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Transcript of ch9
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