In the previous section, we saw the relation between linear velocity and angular velocity. In this section we will see torque
1. Consider a body in pure translational motion
• We know that, for that body, a force →F is required to bring about a change in any of the following two states:
(i) State of rest
(ii) State of uniform motion
2. Now consider a body which is fixed along an axis. We know that, it cannot have translational motion. It can only have pure rotational motion
■ Can we write the same statement that we wrote in (1) here also?
3. Let us elaborate this question:
• The body under consideration can have rotation only. It can have two states:
(i) If it is rotating (with uniform →ω) about the axis, we say that it is in a ‘state of uniform rotation’
(ii) If →ω is zero at any instant, we say that, it is in a ‘state of rest’
■ So the question is:
Can a force →F bring about a change in the above two states?
4. To find the answer, we will see a common situation that we encounter in our day to day life:
(i) Consider the door of a room shown in fig.7.82(a) below:
• The left edge of the door is fixed using hinges
(ii) The door rotates about the hinges. So the hinges can be considered as the 'axis of rotation'
• It is shown in blue color
(iii) To open the door, we have to apply a force →F
• This →F vector is indicated in magenta color
(iv) In fig.a, →F is applied at the hinges
• The door will not rotate. We will not be able to open the door
(v) In fig.b, the →F is applied at a point between the outer edge and the axis
• This time, the door will rotate. But it is not the best way to open a door
(vi) In fig.c, the →F is applied exactly at the outer edge. We see that, →F is inclined (at an acute angle) to the door
• This time also, the door will rotate. But this also, is not the best way to open a door
(vii) In fig.d, the →F is applied exactly at the outer edge. We see that, →F is perpendicular to the surface of the door.
• The door will rotate. And this is the best way to open a door
• →F will be most effective in this case
• In translational motion, a force →F is enough to change the state
• In rotational motion, 'how and where' the →F is applied is also important if we want to change the state
■ In rotational motion, that role is played by moment of force
■ 'Moment of force' is also referred to as torque or couple
• Let us consider a 2-D problem first. We will write it in steps:
1. In fig.7.83(a) below, a spanner is being used to loosen a nut
• A force →F is applied on the spanner
2. →F is applied at a point P on the spanner
• 'P' is shown as a yellow dot
• The vector →F and the spanner lies on the same plane. So this is a 2-D problem
• We want to know the effect caused by →F on the nut
3. Let us assume that, the center of the nut is the origin O of the frame of reference. This is shown in fig.b
• →r is the position vector of P
• →F makes an angle θ with →r
4. We want a perpendicular to →F
• Also, that perpendicular should pass through O
• This is shown by the white dashed line
• Q is the foot of the perpendicular
5. Now we have a right triangle OPQ
• The length of OQ is obviously |→r|×sinθ
• This length OQ is very special:
It is the perpendicular distance of →F from O
6. The product of the applied force →F and this perpendicular distance is called torque
• It is denoted by τ (the Greek letter tau)
• So we can write:
τ=|→F|×(|→r|×sinθ)=|→F|×|→r|×sinθ
• Rearranging this, we get: τ=|→r|×|→F|×sinθ
7. But |→r|×|→F|×sinθ=(→r×→F)
• That means τ is the cross product of →r and →F
• But cross product of two vectors is a vector
• That means, τ is a vector. We must denote it as →τ
8. Thus we get: →τ=(→r×→F)
• So the magnitude of →τ is given by: |→τ|=|→r|×|→F|×sinθ
• We can write:
|→τ|=|→F|×(Perpendicular distance of force from O)=|→F|×|→r⊥|
9. Now we want the direction
• We know that the cross product of two vectors will be perpendicular to the plane containing those two vectors
• So in our present case, →τ is perpendicular to our computer screen
■ The question is:
Is it directed towards us?
OR, Is it directed away from us?
10. To find the answer, we apply the right hand screw rule
(i) Imagine that, a right handed screw is placed perpendicular to the computer screen
(ii) Assume that →F is shifted upwards in such a way that, the tail ends of →F and →r coincide
(iii) Turn the screw in the direction from →r to →F
• The screw will move towards us
(iii) So we can write: →τ is perpendicular to the computer screen and is directed towards us
• If →τ has a high value, the nut will loosen out easily
• If →τ has a low value, the nut may not rotate at all
• So we want a high →τ. Let us see how this can be achieved:
1. We have the magnitude of →τ:
|→τ|=|→r|×|→F|×sinθ
2. The 'sin θ' is causing the problem
• It is a fraction most of the time
Eg: sin 30 = 0.5, sin 45 = 0.7071, sin 60 = 0.8660, sin 75 = 0.9659 etc.
• Since it is a fraction, it will reduce the magnitude (|→r|×|→F|)
3. The only situation where such reduction does not occur, is when 'sin θ' = 1
• This occurs when θ = 90o
■ That is., when →F is perpendicular to →r
• In that case, we get: |→τ|=|→r|×|→F|×sin90
⇒|→τ|=|→r|×|→F|×1
⇒|→τ|=|→r|×|→F|
• This is shown in fig.7.83(c) above
4. Note that, the maximum value possible for sin θ is '1'
• That means, using sin θ, we cannot double or triple (|→r|×|→F|)
(|→r|×|→F|) is the maximum possible value
5. Also note that, if we shift the point of application 'P' to the right, |→r| will increase. This will also cause an increase in |→τ|
• Further more, the magnitude |→F| can also be increased to obtain a higher →τ
• Now we know why the method shown in fig.7.82(d) is the most effective for opening a door
1. We have: |→τ|=|→r|×|→F|×sinθ
Rearranging this, we get: |→τ|=|→r|×(|→F|×sinθ)
2. Now consider fig.7.84(b) below:
• The →F is inclined. But it will have a component perpendicular to →r
• This perpendicular component is shown in green color in fig.b
3. From the right triangle PRS in fig.c, it is clear that, the magnitude of the perpendicular component is:
(|→F|×sinθ)
(Note that, ∠OPR and ∠PRS are alternate angles and hence equal)
4. So the result in (1) can be written as:
|→τ|=(Magnitude of the Perpendicular component of force)×|→r|=|→F⊥|×|→r|
1. In fig.7.85(a) below, a particle 'P' is shown as a small red sphere. it is situated in space
• The position vector of 'P' is →r. It is shown in brown color
• A force →F acts on 'P'. It is shown in magenta color
2. We see that there is a common point for →r and →F
• Because, the head of →r coincides with tail of →F
3. So there must be a plane on which both →r and →F lie
• This plane is shown in fig.7.85(b) above
• A bit of transparency is given to this plane. So we can see the y-axis through it, though in a blurred form
4. Next, we want the perpendicular distance of →F from the origin O
• For that, we extend →F backwards along the same line
• This is shown by the magenta dashed line in fig.7.86(a) below:
• We want a perpendicular to this magenta dashed line. Also it must pass through O
• Such a perpendicular is shown as the yellow dashed line in fig,7.86(b) above
• Q is the foot of the perpendicular
• Now we get a right triangle OPQ
5. Consider the plane in the above fig.7.86. It contains both →r and →F
• Imagine a line perpendicular to that plane. If we look at the plane along that perpendicular line, we will get a 2-D view. It is shown in fig.7.87 below:
• We see the triangle OPQ clearly
• We see that, θ is less than θ1
• So we must take θ as the angle between →F and →r
• In the right triangle OPQ, we get: OQ=|→r|sinθ
• This OQ is the perpendicular distance of →F from O
6. The product of the applied force |→F| and this perpendicular distance is the 'torque exerted by →F at O'
• So we can write:
τ=|→F|×(|→r|×sinθ)=|→F|×|→r|×sinθ
• Rearranging this, we get: τ=|→r|×|→F|×sinθ
• But |→r|×|→F|×sinθ=(→r×→F)
• That means τ is the cross product of →r and →F
• But cross product of two vectors is a vector
• That means, τ is a vector. We must denote it as →τ
• Thus we get:
Eq.7.17: →τ=(→r×→F)
■ So the magnitude of →τ is given by: |→τ|=|→r|×|→F|×sinθ
• We can write:
|→τ|=|→F|×(Perpendicular distance of force from O)=|→F|×|→r⊥|
7. Now we want the direction
• We know that the cross product of two vectors will be perpendicular to the plane containing those two vectors
• We have already drawn that plane
• So in our present case, →τ is perpendicular to that plane
■ The question is:
Is it directed upwards, away from that plane ?
OR, Is it directed downwards, into that plane ?
• To find the answer, we apply the right hand screw rule
(i) In fig.7.88(a) below, a right handed screw is placed perpendicular to the plane
(ii) Imagine that, →F is shifted so that, it's tail end coincide with the tail end of →r
(iii) Turn it in the direction from →r to →F
• The screw will move upwards, away from the plane
(iii) So we can write: →τ is perpendicular to the plane and is directed upwards and away from the 'plane containing →F and →r'
• This →τ is shown in cyan color in fig.7.88(b) above
• Consider the plane containing both →r and →F
• Imagine a line perpendicular to that plane. If we look at the plane along that line we will get a 2-D view. We saw it earlier in fig.7.87. It is shown again in fig.7.89 below:
1. We have: |→τ|=|→r|×|→F|×sinθ
• Rearranging this, we get: |→τ|=|→r|×(|→F|×sinθ)
2. The →F is inclined to →r
• So it will have a component perpendicular to →r
• This perpendicular component can be easily found out by completing the right triangle PRS
• The fact that ∠SPR = ∠OPQ (since they are opposite angles) helps us to complete the right triangle PRS
• Also note that, →r is extended along the same line
3. From the right triangle PRS, we get:
• The magnitude of the perpendicular component is = (|→F|×sinθ)
4. So the result in (1) can be written as:
|→τ|=(Magnitude of the Normal component of force)×|→r|=|→F⊥|×|→r|
1. Consider a body in pure translational motion
• We know that, for that body, a force →F is required to bring about a change in any of the following two states:
(i) State of rest
(ii) State of uniform motion
2. Now consider a body which is fixed along an axis. We know that, it cannot have translational motion. It can only have pure rotational motion
■ Can we write the same statement that we wrote in (1) here also?
3. Let us elaborate this question:
• The body under consideration can have rotation only. It can have two states:
(i) If it is rotating (with uniform →ω) about the axis, we say that it is in a ‘state of uniform rotation’
(ii) If →ω is zero at any instant, we say that, it is in a ‘state of rest’
■ So the question is:
Can a force →F bring about a change in the above two states?
4. To find the answer, we will see a common situation that we encounter in our day to day life:
(i) Consider the door of a room shown in fig.7.82(a) below:
 |
| Fig.7.82 |
(ii) The door rotates about the hinges. So the hinges can be considered as the 'axis of rotation'
• It is shown in blue color
(iii) To open the door, we have to apply a force →F
• This →F vector is indicated in magenta color
(iv) In fig.a, →F is applied at the hinges
• The door will not rotate. We will not be able to open the door
(v) In fig.b, the →F is applied at a point between the outer edge and the axis
• This time, the door will rotate. But it is not the best way to open a door
(vi) In fig.c, the →F is applied exactly at the outer edge. We see that, →F is inclined (at an acute angle) to the door
• This time also, the door will rotate. But this also, is not the best way to open a door
(vii) In fig.d, the →F is applied exactly at the outer edge. We see that, →F is perpendicular to the surface of the door.
• The door will rotate. And this is the best way to open a door
• →F will be most effective in this case
So now we can make a comparison between translational motion and rotational motion. We can write:
• In rotational motion, 'how and where' the →F is applied is also important if we want to change the state
■ We know that, in translational motion, the force →F plays an important role
■ 'Moment of force' is also referred to as torque or couple
• We will first see the role played by moment of force in the case of a single particle
• After that, we will extend it to a system of particles in a rigid body
• After that, we will extend it to a system of particles in a rigid body
1. In fig.7.83(a) below, a spanner is being used to loosen a nut
 |
| Fig.7.83 |
2. →F is applied at a point P on the spanner
• 'P' is shown as a yellow dot
• The vector →F and the spanner lies on the same plane. So this is a 2-D problem
• We want to know the effect caused by →F on the nut
3. Let us assume that, the center of the nut is the origin O of the frame of reference. This is shown in fig.b
• →r is the position vector of P
• →F makes an angle θ with →r
4. We want a perpendicular to →F
• Also, that perpendicular should pass through O
• This is shown by the white dashed line
• Q is the foot of the perpendicular
5. Now we have a right triangle OPQ
• The length of OQ is obviously |→r|×sinθ
• This length OQ is very special:
It is the perpendicular distance of →F from O
6. The product of the applied force →F and this perpendicular distance is called torque
• It is denoted by τ (the Greek letter tau)
• So we can write:
τ=|→F|×(|→r|×sinθ)=|→F|×|→r|×sinθ
• Rearranging this, we get: τ=|→r|×|→F|×sinθ
7. But |→r|×|→F|×sinθ=(→r×→F)
• That means τ is the cross product of →r and →F
• But cross product of two vectors is a vector
• That means, τ is a vector. We must denote it as →τ
8. Thus we get: →τ=(→r×→F)
• So the magnitude of →τ is given by: |→τ|=|→r|×|→F|×sinθ
• We can write:
|→τ|=|→F|×(Perpendicular distance of force from O)=|→F|×|→r⊥|
9. Now we want the direction
• We know that the cross product of two vectors will be perpendicular to the plane containing those two vectors
• So in our present case, →τ is perpendicular to our computer screen
■ The question is:
Is it directed towards us?
OR, Is it directed away from us?
10. To find the answer, we apply the right hand screw rule
(i) Imagine that, a right handed screw is placed perpendicular to the computer screen
(ii) Assume that →F is shifted upwards in such a way that, the tail ends of →F and →r coincide
(iii) Turn the screw in the direction from →r to →F
• The screw will move towards us
(iii) So we can write: →τ is perpendicular to the computer screen and is directed towards us
• If →τ has a low value, the nut may not rotate at all
• So we want a high →τ. Let us see how this can be achieved:
1. We have the magnitude of →τ:
|→τ|=|→r|×|→F|×sinθ
2. The 'sin θ' is causing the problem
• It is a fraction most of the time
Eg: sin 30 = 0.5, sin 45 = 0.7071, sin 60 = 0.8660, sin 75 = 0.9659 etc.
• Since it is a fraction, it will reduce the magnitude (|→r|×|→F|)
3. The only situation where such reduction does not occur, is when 'sin θ' = 1
• This occurs when θ = 90o
■ That is., when →F is perpendicular to →r
• In that case, we get: |→τ|=|→r|×|→F|×sin90
⇒|→τ|=|→r|×|→F|×1
⇒|→τ|=|→r|×|→F|
• This is shown in fig.7.83(c) above
4. Note that, the maximum value possible for sin θ is '1'
• That means, using sin θ, we cannot double or triple (|→r|×|→F|)
(|→r|×|→F|) is the maximum possible value
5. Also note that, if we shift the point of application 'P' to the right, |→r| will increase. This will also cause an increase in |→τ|
• Further more, the magnitude |→F| can also be increased to obtain a higher →τ
• Now we know why the method shown in fig.7.82(d) is the most effective for opening a door
Another aspect:
Rearranging this, we get: |→τ|=|→r|×(|→F|×sinθ)
2. Now consider fig.7.84(b) below:
 |
| Fig.7.84 |
• This perpendicular component is shown in green color in fig.b
3. From the right triangle PRS in fig.c, it is clear that, the magnitude of the perpendicular component is:
(|→F|×sinθ)
(Note that, ∠OPR and ∠PRS are alternate angles and hence equal)
4. So the result in (1) can be written as:
|→τ|=(Magnitude of the Perpendicular component of force)×|→r|=|→F⊥|×|→r|
■ Thus, the magnitude →τ can be calculated in two ways:
(i) |→τ|=|→F|×|→r⊥|
(ii) |→τ|=|→F⊥|×|→r|
(i) |→τ|=|→F|×|→r⊥|
(ii) |→τ|=|→F⊥|×|→r|
Next we will look at a 3-D problem. We will write it in steps:
 |
| Fig.7.85 |
• A force →F acts on 'P'. It is shown in magenta color
2. We see that there is a common point for →r and →F
• Because, the head of →r coincides with tail of →F
3. So there must be a plane on which both →r and →F lie
• This plane is shown in fig.7.85(b) above
• A bit of transparency is given to this plane. So we can see the y-axis through it, though in a blurred form
4. Next, we want the perpendicular distance of →F from the origin O
• For that, we extend →F backwards along the same line
• This is shown by the magenta dashed line in fig.7.86(a) below:
 |
| Fig.7.86 |
• Such a perpendicular is shown as the yellow dashed line in fig,7.86(b) above
• Q is the foot of the perpendicular
• Now we get a right triangle OPQ
5. Consider the plane in the above fig.7.86. It contains both →r and →F
• Imagine a line perpendicular to that plane. If we look at the plane along that perpendicular line, we will get a 2-D view. It is shown in fig.7.87 below:
 |
| Fig.7.87 |
• We see that, θ is less than θ1
• So we must take θ as the angle between →F and →r
• In the right triangle OPQ, we get: OQ=|→r|sinθ
• This OQ is the perpendicular distance of →F from O
6. The product of the applied force |→F| and this perpendicular distance is the 'torque exerted by →F at O'
• So we can write:
τ=|→F|×(|→r|×sinθ)=|→F|×|→r|×sinθ
• Rearranging this, we get: τ=|→r|×|→F|×sinθ
• But |→r|×|→F|×sinθ=(→r×→F)
• That means τ is the cross product of →r and →F
• But cross product of two vectors is a vector
• That means, τ is a vector. We must denote it as →τ
• Thus we get:
Eq.7.17: →τ=(→r×→F)
■ So the magnitude of →τ is given by: |→τ|=|→r|×|→F|×sinθ
• We can write:
|→τ|=|→F|×(Perpendicular distance of force from O)=|→F|×|→r⊥|
7. Now we want the direction
• We know that the cross product of two vectors will be perpendicular to the plane containing those two vectors
• We have already drawn that plane
• So in our present case, →τ is perpendicular to that plane
■ The question is:
Is it directed upwards, away from that plane ?
OR, Is it directed downwards, into that plane ?
• To find the answer, we apply the right hand screw rule
(i) In fig.7.88(a) below, a right handed screw is placed perpendicular to the plane
 |
| Fig.7.88 |
(iii) Turn it in the direction from →r to →F
• The screw will move upwards, away from the plane
(iii) So we can write: →τ is perpendicular to the plane and is directed upwards and away from the 'plane containing →F and →r'
• This →τ is shown in cyan color in fig.7.88(b) above
The other aspect:
• Imagine a line perpendicular to that plane. If we look at the plane along that line we will get a 2-D view. We saw it earlier in fig.7.87. It is shown again in fig.7.89 below:
 |
| Fig.7.89 |
• Rearranging this, we get: |→τ|=|→r|×(|→F|×sinθ)
2. The →F is inclined to →r
• So it will have a component perpendicular to →r
• This perpendicular component can be easily found out by completing the right triangle PRS
• The fact that ∠SPR = ∠OPQ (since they are opposite angles) helps us to complete the right triangle PRS
• Also note that, →r is extended along the same line
3. From the right triangle PRS, we get:
• The magnitude of the perpendicular component is = (|→F|×sinθ)
4. So the result in (1) can be written as:
|→τ|=(Magnitude of the Normal component of force)×|→r|=|→F⊥|×|→r|
So in the 3-D case also, we get the same two methods to find the magnitude →τ:
(i) |→τ|=|→F|×|→r⊥|
(ii) |→τ|=|→F⊥|×|→r|
(ii) |→τ|=|→F⊥|×|→r|
Some interesting results:
1. We have: |→τ|=|→F|×|→r⊥|
• Consider fig.7.87 that we saw earlier
• In that fig., we know that |→r⊥| = OQ
• If the →F pass through O,
OQ = |→r⊥| = 0
• Then |→τ|=|→F|×|→r⊥|=0
• This is the reason why we cannot open a door by applying the force at the hinge
2. We have: |→τ|=|→F⊥|×|→r|
• Consider fig.7.89 above
• In that fig., if θ = 0 or 180o, the →F will be aligned with →r
• There will be no force component perpendicular to →r. That is:
SR = |→F⊥| = 0
• Then |→τ|=|→F⊥|×|→r|=0
• So after reading (1) above, a person decides to apply a force at a point away from the hinge. But if that force is parallel to the surface of the door, it will not open
3. Finally, we all know that, if |→F| = 0, then |→τ|=0
This is like 'just touching the door' with out applying any force. The door certainly will not open
1. We have: |→τ|=|→F|×|→r⊥|
• Consider fig.7.87 that we saw earlier
• In that fig., we know that |→r⊥| = OQ
• If the →F pass through O,
OQ = |→r⊥| = 0
• Then |→τ|=|→F|×|→r⊥|=0
• This is the reason why we cannot open a door by applying the force at the hinge
2. We have: |→τ|=|→F⊥|×|→r|
• Consider fig.7.89 above
• In that fig., if θ = 0 or 180o, the →F will be aligned with →r
• There will be no force component perpendicular to →r. That is:
SR = |→F⊥| = 0
• Then |→τ|=|→F⊥|×|→r|=0
• So after reading (1) above, a person decides to apply a force at a point away from the hinge. But if that force is parallel to the surface of the door, it will not open
3. Finally, we all know that, if |→F| = 0, then |→τ|=0
This is like 'just touching the door' with out applying any force. The door certainly will not open
In the next section, we will see angular momentum
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