Download Finding Torque (ABOUT ?)

Finding Torque (ABOUT ?) –
[The same force, at same point, can cause clockwise, anti-clockwise or no rotation.]
Finding Torque – 4 approaches
1. T = rFsinθ (If r, F and θ are given) – Take precaution about θ.
2. T = r x F (If r and F are given in terms of i, j and k)
3. T = r (F sinθ) – take components of F. Torque is due to perpendicular comp.
4. T = (r sinθ) F – extend the force arm, drop perpendicular on it.
Ex.
1. Which of these forces has maximum turning effect?
2. Find NET torque.
Questions on finding Torque –
f2
N2
T (about A)
l

= Mg  cos    N 2 l sin    f 2 l cos 
2

l
Mg
T (due to P, about A)
l

= P  sin  
2

N1
θ
f1
A
P
L/2

L/2 sin
A
Net Torque 1. Find torque due to all forces
2. Consider them clockwise or anti-clockwise
3. Get the resultant.
Ex.
1. Rod hinged at some point (Rod with and without mass)
2. Torque due to forces on a disc by fixing a nail at various points and seeing
whether the axis has a role to play in deciding whether the disc will rotate
clockwise or anticlockwise.
O
A
About O – anticlockwise
About A – Clockwise
1
A point to be recalled:
Force Equation net ext. F = Ma
a = net ext. F / M is the acceleration of C. M.
P
C
B
O
Q1. (P/M) is the acc. of the point
C, B, O or A
Ans : of point O
Force equation is a general equation. It is always valid.
A
How to make out whether the C.M. moves – Draw FBD. If the net F = 0, the body does
not translate (C.M. does not move) else the C.M. moves.
Torque equation, T = I .
About? Ex. TC.M. = IC.M. 
It is NOT a general equation. It’s validity 
1. In general, Torque equation can be applied only about a) C.M,
b) About a fixed Point. A fixed point is i) Either permanently fixed or ii)
if the body is rotating without slipping, then instantaneous point of
contact. [Point about which the body is rolling | rotating. (The surface
below must be at rest – actually there should be no relative motion
between the two)] else IAOR, in general.
2. T and I are about the same axis
Misconcept: Let us see through examples that if this equation is applied about any other
point then you might get a non-zero value for α even if in actual the disc is not rotating!
A
Ex. 1
F
TA = F  R = IA .
0
Wrong - The body can translate but not rotate.
f=0
Concept:
TCM = 0
ICM  = 0   =0
Correct, the body does not rotate.
F
f=0
2
The Six possible Motions –
1) If Fnet  0
Translation (the C.M. moves)
How to make out whether the C.M. moves – Draw FBD. If the net F = 0, the body does
not translate (C.M. does not move) else the C.M. moves.
2) If TCM  0
Rotation (First see whether torque equation is valid
ABOUT this axis. If valid then the body rotates about that axis)
How to make out whether the body will rotate – Find net torque about C.M. (or else
about a fixed point like nail etc. – the issue is that it should be a valid point). If the net
torque about this point is NOT equal to zero, then the body will rotate about THAT point.
3) If
Pure translation (the body moves but does not rotates)
Fnet  0, TCM = 0
4) If TCM  0, Fnet = 0
Pure Rotation (the body rotates but does not moves)
[There might be a point about which the motion can be considered as pure rotation.
example Instantaneous axis of rotation (IAOR).
If a NAIL is fixed in the body and torque about the nail  0, then body has PURE
ROTATION about the nail.
5) If TCM  0, and Fnet  0 Rotation + Translation [This doesn’t always mean
rolling. For Ex. A pen is struck – there is NO LINK between its ω and V. Even if it is
rolling then V=Rw is true only for pure rolling on Earth’s surface !]
6) Rolling – 6a) Pure Rolling, 6b) Rolling with Sliding. If there is no relative motion
between p.o.c. and surface –Pure Rolling (Rolling without sliding) [VCM ± Rω = VSurface ]
Q. Can a single force cause rotation?
Q. Can a single force cause pure rotation?
F
 0.
M
But a single force CAN cause rotation (It must not pass through C.M.)
net Fext = P a  P = acm
P
A single force can not cause pure rotation as aCM =
m
 C.M. moves with acc. a  P .
M R
m
And for rotation part -
f=0
About the C.M net TExt = P  R  I
 PR   2 MR 2 
5

In the following examples, state which motion will it be (out of above 6 motions)?
1. Single force applied to a sphere along a line passing through C.M.
2. The above force acting horizontally at the highest point
3. Two opposite force acting on a sphere – one at highest point, the other at lowest
point
4. Above forces acting in same direction
3
Which equations to write for each category of motion?
For pure translation (case 3) – ONLY Force equation
For pure Rotation (case 4) – ONLY Torque equation
For case 5 (Rotation + Translation) – BOTH (Force and Torque equation)
For case 6 (Rolling) – As Rolling involves both translation and Rotation, so in
general BOTH force and Torque equations are to be written, BUT if Rolling is on Earth’s
surface, or if Vpoc = 0, then the P.O.C. become IAOR. In such case, it becomes a case of
pure rotation about POC. So only torque equation need to be written.
Let us now study each motion one by one –
Pure translation
Force equation for pure translation  net Fext = m acm. [a= net F / m is the acc. of C.M.]
disp, vel, acc. is same for all points
To find out whether it is Pure Translation
Check if Net Text (about C.M.) = 0
if Net Text (about C.M.)  0
The body will not move in pure translation.
e.g.
(1) Net Text (about C.M.) = 0
f=0
Pure translation
µ=0
u, a
u, a
u, a
N
mg
(2) Net Text (about C.M.)  0
Not Pure Translation
N
f
mg
We will take this example further while studying rolling.
4
Pure Rotation
[If it is about C.M. then C.M. is not moving. If it is pure rotation about any
other point, then that point is at rest, even if it is so for a particular instant.]
To make it out, either a) Net external force is Zero or b) There is a point which is
at rest (even it is only for an instant.
A) For PURE ROTATION about C.M.
acm = 0
 Net force on system = 0
disp = 0
V=0
a=0
e.g.
(1) Pure rotation
as Net Fext = 0
(and this rotation is about C.M.)
F
F
(2) Pure rotation
F
as Net Fext = 0
(will this rotation be necessary about C.M.?)
F/2
F/2
Yes, as net F / M gives acc. of C.M.
B) For PURE ROTATION about any other ‘fixed’ point (even if that point is fixed for an
instant)
i) The point must be fixed (ex. Nail),
ii) There must be net torque about that point
Q1. If the body is rotating about a nail (must be pure rotation about nail), then is the force
exerted by the nail equal to the external applied force?
NO ! The centre of mass is moving, so CM might be having acceleration which
means that there is a NET force of the forces exerted by Nail and external applied force.
Q2. What is the force exerted by the nail? [In some cases, instead of force by the nail, it
might be force of friction that you have to find. Note – force of friction is NOT equal to
µKN unless the body is sliding!]
5
Torque equation for pure rotation, T = Iα
AVOID THE BIG CONFUSION
v = x [If a body purely rotates
with  then velocity of any point
is given by v = x, where ‘x’ is
the distance of that point from the
axis. This is ALWAYS true!]
x
v
For pure rotation v = x, a = x etc
(for any point x away from the axis)
Questions on Pure Rotation –
From HCV (pg. 183 – Ex. No. 1, 2, 3, 4;
pg. 196-197 - Q. No. 17 – 32)
They basically involve
a) Torque equation,
b) 5 variables of the type u, v, a, t, s
c) 3 equations of motion.
Whereas if V is the velocity of
CM and  is the angular velocity
about CM, then V = R is true
only for Pure Rolling!
How to make out whether it is a case of pure rotation or not?
1. Check, If the net force is zero –
In all the cases below the C.M. will remain at rest and the motion will be pure
rotation about C.M.
How do we know that the C.M. will remain at rest?
Net F = Ma, gives acceleration of C.M. As net force is zero in all the cases above
so acceleration of C.M. will be zero. So if C.M. was at rest, it will remain at rest.
2. If the velocity of any point on the body is zero, then the body executes pure
rotation about that point.
Ex. Nail fixed in a body.
The body will go through pure rotation about the nail.
C.M. will move. But about A it will be pure rotation.
Is the force exerted by nail is equal to external applied force? No, as CM is
accelerated, so there is NET force on the body.
.
In this type of case, we will call it Rotation about A (its pure rotation about A, or A is the
IAOR)
6
3. The point at rest (mentioned above) might be only for an instant. In such cases at
that particular instant, the body goes through pure rotation about that point.
Ex. If a stick is thrown so that it rotates too, then there might be a point whose speed
is zero.
Ex. If a body rolls on ground without slipping, then VPOC is zero. (Had it not been
zero, then there would have been some relative velocity – as ground has velocity
zero; and relative velocity would mean that there is slipping).
As VPOC is zero, so about P.O.C. the body can be considered to be going through
pure rotation.
An axis through this point which at an instant has zero velocity is called IAOR
(Instantaneous axis of rotation).
4. The IAOR mentioned above might be outside the body.
Consider the movement shown at right.
The IAOR is passing through A
and is into the plane of paper.
A
What is the big deal if rolling etc. can be considered as a case of pure rotation.
Well, then it becomes a case of 1-D kinematics. 5 terms, 3 relations; and if you know 3 variables
then you can find out remaining terms etc.
Also you have to write only rotational KE term in conservation of energy.
How to find the IAOR?
For this we need to find a point whose velocity is zero.
So first learn to find velocity of a point!
7
Translation + Rotation
Take up question on rod lying on table which has VCM as well as ω (about CM)
which can be useful later while teaching collisions.
1. Sphere rolling on an incline (case 1 -  = 0, case 2 -   0)
2. Figure on the right.
M1, R
Note that for M1 you use only T-equation,
For M2 – only F-equation as for M1 it is pure rotation
and for M2 it is pure translation.
M2
HCV-1,
[pg. 184, Ex. 5, 6], [pg. 190, Ex. 26, 27]
pg. 197, Q. No. 33 to 39 [Q 37 can be considered as if the block is in pure translation, and
the pulley is in pure rotation about A, where A is the POC on the right hand side.]
[1999, 10M] A man pushes a cylinder of mass m1 with the help of a plank of mass m2 as
shown. There is no slipping at any contact. The horizontal component of the force applied
by the man is F. Find:
a) The acceleration of the plank and the centre of mass of the cylinder and
b) The magnitudes and directions of frictional forces at contact points.
F
m2
m1
- No sliding between scale & disc
- Disc is rolling.
N1
(a) FBD
F
f
N1
f
m2g
N2
f2
(b) Scale has translation motion
m1g
The Disc is rolling.
So torque eqn. has to be applied to the disc.
(c) Choose a pt. about which torque due to max. no. of forces is zero.
So find T about lowest pt.
1

F (2R) =  MR 2  MR 2 
2

3
2 fR = MR 2
2
8
Finding Velocity of any point for a body undergoing Translation, Rotation,
Translation+Rotation and thus if it is zero, then to find IAOR –
How to find velocity of any point –
V of any point is the vector resultant of VCM and xω.
Velocity of any point of a rolling body
B
,
VA = VCM ± x
= Vo - R.
VB = VCM + R
VC = Vo 2   R 
C
2
VCM = Vo,
A
For No slipping VPOC = VSurface
So VCM - R = VSurface. This is a general result.
If the body is rolling on ground, then VSurface = 0
So VCM = Rω
So, if you say that for pure rolling VCM = Rω then
the result is true only for rolling on ground.
Else in general, VCM - R = VSurface
Q1. Find velocities of various points (A, B, C, D, O, E) on the sphere if it rolls without
slipping with the COM having velocity Vo.
Q2. A stick is hit in the lower half in such a way that COM has velocity Vo and about
C.M. the angular velocity is ωo anticlockwise. Find velocities of various points.
Q3. A cylinder rolls without slipping on a plank with angular velocity ω. The velocity of
the plank is Vp. Find the velocity of point A, B and O of the cylinder [Note that the POC
is not at rest so cylinder cannot be considered to be in pure rotation about POC]
Q4. A cylinder is placed on a plank. The plank is pulled with speed VP such that the
cylinder does not slides. Find velocity of point A, B and O.
9
Finding IAOR (Instantaneous axis of rotation) and IC (Instantaneous centre of zero
velocity)
The combined effects of translation of the centre of mass and rotation about
an axis through the centre of mass are equivalent to a pure rotation with the same angular
speed about an axis passing through a point of zero velocity. Such an axis is called the
IAOR.
If a rod – Vo, Wo. There are point for which Vo – xWo = 0. Find x. The IAOR is x away
from the point having velocity Vo
Methods of Locating IC –
1. If velocity of a point on the body is given and also the ω (normally this point is
the CM, whose velocity is found from conservation of momentum)
P
Draw  to V
r = V/ω
gives position of IC.
Note the side on which IC lies!
V
ω
r
IC
Question DCP pg 25
A rotating disc moves along +X axis.
V(initial) of centre of disc = V
ω (initial) of disc = 0
angular acceleration of disc =  (anticlockwise).
Find the equation [y=y(x)] of the IAOR.
V
The point of IAOR will lie above centre, given by
x
V = y ω = y ( t) = y ( )
V
2
V
So, xy =

2. If the velocities of two points is given and lines of action are non-parallel.
A
Draw  to VA and VB at A and B.
The point of intersection of these s
as shown gives the location of IC.
VA
VB
IC
10
DCP, pg. 26
A uniform rod (M, L) standing on a smooth surface slips on it. Find VCM at the instant
when it makes angle  with the horizontal.
The CM does not move horizontally, as
there is no horizontal force.
Draw  to VA and VB to mark IAOR.
IC
VC

ω
VB
Conservation of energy
1
mgh = I IAOR 2
2
L
1
L
mg (1  sin  )  {I CM  M ( cos  ) 2 } 2
2
2
2
2
1 ML
L2
= {
 M cos2  } 2
2 12
4
L
VIC = ( cos  )ω
2
3. If given is the magnitude and direction of two parallel velocities.
vA
IC
rA
rA
d
ω
IC
vA
rB
rB
d
vB
vB
vA

v
rB  B

rA  rB  d
vA

v
rB  B

rB  rA  d
rA 
rA 
11
Ex. 1 Find angular velocity when velocity of 2 pts is given.
v1
v2
L
x
O
Let us say that v1 > v2 . Then IAOR is supposed to be at O.
v
v1  ( L  x) => L  x  1

v2
v2  x
=> x 

There are 2 variables (x and  ) and 2 equations. Solving them,
v v
v v
L = 1 2 or   1 2

L
v2 L
and x 
v1  v2
Ex. 2
=
3v  v
2R
v
v
(Anti clock wise)
R
O’
3v
P
Q. Find velocity (magnitude and direction) of the point P in above diagram.
Ans - 5 V. As O’ is the IAOR, so it is pure rotation about O’ and the velocity of P is 
to the line joining the IAOR and the point P.
12
Rolling
Pure Rolling - When point of contact has the same velocity and acc. as that of the
surface, then it is rolling without sliding
 V = VS, a = aS.
If VS = 0.
then acm = R 
and Vcm = R 
, 
V, a
VS, aS
If there is Sliding | skidding
Point of contact of the surface has a relative velocity w. r. t. the surface.
Pure rolling is equivalent to pure rotation about p.o.c. with same angular velocity.
Rolling – sphere down the incline (no friction and with friction) – acc/F approach
Alternate approaches - rotation about point of contact or else  about CM + VCM
Questions on rolling –
Sphere/cylinder/ring etc. rolling down the incline
Cylinder (1 or 2 of them) unwrapping down
Conservation of energy approach –
Examples of conservation of energy –
sphere rolling within other sphere, fraction of energy associated with rolling,
Stick rotating down about a fixed point in vertical plane,
carpet unfolding, etc.
[1990, 8M] A carpet of mass M made of inextensible material is rolled along its length in
the form of a cylindrical of radius R and is kept on a rough floor. The carpet starts
unrolling without sliding on the floor when a negligibly small push is given to it.
Calculate the horizontal velocity of the axis of the cylindrical part of the carpet when its
radius reduces to R / 2.
13
Acceleration of any point of a rolling body
B
,
C
acm = a
A
Acceleration of any point is the vector resultant of acm and x
F-eqn  Net F = ma
acm = net F / m and it is in the direction of net F.
Find net torque about C.M.
net T = I 
 = net T / I
a A = acm ± x
= acm - r.
a B = acm + r
aC=
2
a 2  r  .
Find the value of x above C.M. where if an external force is applied, the body rolls
without sliding, even on smooth surface.
Method – apoc must be zero.
Result –
For sphere – x = 2R/5 etc.
14
Finding acceleration of poc, & condition for this to be zero
F
net Fext  0
net Text (about C.M.)  0,
So both translation & rotation
(1) First find net Fext
F
[This is acceleration of C.M.]
a
M
(2) find net Text (about C.M.)
T
  CM
I CM
Now acc. of any point (as given earlier) is
 
a  aCM  r
[Note: This is VECTOR ADDITION]
Ex. (1)
F
aCM =
M
TCM = F  R   1 MR 2  
2
2
F
R/2

F and this is clockwise
 
.
MR
alowest pt = aCM   R. For point A ‘R’ is towards left, so
F  F 
=

R  0
M  MR 
Check out –
When will the lowest pt. have non – zero acceleration!
Result – (co-efficient of MI) x R above the centre. We will find it in the coming pages.
Ex (2)
F
2F
aCM =
M
F
1

TCM = F  R   MR 2 
2

2F

.
MR
a (of lowest pt.) = aCM + x
15
2F  2F 

R0
M  MR 
How far up from O should the force be applied for acc (poc)  0
=
x
1

TCM = F  x =  MR 2 
2

2 Fx
F
 
aCM 
MR 2
M
apoc = aCM - R
F  2x 
F 2 Fx
=

=
1    0 [for apoc = 0]
M MR M 
R
R
x .
2
R
So for x x  ,  a poc  0. So if x = R/2, acc (poc) = 0, else not. We will use it
2
to find the direction of friction.
Friction does not necessarily slows the motion.
= 1N
F
= 1N
M = 1 kg
a = 1 m/s2
M = 1 kg.
1 
F + f =  M a 
2 
F – f = Ma
3
2F = Ma
2
4F 4
a=
= m / s 2  1m / s 2 .
3M 3
Misconcept: - ‘Friction opposes motion’.
Concept: - ‘Friction opposes the motion of point of contact’.
16
Finding out direction and magnitude of friction for rotating bodies.
[Find the direction in which the poc has the tendency to move – this is known by
finding the acceleration of poc. Now as friction opposes this motion, so the direction of
friction is opposite to the direction of this acceleration of poc]
Ex 1
(1) Assume no friction
(2) net Fext = mg sin
acm = mg sin 
N
mg cos
mg sin
m
= g sin.

(3) Tcm = 0
=0
(4) a (of pt. of contact) = acm + 0
And is down the slope.
So pt. of contact has tendency to move downwards
So friction will act upwards (up the slope)
[Friction will oppose the motion of point of contact]
Ex 2
F
2F
aCM =
F
M
2F
1

TCM = F  R =  MR 2     
.
MR
2
 CM
a (of poc) = acm  x => acm - R = 0
So ‘poc’ has no acc. [no tendency to move] hence NO FRICTION acts
Ex 3
F
1

aCM =
, T  F   MR 2 
M
2

F  2F 
apoc

R
M  MR 
F
= 
[apoc , so friction is
]
M
Ex 4
apoc = 0
f=0
R/2
Ex 5
F
, TCM  0
M
 0
F
apoc =
i.e. towards 
M
So friction is .
aCM =
17
Direction of friction in Rolling
For block kept on floor.
P
fS(MAX.) = S N
fk = k N
But for Disc etc
‘f’ is unknown.
so take it whether towards right or towards left
you get the right answer.
If ‘f’ taken backwards
4F
f
a=
,f  .
3m
3
If ‘f’ taken along motion
4F
f
a=
,f  .
3m
3
In Rotation
f<N
For “just starts slipping”
Or “just stops rolling”
We can use
f=N
for this, take proper direction.
f=N
F
 If given that ‘poc’ has velocity relative to surface [i.e. sliding]
then f = kN
and it’s direction is to be found using techniques given earlier [see the tendency of
motion of poc, friction opposes that motion]
 If said – “just slides” or “just rolls” take f = SN
 If it is not specified whether the body rolls or slides, then
f  variable (unknown) so any direction.
Find ‘f’ [if +ve sign, then our assumed dirn is right, else opposite (-ve sign)]
If value of ‘f’ is  SN, then the body is rolling.
Else it is sliding & f = kN.
Ex.
1

5 – f = 1  a [Tram] & (5  R) + (f  R) =  MR 2  CM
2

MR CM
=5+f=
2
 MR  a
?
=5+f= 

 2 R
5–f=a
a
5+f= .
2
18
3a
20
a
m / s2
2
3
5
f=- N
3
mean it is towards right 
SN = 04  1  10 = 4 N
5
f =  1  66 N
3
So body rolls and f = 166 N.
10 =
Applying Torque equation
Ex 1
N
(a) Draw all forces.
(b) Choose a pt. about
F
Which T = 0 for most
A
forces. [here it is pt. B]
O
(c) F  (2R) = IB 
B
2

=  MR 2  MR 2  ,
mg f
5

7

F (2R) =  MR 2 
5

10 F

.
7 MR
Now ‘a’ for any pt. ‘x’ away from B is
ax = aB + x.
19
We have seen only one force being applied on the body (ex.1) and then two
forces applied (ex.2)
Now let’s take the case when the two forces are
F
as follows (due to presence of friction)
M
R
How much is friction?
& how much the acc. of poc when friction is taken into account?
Ff
aCM =
F
M
1

FR  fR   MR 2 
2

a
1


=  MR  CM
-? Is it right.
M
F
2
 R
Ma
So F – f =
2
F + f = Ma.
F – 3f = 0
F
f=
3
+ve sign indicate rightly assumed dirn.
F
will have some value
3
Compare it with ‘S mg’.
F
F
If
< S mg , Rolling & f =
.
3
3
F
> S mg , sliding & f = S mg.
3
This
Toppling –
20
For ‘F’ applied below
point friction is
backwards.
!!!
R
3
Insert them somewhere (Good questions or variations thereof)
A
What is the force exerted by the nail (at A) on the rod?
O
P
If the force by the nail is F, then
P – F = M x acm
(1)
As A is fixed (pure rotation about A)
P x L = IA α
So α = PL/IA
from (1) P – F = M [α (L/2)]
= (ML/2)(PL/IA)
Thus F can be calculated.
[IIT xxxx]
F
21