Torque in rotation corresponds to force in translation. It is the "cause" that accelerates or decelerates a particle in rotation. Quantitatively, it is defined as a vector given by :
t
=
r
x
F
t
=
r
x
F
The important aspect of this relation for rotation is that vector "r" is constrained to be perpendicular to the axis of rotation as every particle of the body moves around a circular path in the plane perpendicular to the axis of rotation. In the general case of a particle, however, this restriction may not exist at all. First of all, there may not be a fixed axis of rotation. In this sense, we can realize that we earlier had a restricted interpretation of torque as far as rotation about a fixed axis is concerned.
A general interpretation of the vector equation as above tells us that torque may be interpreted even without an axis or a line. An external force on a particle shall constitute a torque with respect to a point. Only condition is that the point should not lie on the line of force, in which case torque is zero. The important aspect of this interpretation is that we are referring torque on the particle with respect to a point - not with respect to a line like axis of rotation. This is the difference. This is the generalized interpretation of torque.
Mentally, we should develop techniques to quickly interpret the vector product on the right hand side of the equation. We employ right hand vector product rule to determine the direction of resulting product vector. However, the "in place" orientations of vector operands make it difficult to apply this rule. Incidentally, the position vector has its tail at the point about which torque is to be calculated. This fact simplifies to find the direction of the torque.
We shift the second vector (F) so that tails of two operand vectors meet at the point about which torque is calculated. The two operand vectors define a plane. The torque (t) is, then, acting along a line perpendicular to this plane and passing through the meeting point. The direction of the orientation is then found by applying right hand rule.
To illustrate the technique, we consider the situation as shown in the figure. Let "r" and "F" vectors lie in the xy-plane. We shift the force vector as shown. Now, we apply right hand rule of vector product to determine the direction of torque. For this, we sweep the closed fingers of the right hand from position vector to force vector. The outstretched thumb, then, indicates that torque is acting in the positive z-direction.
While interpreting the vector product, we must be careful about the sequence of operand. The vector product "
F
x
r
F x r
" is negative or opposite in sign to that of "
r
x
F
r x F
".
As explained earlier module, we can find the magnitude of torque, using any of the following relations :
1:
τ
=
r
F
sin
θ
1:
τ
=
r
F
sin
θ
2:
τ
=
r
F
⊥
2:
τ
=
r
F
⊥
3:
τ
=
r
⊥
F
3:
τ
=
r
⊥
F
Problem : A projectile of mass "m" is projected with a speed "v" at an angle "θ" with the horizontal. Calculate the torque on the particle at the maximum height in relation to the point of projection.
Solution : The magnitude of the torque is given by :
τ
=
r
⊥
F
=
product of moment arm and magnitude of force
τ
=
r
⊥
F
=
product of moment arm and magnitude of force
where
r
⊥
r ⊥
is moment arm. This moment arm is equal to half of the horizontal range of the flight,
r
⊥
=
R
2
=
v
2
sin
2θ
2
g
r
⊥
=
R
2
=
v
2
sin
2θ
2
g
Now, the force on the projectile is "m" is the force due to gravity :
F
=
m
g
F
=
m
g
Putting in the expression of torque, we have :
⇒
τ
=
m
v
2
sin
2θ
2
⇒
τ
=
m
v
2
sin
2θ
2
Application of right hand rule indicates that torque is clockwise and is directed in to the page.
Problem : A particle in xy plane is acted by two forces 5 N and 10 N in xy plane as shown in the figure. Find the net torque on the particle.
Solution : In this situation, calculation of the magnitude of torque suits to the form of expression that uses moment arm. Here, moment arms are :
For 5 N force, the moment arm "AB" is :
AB
=
OA
cos
30
0
=
5
x
0.866
=
4.33
m
AB
=
OA
cos
30
0
=
5
x
0.866
=
4.33
m
For 10 N force, the moment arm "AC" is :
AC
=
OA
sin
30
0
=
10
x
0.5
=
5
m
AC
=
OA
sin
30
0
=
10
x
0.5
=
5
m
The torques due to 5 N is :
⇒
τ
1
=
=
4.33
x
5
=
21.65
N
-
m
(positive being anti-clockwise)
⇒
τ
1
=
=
4.33
x
5
=
21.65
N
-
m
(positive being anti-clockwise)
The torques due to 10 N is :
⇒
τ
2
=
-
5
x
10
=
-
50
N
-
m
(negative being clockwise)
⇒
τ
2
=
-
5
x
10
=
-
50
N
-
m
(negative being clockwise)
Net torque is :
⇒
τ
net
=
τ
1
+
τ
2
=
21.65
-
50
=
-
28.35
N
-
m
(negative being clockwise)
⇒
τ
net
=
τ
1
+
τ
2
=
21.65
-
50
=
-
28.35
N
-
m
(negative being clockwise)
We can also first calculate the resultant force and then apply the definition of torque to obtain net torque. Further, we can also use vector notation with unit vectors to solve this problem.
Problem : A particle in xy plane is acted by a force 5 N in z-direction as shown in the figure. Find the net torque on the particle.
Solution : In this case, moment arm is equal to the magnitude of position vector. Hence, torque is :
τ
=
10
x
5
=
50
N
-
m
τ
=
10
x
5
=
50
N
-
m
However, finding its direction is not as straight forward here. The situation here differs to earlier example in one important respect. The plane containing position vector and force vector is a plane defined by zOA and is perpendicular to xy - plane. Now, torque is passing through the origin and is perpendicular to the plane zoA. Thus, torque is shifted by 30 degree from the y-axis and lies in xy-plane as shown in the figure here.
