two parallel axes.

K2=K2C+d2

where KC=radius of gyration about the axis passing through centre of massK = radius of gyration about the parallel axis.

22

FORMULA

Perpendicular axes Theorem:Ix+Iy=Iz

where Ix,Iy are moments of inertia of a body about two mutually perpendicular axes in the plane of body (say X,Y axes) and Iz is moment of inertia of same body about an axis perpendicular to the plane of the body and passing through the point of intersection of first two axes (say z-axis).

K2x+K2y=K2z

where Kx,Ky,Kz are radii of gyrations about X,Y,Z axes respectively.

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RESULT

Expressions for Moment of Inertia of Regular Rigid Bodies:

Rigid Body Axis of Rotation Moment of Inertia (I)

Radius of Gyration (K)

Circular ring of mass M and radius R

⊥r to the plane of ring and passing through its centre

⊥r to the plane of ring and passing through its rim(or) passing

MR2

2MR2

R

2√R

through any tangent⊥r to the plane of ringIn the palne of the ring and passing through its centre (or) passing through any diameter of ring

In the plane of the ring and passing through its edge (or) passing through any tangent of ring in its plane.

MR2/2

3MR2/2

R/2√

3/2−−−√R

Thin circular plate of mass M and radius R

⊥r to the plane of plate and passing through its centre

⊥r to the plane of plate and pasing through its edge(or ) passing through any tangent ⊥r to its plane.

In the plane of plate and passing through its centre(or) passing through any diameter of plate

In the plane of the plate and passing through its edge(or) passing through any tanget of plate in its plane.

MR2/2

3MR2/2

MR2/4

5MR2/4

R/2√

3/2−−−√R

R/2

5√R/2

Thin hollow sphere of mass M and radius R

Passing through its centre or any diameter

Passing through any tangent

2MR2/3

5MR2/3

2√R/3√

5√R/3√

Solid sphere of mass M and radius R

Passing through its centre or any diameter

Passing through any tangent

2MR2/5

7MR2/5

2√R/5√

7√R/5√

Thin uniform rod of mass M and L

⊥r to the length of rod and passing through its centre

⊥r to the length of rod and passing through its end

ML2/12

ML2/3

L/23√

L/3√

Thin uniform rectangular plate of mass M Length L and Breadth B.

⊥r to the plane and passing through its centre

⊥r to the plane of plate and passing through a corner

In the plane of plate ⊥r to breadth and passing through centre of plate.

In the plane of plate ⊥r to breadth and passing through edge of plate

M12(L2+B2)

M3(L2+B2)

MB2/12

MB2/3

ML2/12

L2+B2√23√

L2+B2√3√

B/23√

B/3√

L/23√

L/3√

In the plane of plate ⊥r to length and passing through centre of plate.

In the pale of plate ⊥r to length and passing through edge of plate

ML2/3

Thin square plate of mass M and side length L.

⊥r to the plane of plate and passing through its centre

⊥r to the plane of plate and passing through a corner

In the plane of plate parallel to any side and passing through centre of plate

In the plane of plate and passing through any two opposite corners

ML2/6

2ML2/3

ML2/12

ML2/12

L/6√

2√L/3√

L/23√

L/23√

Thin hallow cylinder of mass M radius R and Length L

About geometrical or natural axis

Parallel to the length of cylinder and touching its surface (or) passing through line of contact of cylinder with floor when it is

MR2

2MR2

M(L212+R22)

R

2√R

L212+R22−−−−−−−√

rolling.

⊥r to the axis of cylinder and passing through its centre

Solid cylinder of Mass M radius R and length L.

⊥r to the axis of cylinder and passing through one end

About geometrical or natural axis

Parallel to the length of cylinder and touching its surface (or) passing through line of contact of cylinder with floor when it is rolling.

⊥r to the axis of cylinder and passing through its centre 

⊥r to the axis of cylinder and passing through one end

M(L23+R22)

MR2/2

3MR2/2

M(L212+R24)

M(L23+R24)

L23+R22−−−−−−−√

R/2√

3√R/2√

L212+R24−−−−−−−√

L23+R24−−−−−−−√

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FORMULA

Angular MomentumAngular momentum = moment of inertia x angular velocity

L→=Iω→Note: moment of linear momentum is also equal to angular

momentum

L→=r→×P→     (wherep→=mv→)L = mvr sin\thetawhere θ is angle between r→ and p→

L = mvd where d is perpendicular distance between v→ and axis of rotation.

If a particle of mass m is moving along a circle of radius r at a speed v then L = mvrDirection of angular momentum is parallel to axis of rotation or perpendicular to plane of rotation.

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LAW

Law of Conservation of Angular Momentum:Angular momentum of a rotating body remains constant when no external torque acts on it.

i.e., Iω=constant when τext=0(or) I1ω1=I2ω2

Consequences of Conservation of Angular Momentum :A person holding dumb-bells in his stretched hands is on a turn table. If the person folds his hands bringing dumb-bells close to his axis of rotation due to which his moment of inertia decreases and his angualr velocity increases because angular momentum Iω is constant.A diver while diving down at highest position, folds his body close to his axis of rotation due to which his moment of inertia decreases and his angular velocity increases since angular momentum is constant. Hence the diver can make more number of somersaults or acrobats (rotations)2Circular Motion

1

FORMULA

If acceleration is not constant then use the following relationsdθdt=ωd2θdt2=αωdωdθ=α

2

FORMULA

For non uniform circular motion,a2=V2r=rω2 ar: radial accelerationat−rα=dvdt at: tangential accelerationa=a2r+a2t−−−−−−√

3

FORMULA

In uniform circular motion, at=0;a=ar=rω2=V2r

ω=2πTa→=α→×r→V→=ω→×r→ar=ω→×V→

4

FORMULA

Banking of tracks : tanθ=v2Rg

5

FORMULA

Conical Pendulum

time period=2πhg−−√

6

FORMULA

Basic Definitions in Circular MotionAngular Displacement (θ):θ=arc lengthradius(r)Note : a) Small angular displacements are vectors and Large angular displacements are not vectors S.I unit : Radian

Angular Velocity:(ω)ωinst=dθdt  S.I Unit : radian sec−1

ωavg=ΔθΔt 

ωavg is the slope of the secant and ωinst is slope of the tangent for θ vs t graph. 

Angular acceleration (α):αinst=dω dt S.I unit : radian/sec2

αavg=ΔωΔt

αavg is the slope of the secant and αinst is slope of the tangent for θ vs t graph. 

Direction of angular Quantities:all Angular quantities are directed along the axis of rotation and perpendicular to the plane of rotation.

7

FORMULA

Motion of a Body along Vertical Circle with non Uniform Speed:L is length of a simple pendulum and m is mass of its bob. Let the bob is given a velocity V along horizontal direction

The bob performs oscillations if V≤2gL−−−−√The bob describes vertical circle of radius L if V≥5gL−−−−√The bob neither performs oscillations nor describe vertical circle if

2gL−−−−√<V<5gL−−−−√The bob starts to move in parabolic path from a point where tension in the string is zero and retains circular path at same level on the other side.If θ is angular displacement and h is height of the bob from lowest point of vertical circle where tension in the string is zero then V=gL(2−3cosθ )−−−−−−−−−−−−√

V=g(3h−L)−−−−−−−−√3Rotational Mechanics

1

FORMULA

Equation of constant Angular accelerationw=w0+αtw2=w20+2αθθ=w0t+12αt2

2

FORMULA

Torque

τ→=r→×F→τ→=rFsinθτ=Iα

3

FORMULA

Angular momentum about an axisL=Icomw+mVcomr

4

FORMULA

Total Kinetic EnergyK=KR+KT=12Icomw2+12mV2com

5

FORMULA

Combined Rolling and translational motion is superposition of translational and Rotational motion.

6

FORMULA

Non Uniform Rolling

acom+αR=a1acom−αR=a2

7

FORMULA

Pure rolling on inclined Plane

fs=(IcomMR2+Icom)mgsinθfs=μmgcosθ

8

RESULT

Conical Pendulum:

The bob is given a horizontal push a little through angular displacement θ and arranged such that the bob describes a horizontal circle with uniform angular velocity in such a way that the string always makes an angle with the vertical. As the string traces the surface of the cone, the arrangement is called a conical pendulum 

Let T be the tension in the string of length and r the radius of circular path. The vertical component of tension T balances the weight of the bob and horizontal component provides the necessary centripetal force.

T cosθ=Mg        -(1)

From (1) and (2), we get

tanθ=rω2gi.e.,ω=g tanθ r−−−−−−−√

But r=l sinθ and ω=2πT

Time period of revolution

T=2πl cosθ g−−−−−−√