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PHYSICS 36LOCUS
In terms of moment of inertia, equation (7.18) can be written
as...(7.19 a)
The vector form of the above equation is
...(7.19 b)The angular acceleration produced is along the
direction of applied net external torque. The magnitude of
theproduced angular acceleration is directly proportional to the
magnitude of the net external torque and inverselyproportional to
the moment of inertia of the body. The above relation looks the
translational equation net .F maHere, you should not forget that I
is not an independent rule. It is derived from F = ma only. We can
establishan analogue between translational and rotational
variables. By doing so concept developed so far for
translationalmotion would help to solve the problems involving
rotational motion. The possible analogue is as follows
/ /( ) distance traversed, angle turned,
s( ) average speed, average angular speed,t t
( ) intantaneous speed, instantaneous a
i S
ii v
dsiii vdt
ngular speed,
( ) average acceleration, average angular acceleration,
( ) instantaneous acceleration, instantaneous angular
acceleration,
( ) mass, moment of inertia, (
ddt
dviv adt t
dv dv adt dt
vi m Iv ) force, torque,
( )ii F
viii F ma I( ) linear momentum, Angular momentum, ( )
( )
( ) conservation of linear momentum: conservation of angular
momentum:When 0, constant. When 0, constant.translational ki( )
ix p lx p mv l I
dp dlxi Fdt dt
xiiF p l
xiii2 21 1
2 2
netic energy, rotational kinetic energy,
( ) work done, work done,
k mv k I
xiv d F ds d d
Angular quantities involved in analogues (ix) to (xiv) would be
discussed later in this topic.
Find the angular acceleration of the rod given in example 4 at
the moment
(a) when it is released from rest in the horizontal
position;
(b) when it makes an angle with the horizontal.
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PHYSICS 37LOCUS
After the moment when the rod is released from the restin the
horizontal position , it would rotate in the vertical plane abouta
horizontal axis passing through the hinge and perpendicular to
thelength of the rod. Initial angular velocity of the rod is zero
but dueto nonzero torque of gravity it has some angular
acceleration andhence, it will acquire some angular speed as it
rotates. Asdiscussed in example 4, the hinge force does not provide
anytorque about the axis under consideration and the weight of the
rod tries to rotate it in the clockwise sense, i.e.,it provides a
torque perpendicularly, inward to the plane of the paper. An
approach using r F to find the torquewould also give the same
result. Hence, angular acceleration of the rod,
gravitynet
I I
22
3mg lml
32
gl
(b) When the rod makes an angle with the horizontal, its angular
acceleration,
mgnet
I I
mg rI
/
2
cos2
3
lmg
ml[from figure 7.40 (b)]
3 cos2
gl
In the previous case, find the angular velocity of the rod when
it has turned through an angle after the moment whenit was released
from rest in the horizontal position. Also find the angular
velocity when the rod becomes vertical.
From the result obtained in part (b) of the previous example, at
some angle , the angular acceleration ofthe rod,
3 cos2
gl
3 cos2
d gdt l
3 cos2
d d gd dt l [Using chain rule.]
0 0
3. cos2
gd dl at = 0, = 0
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PHYSICS 38LOCUS
2
00
3sin
2 2gl
2 3 sin2 2
gl
3 singl
When the rod becomes vertical, 2, and hence, angular
velocity,
3g l
In the previous example, find the hinge force on the rod at =
0.
Just after the moment when the rod was released from the rest in
the horizontal position, it is shown infigure 7.41(a). Let the
vertical component of the force on the rod from the hinge be 1R and
the horizontal componentof the same be 2R , as shown in figure. The
subsequent motion of the centre of mass of the rod is a
nonuniformcircular notice on the vertical circular path of radius
l/2 with the centre at the hinge, as suggested in the figure.
Initially the rod is at rest and hence radial component of the
acceleration of the centre of mass of the rod, ²r, is zero.Hence,
applying net cmF ma along the radial direction, we get
net, radial cm, radialF ma
2 0R m0
Applying the same along the tangential direction, we have, =
.
net, tangential cm, tangentialF ma
1 2lmg R m
1 2lR mg m
34
mg mg
14
mg
Net hinge force 2 21 21 .4
R R mg 2[ 0]R
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PHYSICS 39LOCUS
In the previous example, find the magnitude of the net hinge
force on the rod when the rod has turned through anangle .
If be the angular velocity of the rod when it has turned through
an angle , the centre of mass of the rod
has 2 2l
and 2l
as radial and tangential components of its acceleration,
respectively, as shown in figure 7.41(b).
Applying ext cmF Ma on the rod along the radial direction, we
have,
22 sin 2
lR mg m
23sin sin2
R mg mg
.
3 sin
3 cos2
gl
gl
=
=
Rod when it makes an angle with thehorizontal. , are
perpendicular and radial components, respectively, of the reaction
force acting
R R
on the rod from the hinge.
You can also assume reaction force as andacting at some angle
with the rod.
R
5 sin2
mg
Applying the same along the tangential direction, we have,
1cos 2lmg R m
1 cos 2lR mg m
3cos cos4
mg mg
1 cos4
mg
Therefore, net force on the rod from the hinge can be obtained
by solving2 21 2R R R
If the disc given in example 5 has mass M and it is free to
rotate about its symmetrical axis passing through O, findits
angular acceleration.
If be the angular acceleration of the disc, then, using net ,I
we have,
net
I
23
( 2)FR
MR
6FMR
As the net torque is in clockwise sense, has the same sense of
rotation.
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PHYSICS 40LOCUS
A uniform disc of radius 0.12 m and mass 5 kg is pivoted so that
itrotates freely about its axis. A thin, massless and inextensible
stringwrapped around the disc is pulled with a force of 20 N, as
shownin figure 7.42(a) .
(a) What is the torque exerted on the disc about its axis? (b)
What is the angular acceleration of the disc?(c) If the disc starts
from rest, what is the angular velocity after 3s?
It is obvious that the string force gives a torque to the disc
in the clockwise direction. As the torque givenby the force from
the axle is zero. Net torque on the disc is,
net torque of the string force
F rF R(20 N) (0.12 m)2.4 N-m.
As the net torque on the disc is in clockwise direction, the
disc has angular acceleration in the same direction. If be the
magnitude of the angular acceleration,
net net2( 2)I MR
22
2.4 2 rad/s5 (0.12)
266.66 rad/sAt t = 0 if the disc has zero angular velocity,
then, at some time t, its angular velocity,
in0
t
t
At t = 3 s,
200 rad/s
A uniform disc of radius R and mass M is mounted on an
axissupported in fixed frictionless bearing. A light string is
wrappedaround the rim of the disc and a body of mass m is supported
bythe string, as shown in figure 7.43(a).
(a) find the angular acceleration of the disc;(b) find the
magnitude of the tangential acceleration of
the point on the rim where the string separatesfrom the rim.
(c) if the system is released from rest at t = 0, find thespeed
of the block at some time t (>0).
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PHYSICS 41LOCUS
Analyze the situation according to the information provided in
the figure 7.43(b). You should also note thefollowing points:• Only
tension force of the string, T, produces a torque on the disc about
its centre O. Torque of the weight of
the disc and that of the reaction force from the bearing are
zero about O.• If be the angular acceleration of the disc (in the
clockwise direction) then the point P on the disc has a
tangential acceleration R in the vertically downward direction
at the moment shown in figure. The stringunwinds at the same
acceleration and the block has the same acceleration in the
vertically downwarddirection. Therefore, if a be the acceleration
of the block, then,
a R ...(i)Now, applying net I on the disc about its symmetrical
axis, we have,
net I
2.
2MRT R
2M RT
2MaT [Using (i) ...(ii)
Using netF ma for the block in the vertical direction, we
have,
netF ma
mg T ma ...(iii)Adding (ii) and (iii), we get,
2Mmg m a
2ma g R
m MIf v be the speed of the block at some time t, then, we
have,
v u at is constanta
at 0u
2m gt
m M
Find the acceleration of 1m and 2m in an Atwood’s Machine,shown
in figure 7.44(a), if there is friction present between thesurface
of pulley and the thread does not slip over the surface ofthe
pulley. Moment of inertia of the pulley about its symmetricalaxis
is I and its radius is R. The pulley can rotate freely about
itssymmetrical axis.
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PHYSICS 42LOCUS
Due to friction between the pulley and the thread tensions in
the parts of the thread on the two sides of thepulley are
different. Let that in the right part it is 1T and that in the left
part is 2 ,T as shown in figure 7.44(b). Forcesacting on the two
blocks and the pulley are also shown in figure 7.44(b). Force on
the pulley from the support andits weight are not shown because
they do not produce torque on the pulley about its symmetrical axis
of rotation. Ifthe block 1m comes down with an acceleration a then
2m would go up with the same acceleration because theyare connected
by the same string, as shown in the same figure.7.44(b).
If we assume that the pulley gets an angular acceleration in the
clockwise sense then the torque of 1T would bepositive and that of
2T would be negative, as suggested in figure 7.44(c).Again, as any
point on the rim of the pulley has a tangential acceleration R, the
block 1m comes down and theblock 2m goes up with the same
acceleration, as shown in figure 7.44(d).Therefore, we can
write,
a R ...(i)
Using netF ma for the two blocks, we have,
1 1 1m g T m a ...(ii) [for 1m ]
2 2 2T m g m a ...(iii) [for 2m ]
Using net I for the pulley, we have,
1 2T R T R I ...(iv)torque of support forceand weight are
zero
Substituting from (i) in (iv), we get,
1 2 2aT T I
R...(v)
Adding (ii), (iii) and (v), we get,
1 2 1 2 2( )Im m g m m a
R
1 22
1 2
m ma gm m I R
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PHYSICS 43LOCUS
A thin uniform rod AB of mass m = 1.0 kg moves
translationallywith acceleration a = 2.0 m/s² due to two
antiparallel force 1F and
2F acting on it perpendicularly to its length, as shown in
figure7.45. The distance between the points at which these forces
areapplied is x = 20 cm. Besides, it is known that 2 5.0 N.F
Findthe length of the rod.
Before analyzing the details of the given situation, let us
analyze the rotational effect of two antiparallelforces. Consider
the situations shown in figure 7.45.
and are producing torques about in opposite directions.
F FA
21 and are producing torques about in opposite directions.
F FB
21 and are producing torques about in the same direction
F FC
21
If we analyze the torques of the two forces about every point in
their plane containing them, then, we arrive at theconclusion that
if the point lies between the lines of action of 1F and 2F then
torques of the forces about that pointadd up together otherwise
they are in opposite directions.If the magnitudes of the two forces
are equal then such a pair is called as a couple. If the magnitude
of each force isF and the distance between their lines of
application is d, then, the net torque about any point in their
plane is F.d,as shown in figure 7.47.
net 1 2
( )F l d FlF d
net 2 1
( )F l d FlF d
net 1 2
1 1 2 2
1 2( )F d F dF d dF d
Torque of a couple.
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PHYSICS 44LOCUS
Now, let us discuss the given case. As the rod is in pure
translationmotion, net torque on it about any point must be zero.
Therefore,the centre of mass of rod can not lie between the lines
of action ofthe forces because in that case torques produced by
then aboutthe centre of mass do not cancel each other.
Let us assume that the centre of mass of the rod lies at a
distance yaway from the line of action of 2 ,F as shown in figure
4.48. As therod translates towards right, 2F must have a greater
magnitudethan 1.FUsing net ,F ma we have
2 1F F ma
1 2F F ma(5 1 2) N
= 3 NAgain, as the net torque on the rod about C must be
zero,
1 2
magnitude of the torque magnitude of the torqueproduced by about
produced by about F C F C
the two torques haveopposite directions
1 2( )F x y F y
2 1 1( )F F y F x
1
2 1
Fy xF F
3 205 3
cm
= 30 cmLength of the rod, 2( ) 1.0 ml x y
A force ˆ ˆF Ai Bj is applied to a point whose radius vector
relative to the origin of coordinates O is equal toˆ ˆ,r ai bj
where a, b, A, B are constants, and ˆ ˆ,i j are the unit vectors of
the x and y axes. Find the torque
and the arm length l of the force F relative to the point O.
Torque of F about O is
r Fˆ ˆ ˆ ˆ( ) ( )ai bj Ai Bj
ˆ( )aB bA kArm length of F with respect to O is
sinl r is the distance of the point of application of
from and is the angle between and .r F
O r F
r Fr sinr F r F
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PHYSICS 45LOCUS
2 22 2 2
aB bAa b
a b A B
2
aB bA
A B
A uniform cylinder of radius R is spun about its axis to the
angularvelocity 0 and then placed into a corner, as shown in figure
6.50(a).The coefficient of kinetic friction between the corner
walls and thecylinder is equal to k. How many turns will the
cylinder accomplishbefore it stops?
All forces acting on the cylinder are shown in figure6.50(b). As
the cylinder rotates, its surface slips over the cornerwalls and
hence frictional forces acting on it, 1f and 2 ,f are kineticin
nature. Normal contact forces acting on the cylinder from thecorner
walls, 1N and 2 ,N and the weight of the cylinder, mg,pass through
the centre of the cylinder and hence, these forcesproduce no torque
about the centre C. Only frictional forcesproduce torque about C
and the torques produced by them are inopposite direction of the
direction of the angular velocity of thecylinder and hence, they
retard the rotational motion of the cylinder
As the cylinder does not translate, net force on it in both
vertical and horizontal directions must be zero. Therefore,
1 2N f mg
1 2N N mg ...(i) 2 2[ ]f Nand 2 1N f
2 1N N ...(ii) 1 1f NSubstituting for 2N in equation (i) from
equation (ii), we have,
21 1N N mg
1 21mgN ...(iii)
Substituting for 1N in equation (ii) from equation (iii), we
have,
2 21mgN ...(iv)
If we define the anticlock wise sense of rotation as the +ve
direction of rotation, then, the clock wise sense becomesthe –ve
direction for the same. Hence, angular acceleration in the present
case becomes negative for this choice ofrefrence direction. The
angular acceleration,
1 2net torque due to and about moment of inertia about the axis
of rotation
f f C
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PHYSICS 46LOCUS
1 22
2
f R f RmR
1 22( )N NmR
1 22 ( )N NmR
22 1
1mg
mR
22 1
1g
R
If the cylinder had the angular velocity 0 at t = 0and at some
time t it has an angular velocity , and in this durationit has
turned through an angle , then,
2 20 2
20 2
4 11
gR
If the cylinder stops having rotated through an angle 0 , then
at 0 , = 0. Therefore,
2 20 02
4 101
gR
2 20
0(1 )
4 (1 )Rg
Therefore, the number of rotations accomplished by the cylinder,
before it stops,
0
2n
2 20 (1 )
8 (1 )Rg