Chapter 9

Rotational Dynamics

Torque

Fr

sinrF

Direction : RHR

Note : The torque of a force depends upon the reference point. One can choose the reference point such that the torque is zero even though the force is not.

sinrF

Rotational Inertia

Rotational inertia is the measure of resistance a rigid body offers to the change of its angular velocity.

Less effort is required to produce the same change of angular velocity in (a) than in (b)

Rotational Inertia of a single particle

Consider a single particle as a rigid body, constrained to rotate in the x-y plane, about the z-axis.

Massless rigid rod

The net tangential force on the particle is :

sinFFT

Equating the tangential force to the tangential acceleration,

zT mrmaF sin

Now, the torque of the force about O is :

zz2

z ImrrF sin

inertiarotationaltheismrIwhere 2,

x

yF

TF

RF

zF

zTT mrmaF

z2

Tz mrrF

Two-Particles Rigid Body Rotating about a Fixed Axis

r222r111 TTFTTPF

;

r2T22r1T11 FFFFFF

;

It is only the tangential components of the forces, that produce torques about O :

z2

11T111T11z1 rmamrFr

z2

22T222T22z2 rmamrFr

zz222

211z2z1z Irmrm

inertiarotationaltheisrmrmIwhere 222

211 ,

iTizi Fr ,

iTii amr

z2iirm

Or

rTFm

zThree Dimensional Rigid Body

Equation of Rotational Dynamics

N

1izziz I,

Where, is the rotational

inertia about the axis of rotation.

N

1i

2iirmI

ofcetandislarperpendicuri

the ith particle from the axis of rotation

m1

m21r

2r12F

21F

Contribution of the two internal forces to the net torque is :

212121 FrFr

Since all internal forces form pairs as above, the net torque of all the internal forces is zero

The torque of the internal forces is zero

O

1221 Frr

0

Equation for rotational dynamics can, therefore, be written as :

zzext I ,

Rotational inertia of continuous mass distribution

dmrI 2

Where,ondistributilinedldm

ondistributisurfaceda ondistributivolumedv

and r is the perpendicular distance of dm from the axis of rotation

Rotational Inertia of a discdm

r

Rdr

rdr2rdmrdI 22

drrR

M2 32

2R

0

32

RM2

1drr

R

M2dII

1.

dm

r

ddrr

R

MdmrdI 3

22

R

0

2

0

232

RM2

1ddrr

R

MI

2.

Rotational Inertia of a rectangular plate

2b

2b

22a

2a

x2

x dyydxab

MIdxdyydI

/

/

/

/x

a

b

y

xy

2Mb12

1

dxdyyxab

MdI 22

z

2b

2b

2a

2a

2222z ba

12

Mdxdyyx

ab

MI

/

/

/

/

The two theorems of rotational inertia

Prob 9.10 The Perpendicular Axis Theorem (True only for lamina)

x

y

z

yxz III

If x, y & z are mutually perpendicular axes as shown, then

dm

y2x2

x2+y2

Proof : ;dmydI 2x dmxdI 2

y

yx22

z dIdIdmyxdI )(

Summing (Integrating) over the entire lamina,

yxz III

2. Parallel Axis Theorem (True in general)

The rotational inertia about any axis is the sum of rotational inertia about a parallel axis through the CM and total mass of the body times square of the distance between the axes

2CM MRII

dmyxdI 22

dmyxdI 22cm

cmxxxNow ,

cmyyy

)( cmcm22222 yyxx2Ryxyx

dmyyxx2dmRdIdIOr cmcm2

cm )(,

x

z z

y

x

y

dm

cm

Integrating over the entire body,

dmyy2dmxx2MRII cmcm2

cm

0xMdmxNow cm,

Similarly, the last integral is also zero

2cm MRII

Torque due to gravity

O

mi

ir mig

)gmr(N

1iii

grmi

ii

Wr)gM(r cmcm

cmr

w

The torque of the gravitational forces on a body about any point is the same as the torque of the net weight of the body as if it were acting at the centre of mass.

Corollary : The gravitational torque about the centre of mass is zero.

Centre of Gravity

Suppose acceleration due to gravity is not constant.

i

iii gmr

The gravitational torque is then :

cgr

wrcg

iiigmwwhere,

A point , such that the above torque can be expressed as

is defined as the Centre of Gravity.

Suppose acceleration due to gravity is constant in direction.

WW

rwnrgm i

ii

iiii

ˆ

W

rwr i

ii

cg

Clearly, CG coincides with CM, when g is uniform

Equilibrium Applications of Newton’s Laws

For equilibrium, both translational and rotational, the net force and the net torque must be zero.

0Fext

0ext

Note : Under the above conditions, if torque is zero about one point, it is zero about every other point.

O

O

1F

3F

2F

1r

1r

R

i

extiiext Fr ,

extii

i FrR ,)(

i

extiextii

i FRFr ,,

ext

Proof :

00 extext

Ex. 9.32 A 274-N plank, of length L = 6.23 m, rests on a frictionless roller at the top of a wall of height h = 2.87 m. The centre of gravity of the plank is at its centre. The plank remains in equilibrium for any value of θ > 68.00 but slips if θ < 68.00. Find the coefficient of static friction between the plank and the ground.

The net torque about O is :

cossin 2

wLFh

Equating this to zero,

2h4

wLF sin

Equating the net vertical force to zero,

h4

2L1wFwN

sincoscos

w

F

N

fO

Equating the net horizontal force to zero,

sinFf

sin, FNOr

40.

Weight of each ball : W

Radius of each ball : R

Find the force exerted on the spheres by :

Prob. 9.6

1) Container bottom

2) Container sides

3) One another

F

N

N

W

W

System : The two spheres

1. F = 2W

2. The forces at the container sides are equal

Equating the torque about the centre of the bottom sphere to zero :

)cosR2(W)sinR2(N

cotWN,Or

F

N

W

f

c) System : Bottom sphere

Equating the horizontal component of the net force to zero :

Ncosf

eccosWf,Or

Prob. 9.8

L = 2.76m

W = 315 N

θ = 320

Tbreak= 520 N

a) Find the tension in the wire

b) Horizontal and vertical components of force at A

c) Find x at which the wire breaks

T

WFh

Fv

a) Equating the torques of T and W about the hinge

sinL

WxT

b) Equating the horizontal and vertical components of Fext to zero,

sin;cos TWFTF vh

System : The horizontal bar

c) xbreak is obtained from the first eq. by putting T = Tbreak

Non-equilibrium applications of Newton’s equations (Fixed axis rotation)

cmext aMF

zzext I ,(z is the fixed axis of rotation)

For general rotation,

)generalin(I

Torque-free rotation of a spheroid

Rolling Motion

Rolling is a combination of pure rotation and pure translation

a) Pure translation

b) Pure rotation

c) Rolling

In rolling :

Rvcm

Velocity distribution in a rolling disc

Rolling with slipping and sliding

ω

Pure Rolling

0vcp Rvcm

Rolling with slipping

0vcp (Directed backwards)

ω ω

Rolling with sliding

Rvcm Rvcm 0vcp

(Directed forward)

Friction in Rolling

1. Rolling with slipping or sliding :

Nf k2. Pure rolling

Relative motion exists between the two surfaces (ground and bottom of the wheel) in contact. So friction is kinetic friction.

No relative motion between the two surfaces. So friction is static friction.

Nf0 s

Moreover, the direction of the frictional force can either be the same as or opposite to the direction of linear motion.

Fapp

f

Fapp

f

Friction in the absence of relative motion

Direction of friction on the body on top is opposite to the direction of attempted relative motion

Rolling without friction

As ω is increased, friction develops forward

As ω is reduced, friction acts backward

fmg

N

Prob. 9.23

Condition for slipping : μs < tanθ/3

Ans.

Eq. of translational motion along the incline :

mafmg sin

Eq. of rotation about its axis :

2mR2

1fR

For rolling, α = a/R, and solving for f,

sinmg3

1f

Since the condition for rolling is :

cosmgNf ss

cossin mgmg3

1s

tan3

1s

If μs < tanθ/3, the equations of motion are :

mamgmg k cossin

2k mR

2

1mgR cos

&

)cos(sin kga cos& g2R k

)cos(sin k3gRa

03g s )cos(sin

The cylinder will slide

Angular Impulse

2

1

2

1

t

t

t

t

12z Idt

dIdtN

N is the angular impulse

If the torque is due to a force F with moment arm r,

,rJrFdtN2

1

t

t

where J is the linear impulse.

Prob. 10.5

A billiard ball is given a sharp impulse by a cue as shown. Its leaves the cue with an initial velocity of v0 and because of “forward English” acquires a final velocity of 9v0/7. Show that h = 4R/5

F

R

h

Ans

0

t

0

mvFdtJ

0

t

0

0 hmvhJdtI

20

0 R2

hv5Or ,

00 vR5

R2hIf ,

and the ball will roll from the start

(Prob. 10.3)

ω

ωR > v0 f

Let it take time T for the ball to start rolling after it was hit.

7

mv2vmfT 0

7

Rmv2RfTTI 0

R7

v5Or 0,

R7

v5

R2

hv5 020 (When rolling starts)

However, when the ball starts rolling, 9v0/7 = ωR

Solving for h,

5

R4h